Gravitation

(a) No. A body cannot be shielded from the gravitational influence of nearby matter by any means. Electrical shielding is possible because charges are of two kinds and the induced charges on a hollow conductor cancel the external field inside it. Gravitational mass, however, is of only one kind and gravitation is completely independent of the presence of other bodies and of the properties of the intervening medium. Hence no arrangement of matter can cancel the gravitational field inside it.

(b) Yes. The gravitational pull of the earth on the spaceship provides the centripetal force needed to keep it in orbit, so the astronaut and the spaceship both fall freely together and the astronaut feels weightless with respect to the earth. He can, however, detect the gravitational field of the spaceship itself. If the space station is small its own gravitational force is negligible, but if it is massive or of large size, the astronaut will experience its gravity and will be able to detect it.

(c) The tidal effect is produced not by the gravitational force itself but by the variation of that force across the diameter of the earth. The gravitational force varies inversely as the square of the distance, whereas the tidal force varies inversely as the cube of the distance. The moon is very much nearer to the earth than the sun is, so although the sun's pull on the earth is much greater than the moon's pull, the moon's pull changes far more sharply from the near face of the earth to the far face. This larger difference in pull stretches the oceans more effectively, and therefore the tidal effect of the moon's pull is greater than that of the sun.

(a) Decreases. As we go above the surface of the earth the distance from the centre increases, and

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

so that g' < g.

(b) Decreases. At a depth h below the surface, only the inner solid sphere of radius (R_e − h) attracts the body, and

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

so that g' < g.

(c) Mass of the body. Since $\text g = \frac{\text{GM}_e}{\text R_e^2}$, the expression is free from the mass of the body. Hence the earth attracts all bodies with the same acceleration, whatever their masses be.

(d) More. The formula $-\text{GMm}\left(\frac{1}{\text r_2} - \frac{1}{\text r_1}\right)$ is exact for all separations, while m g (r₂ − r₁) assumes that g has the same value at r₁ and at r₂. Since g is not the same at the two distances, the first formula is more accurate.

Given,

• Period of the planet, $\text T_{planet} = \frac{\text T_{earth}}{2}$ (it goes round the sun twice as fast)

By Kepler's law of periods, the square of the period of revolution of a planet is directly proportional to the cube of its mean distance from the sun,

$\text T^2 \propto \text r^3$

Applying this to the planet and the earth,

$\left(\frac{\text T_{planet}}{\text T_{earth}}\right)^2 = \left(\frac{\text r_{planet}}{\text r_{earth}}\right)^3$

Substituting $\text T_{planet} = \frac{\text T_{earth}}{2}$,

$$\begin{aligned}\left(\frac{1}{2}\right)^2 = \left(\frac{\text r_{planet}}{\text r_{earth}}\right)^3 \\[1em] \frac{1}{4} = \left(\frac{\text r_{planet}}{\text r_{earth}}\right)^3\end{aligned}$$

$\frac{\text r_{planet}}{\text r_{earth}} = \left(\frac{1}{4}\right)^{1/3} = 0.63$

Hence, the orbital size of the planet is about 0.63 times that of the earth.

Given,

• Orbital period of the satellite, T = 1.769 days = 1.769 × 24 × 60 × 60 = 1.528 × 10⁵ s
• Radius of the orbit, r = 4.22 × 10⁸ m
• G = 6.67 × 10^-11 N m² kg^-2
• Mass of the sun, M_s = 1.99 × 10³⁰ kg

The gravitational pull of Jupiter on its satellite supplies the necessary centripetal force. Hence the period of revolution of a satellite in an orbit of radius r around Jupiter (mass M_J) is

$\text T = 2\pi \sqrt{\frac{\text r^3}{\text{GM}_J}} \quad \Rightarrow \quad \text M_J = \frac{4\pi^2 \text r^3}{\text T^2 \text G}$

Substituting the values,

$$\begin{aligned}\text M_J = \frac{4 \times (3.14)^2 \times (4.22 \times 10^8)^3}{(1.528 \times 10^5)^2 \times (6.67 \times 10^{-11})} \\[1em] = \frac{39.44 \times 7.515 \times 10^{25}}{2.335 \times 10^{10} \times 6.67 \times 10^{-11}}\end{aligned}$$

$\text M_J = \frac{2.964 \times 10^{27}}{1.557} = 1.90 \times 10^{27}\ \text{kg}$

Comparing this with the mass of the sun,

$\frac{\text M_J}{\text M_s} = \frac{1.90 \times 10^{27}}{1.99 \times 10^{30}} \approx \frac{1}{1000}$

Hence, the mass of Jupiter is about one-thousandth that of the sun.

Given,

• Number of stars in the galaxy = 2.5 × 10¹¹
• Mass of one star (one solar mass) = 1.99 × 10³⁰ kg
• Distance of the star from the galactic centre, r = 50,000 ly
• Diameter of the Milky Way = 10⁵ ly
• 1 ly = 9.46 × 10¹⁵ m

The total mass of the galaxy is

$\text M = 2.5 \times 10^{11} \times 1.99 \times 10^{30} = 4.97 \times 10^{41}\ \text{kg}$

The diameter of the galaxy is 10⁵ ly, which is twice the distance of the star from the centre. Hence the star lies at the very edge of the galactic disk, and by Newton's shell theorem the whole mass of the galaxy enclosed within its orbit acts as if concentrated at the centre.

The radius of the star's path is

$\text r = 50{,}000 \times 9.46 \times 10^{15} = 4.73 \times 10^{20}\ \text m$

The gravitational pull of the galaxy provides the centripetal force, so the period of revolution is

$\text T = 2\pi\sqrt{\frac{\text r^3}{\text{GM}}}$

Substituting the values,

$$\begin{aligned}\text T = 2 \times 3.14 \times \sqrt{\frac{(4.73 \times 10^{20})^3}{(6.67 \times 10^{-11}) \times (4.97 \times 10^{41})}} \\[1em] = 6.28 \times \sqrt{\frac{1.058 \times 10^{62}}{3.315 \times 10^{31}}}\end{aligned}$$

$\text T = 6.28 \times \sqrt{3.192 \times 10^{30}} = 6.28 \times 1.787 \times 10^{15} = 1.12 \times 10^{16}\ \text s$

Expressing this in years (1 year = 3.156 × 10⁷ s),

$\text T = \frac{1.12 \times 10^{16}}{3.156 \times 10^7} = 3.55 \times 10^8\ \text{year}$

Hence, the star takes about 3.55 × 10⁸ years to complete one revolution.

(a) Kinetic energy. For a satellite revolving round the earth in a circular orbit of radius r, the gravitational potential energy and the kinetic energy are

$\text U = -\frac{\text{GM}_e \text m}{\text r} \quad \text{and} \quad \text K = \frac{1}{2}\frac{\text{GM}_e \text m}{\text r}$

so that the total energy is

$\text E = \text U + \text K = -\frac{1}{2}\frac{\text{GM}_e \text m}{\text r} = -\text K$

Hence the total energy of an orbiting satellite is negative of its kinetic energy.

(b) Less. An orbiting satellite already possesses kinetic energy $\frac{1}{2}\text{mv}_o^2$, and only the additional energy needed to raise its total energy to zero has to be supplied. The energy required to launch the orbiting satellite out of earth's gravitational field is

$\frac{1}{2}\text{mv}_e^2 - \frac{1}{2}\text{mv}_o^2 = \frac{1}{2}\text m \times \frac{2\text{GM}_e}{\text r} - \frac{1}{2}\text m \times \frac{\text{GM}_e}{\text r} = \frac{\text{GM}_e \text m}{2\text r}$

whereas a stationary object at the same height has no kinetic energy, and the energy required to project it out of earth's field is

$\frac{1}{2}\text{mv}_e^2 = \frac{\text{GM}_e \text m}{\text r}$

Hence, the energy required to launch an orbiting satellite is less, in fact half, of that required for a stationary object at the same height.

The escape velocity of a body from the earth is

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} = \sqrt{2\text{gR}_e}$

(a) No. The expression is free from the mass m of the body, so the escape velocity is independent of the mass of the body.

(b) Yes. The escape velocity depends upon the gravitational potential at the point of projection, and this potential varies slightly with the latitude of the place. Hence it depends slightly on the location from where the body is projected.

(c) No. The escape velocity is the minimum speed required to leave the gravitational field, and it is independent of the angle, that is, the direction of projection.

(d) Yes. The escape velocity depends upon the distance of the body from the centre of the earth. At a height h the escape velocity becomes $\sqrt{\frac{2\text{GM}_e}{\text R_e + \text h}}$, which decreases as h increases.

Hence, the escape velocity is independent of the mass of the body and the direction of projection, but depends on the location and the height of the point of projection.

(a) No. The comet moves fastest when it is nearest the sun (perihelion) and slowest when it is farthest (aphelion), so the linear speed is not constant.

(b) No. By Kepler's law of areas the comet sweeps out equal areas in equal times. When it is near the sun the radius vector is short, so it must sweep through a larger angle in a given time. Hence the angular speed is greater near the sun and smaller far from it.

(c) Yes. The gravitational force on the comet is a central force directed towards the sun, so it exerts no torque about the sun. Hence the angular momentum of the comet remains constant throughout the orbit.

(d) No. The kinetic energy $\frac{1}{2}\text{mv}^2$ depends on the speed, and since the speed varies along the orbit, the kinetic energy also varies.

(e) No. The potential energy $\text U = -\frac{\text{GMm}}{\text r}$ depends on the distance r from the sun. At perihelion r is least and U is most negative, while at aphelion r is largest and U is less negative.

(f) Yes. Gravitation is a conservative force, so the total mechanical energy E = K + U remains constant throughout the orbit. For a bound elliptical orbit this constant total energy is negative.

(b), (c) and (d).

Inside an orbiting satellite the astronaut is in a state of weightlessness, since the gravitational pull provides exactly the centripetal force and the reaction of the floor on him becomes zero.

Because of this weightlessness the blood and other body fluids no longer collect in the lower part of the body but are redistributed towards the upper part. This causes the swollen face and the headache. The absence of any 'up' and 'down' direction also produces the orientational problem.

Swollen feet, however, are caused by the pooling of blood in the feet under the action of gravity, which does not occur in the state of weightlessness.

Hence, the astronaut is likely to be afflicted by swollen face, headache and orientational problem, but not by swollen feet.

(iii) c

Consider the mass elements of the hemispherical shell. Because of the symmetry of the shell about its axis, the horizontal components of the gravitational intensity due to all the mass elements cancel out in pairs.

The vertical components, however, due to every element of the curved hemispherical surface, all point in the same direction and add up. The resultant gravitational intensity at the centre C of the flat face is therefore directed vertically downwards, perpendicular to the flat face, into the shell along the axis of symmetry.

Hence, the direction of the gravitational intensity at the centre is that of the arrow c.

(ii) e

The point P is not on the axis of symmetry of the hemispherical shell, so the horizontal components of the gravitational intensity due to the mass elements no longer cancel completely.

The resultant intensity at P is therefore directed towards the mass distribution of the curved hemispherical surface, that is, it points inward and is also tilted towards the shell instead of being vertically downward. The arrow e represents exactly this direction, being inclined inward and slightly to the side.

Hence, the direction of the gravitational intensity at the arbitrary point P is that of the arrow e.

Given,

• Mass of the sun, M_s = 2 × 10³⁰ kg
• Mass of the earth, M_e = 6 × 10²⁴ kg
• Distance between the earth and the sun, D = 1.5 × 10¹¹ m

Let the gravitational force on the rocket be zero at a distance x from the centre of the earth, so that its distance from the sun is (D − x). At this point the pull of the earth on the rocket is balanced by the pull of the sun.

The gravitational force on the rocket (mass m) due to the earth is

$\text F_e = \frac{\text{GM}_e \text m}{\text x^2} \qquad \text{...(i)}$

and that due to the sun is

$\text F_s = \frac{\text{GM}_s \text m}{(\text D - \text x)^2} \qquad \text{...(ii)}$

For the net force on the rocket to be zero, the two forces must be equal,

$\frac{\text{GM}_e \text m}{\text x^2} = \frac{\text{GM}_s \text m}{(\text D - \text x)^2} \quad \Rightarrow \quad \left(\frac{\text D - \text x}{\text x}\right)^2 = \frac{\text M_s}{\text M_e}$

Substituting the values,

$\frac{\text D - \text x}{\text x} = \sqrt{\frac{2 \times 10^{30}}{6 \times 10^{24}}} = \sqrt{3.333 \times 10^5} = 577.4$

$\frac{\text D}{\text x} = 1 + 577.4 = 578.4$

$\text x = \frac{1.5 \times 10^{11}}{578.4} = 2.59 \times 10^8\ \text m$

Hence, the gravitational force on the rocket is zero at a distance of 2.59 × 10⁸ m from the earth's centre.

Given,

• Mean orbital radius of the earth, r = 1.5 × 10⁸ km = 1.5 × 10¹¹ m
• Period of revolution of the earth, T = 1 year = 3.15 × 10⁷ s
• G = 6.67 × 10^-11 N m² kg^-2

The earth completes one revolution of the sun in one year, and the gravitational force exerted by the sun on the earth provides the necessary centripetal force for this revolution. Therefore,

$\frac{\text{GM}_s \text M_e}{\text r^2} = \text M_e \text r \omega^2 = \text M_e \text r \left(\frac{2\pi}{\text T}\right)^2$

Cancelling M_e and rearranging, the mass of the sun is

$\text M_s = \frac{4\pi^2 \text r^3}{\text{GT}^2}$

Substituting the values,

$$\begin{aligned}\text M_s = \frac{4 \times (3.14)^2 \times (1.5 \times 10^{11})^3}{(6.67 \times 10^{-11}) \times (3.15 \times 10^7)^2} \\[1em] = \frac{39.44 \times 3.375 \times 10^{33}}{(6.67 \times 10^{-11}) \times (9.92 \times 10^{14})}\end{aligned}$$

$\text M_s = \frac{1.331 \times 10^{35}}{6.617 \times 10^4} = 2.0 \times 10^{30}\ \text{kg}$

Hence, the mass of the sun is 2.0 × 10³⁰ kg. This is how the sun is 'weighed'.

Given,

• Period of Saturn, T₂ = 29.5 times the earth year, that is, T₂ = 29.5 years
• Period of the earth, T₁ = 1 year
• Distance of the earth from the sun, r₁ = 1.50 × 10⁸ km

By Kepler's third law, T² = K r³, where T is the period of a planet and r is its distance from the sun. For the earth and Saturn,

$\frac{\text T_1}{\text T_2} = \left(\frac{\text r_1}{\text r_2}\right)^{3/2}$

Substituting the values,

$\frac{1}{29.5} = \left(\frac{1.50 \times 10^8\ \text{km}}{\text r_2}\right)^{3/2}$

$$\begin{aligned}\text r_2 = (1.50 \times 10^8\ \text{km}) \times (29.5)^{2/3} \\[1em] = 1.50 \times 10^8 \times 9.547\end{aligned}$$

$\text r_2 = 1.43 \times 10^9\ \text{km}$

Hence, Saturn is 1.43 × 10⁹ km away from the sun.

Given,

• Weight of the body on the earth's surface, mg = 63 N
• Height above the surface, $\text h = \frac{\text R_e}{2}$

The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Hence the weight of the body of mass m at this height is

$\text{mg}' = \frac{\text{mg}}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Putting $\text h = \frac{\text R_e}{2}$,

$\text{mg}' = \frac{\text{mg}}{\left(1 + \frac{\text R_e/2}{\text R_e}\right)^2} = \frac{\text{mg}}{\left(\frac{3}{2}\right)^2} = \text{mg} \times \frac{4}{9}$

Substituting mg = 63 N,

$\text{mg}' = 63 \times \frac{4}{9} = 28\ \text N$

Hence, the gravitational force on the body at that height is 28 N.

Given,

• Weight of the body on the earth's surface, W = mg = 250 N
• Depth below the surface, $\text h = \frac{\text R_e}{2}$

The acceleration due to gravity at a depth h below the earth's surface varies as

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right) \qquad \text{...(i)}$

Putting $\text h = \frac{\text R_e}{2}$ in equation (i),

$\text g' = \text g\left(1 - \frac{\text R_e}{2\text R_e}\right) = \frac{\text g}{2}$

The weight of the body at this depth is

$\text W' = \text{mg}' = \frac{\text{mg}}{2} = \frac{250}{2} = 125\ \text N$

Hence, the body would weigh 125 N half way down to the centre of the earth.

Given,

• Speed of projection, v = 5 km s^-1 = 5 × 10³ m s^-1
• Mass of the earth, M_e = 6.0 × 10²⁴ kg
• Radius of the earth, R_e = 6.4 × 10⁶ m
• G = 6.67 × 10^-11 N m² kg^-2

The gravitational potential energies of the rocket (mass m) at the surface of the earth and at a height h from the surface are

$\text U = -\frac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text U_h = -\frac{\text{GM}_e \text m}{\text R_e + \text h}$

Therefore the increase in gravitational potential energy is

$\text U_h - \text U = \text{GM}_e \text m\left(\frac{1}{\text R_e} - \frac{1}{\text R_e + \text h}\right) = \frac{\text{GM}_e \text{mh}}{\text R_e(\text R_e + \text h)}$

This increase is obtained from the initial kinetic energy given to the rocket. Hence

$\frac{\text{GM}_e \text{mh}}{\text R_e(\text R_e + \text h)} = \frac{1}{2}\text{mv}^2$

$\text R_e(\text R_e + \text h)\text v^2 = 2\text{GM}_e \text h \quad \Rightarrow \quad \text h = \frac{\text R_e^2 \text v^2}{2\text{GM}_e - \text R_e \text v^2}$

Substituting the values,

$$\begin{aligned}\text h = \frac{(6.4 \times 10^6)^2 \times (5 \times 10^3)^2}{2 \times (6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) - (6.4 \times 10^6) \times (5 \times 10^3)^2} \\[1em] = \frac{(4.096 \times 10^{13}) \times (2.5 \times 10^7)}{(8.004 \times 10^{14}) - (1.60 \times 10^{14})}\end{aligned}$$

$\text h = \frac{1.024 \times 10^{21}}{6.404 \times 10^{14}} = 1.6 \times 10^6\ \text m$

Hence, the rocket goes to a height of 1.6 × 10⁶ m from the earth's surface before returning.

Given,

• Escape velocity of the projectile, v_e = 11.2 km s^-1
• Initial speed of projection, v_i = 3v_e

Let m be the mass of the projectile and v_f its final speed at infinity. The initial kinetic energy is $\frac{1}{2}\text{mv}_i^2$ and the initial gravitational potential energy on the earth's surface is $-\frac{\text{GM}_e \text m}{\text R_e}$. At infinity the potential energy is zero and the kinetic energy is $\frac{1}{2}\text{mv}_f^2$.

By the conservation of energy,

$\frac{1}{2}\text{mv}_i^2 - \frac{\text{GM}_e \text m}{\text R_e} = \frac{1}{2}\text{mv}_f^2 \qquad \text{...(i)}$

The escape velocity v_e of the projectile is given by

$\frac{1}{2}\text{mv}_e^2 = \frac{\text{GM}_e \text m}{\text R_e}$

Making this substitution in equation (i),

$\text v_i^2 - \text v_e^2 = \text v_f^2 \quad \Rightarrow \quad \text v_f = \sqrt{\text v_i^2 - \text v_e^2}$

Now v_i = 3v_e, therefore

$$\begin{aligned}\text v_f = \sqrt{9\text v_e^2 - \text v_e^2} = \sqrt8\ \text v_e = 2.83 \times 11.2 \\[1em] = 31.7\ \text{km s}^{-1}\end{aligned}$$

Hence, the speed of the body at infinity is 31.7 km s^-1.

Given,

• Mass of the satellite, m = 200 kg
• Height of the orbit, h = 400 km = 0.4 × 10⁶ m
• Mass of the earth, M_e = 6.0 × 10²⁴ kg
• Radius of the earth, R_e = 6.4 × 10⁶ m
• G = 6.67 × 10^-11 N m² kg^-2

The total energy of a satellite of mass m orbiting the earth in a circle of radius r is

$\text E = -\frac{\text{GM}_e \text m}{2\text r}$

For a satellite at a height h above the surface, r = R_e + h, so that

$\text E = -\frac{\text{GM}_e \text m}{2(\text R_e + \text h)}$

Here,

$\text R_e + \text h = 6.4 \times 10^6 + 0.4 \times 10^6 = 6.8 \times 10^6\ \text m$

Substituting the values,

$$\begin{aligned}\text E = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 200}{2 \times (6.8 \times 10^6)} \\[1em] = -\frac{8.004 \times 10^{16}}{1.36 \times 10^7} = -5.89 \times 10^9\ \text J\end{aligned}$$

To send the satellite out of the earth's gravitational field its total energy must be raised to zero. Hence the energy to be expended is

$\text W = 0 - \text E = 0 - (-5.89 \times 10^9) = 5.89 \times 10^9\ \text J$

Hence, 5.89 × 10⁹ J of energy must be expended to rocket the satellite out of the earth's gravitational influence.

Given,

• Mass of each star, M = 1 solar mass = 2 × 10³⁰ kg
• Radius of each star, R = 10⁴ km = 10⁷ m
• Initial separation, r = 10⁹ km = 10¹² m
• G = 6.67 × 10^-11 N m² kg^-2

The initial gravitational potential energy of the star-system is

$$\begin{aligned}\text U_i = -\frac{\text{GMM}}{\text r} = -\frac{(6.67 \times 10^{-11}) \times (2 \times 10^{30}) \times (2 \times 10^{30})}{10^{12}} \\[1em] = -2.67 \times 10^{38}\ \text J\end{aligned}$$

Since their speeds at this separation are negligible, the initial kinetic energy is zero and the initial total energy is

$\text E_i = -2.67 \times 10^{38}\ \text J$

When the stars collide, the separation between their centres is 2R, so the final potential energy is

$$\begin{aligned}\text U_f = -\frac{\text{GMM}}{2\text R} = -\frac{(6.67 \times 10^{-11}) \times (2 \times 10^{30}) \times (2 \times 10^{30})}{2 \times 10^7} \\[1em] = -1.33 \times 10^{43}\ \text J\end{aligned}$$

If v be the speed of each star at collision, the kinetic energy acquired by the system is

$\text K_f = 2 \times \frac{1}{2}\text{Mv}^2 = (2 \times 10^{30})\text v^2$

By the conservation of energy, E_i = E_f,

$-2.67 \times 10^{38} = -1.33 \times 10^{43} + (2 \times 10^{30})\text v^2$

$(2 \times 10^{30})\text v^2 = 1.33 \times 10^{43} - 2.67 \times 10^{38} \approx 1.33 \times 10^{43}\ \text J$

$\text v = \sqrt{\frac{1.33 \times 10^{43}}{2 \times 10^{30}}} = 2.58 \times 10^6\ \text{m s}^{-1}$

Hence, the stars collide with a speed of 2.58 × 10⁶ m s^-1 each.

Given,

• Mass of each sphere, M = 100 kg
• Radius of each sphere = 0.10 m
• Distance between the centres = 1.0 m
• G = 6.67 × 10^-11 N m² kg^-2

Since the spheres are uniform, their masses may be taken to be concentrated at their centres. The distance of the mid-point from the centre of either sphere is

$\text r = \frac{1.0}{2} = 0.5\ \text m$

Gravitational field : The gravitational field intensities at the mid-point due to the two spheres are each of magnitude $\frac{\text{GM}}{\text r^2}$ but are directed opposite to each other, since the mid-point lies between the two spheres. Hence the resultant gravitational field at the mid-point is zero.

Gravitational potential : The potential at the mid-point due to each sphere is $-\frac{\text{GM}}{\text r}$. Potential is a scalar quantity, so the two potentials are added algebraically,

$$\begin{aligned}\text V = -\frac{2\text{GM}}{\text r} = -\frac{2 \times (6.67 \times 10^{-11}) \times 100}{0.5} \\[1em] = -2.67 \times 10^{-8}\ \text{J kg}^{-1}\end{aligned}$$

Equilibrium : An object placed at the mid-point is attracted equally by the two spheres in opposite directions, so the net force on it is zero and it is in equilibrium.

This equilibrium is, however, unstable. If the object is displaced slightly towards either sphere, it comes nearer to that sphere and farther from the other, so the attraction towards the nearer sphere becomes greater. The net force then pulls it further towards that sphere and it does not return to its original position.

Hence, the gravitational field at the mid-point is zero, the potential is −2.67 × 10^-8 J kg^-1, and an object placed there is in unstable equilibrium.

Given,

• Acceleration due to gravity on the planet, $\text g_p = \frac{\sqrt6}{11}\text g$
• Average mass density of the planet, $\rho_p = \frac{2}{3}\rho_e$
• Escape speed on the earth, v_e = 11 km s^-1

The acceleration due to gravity on a body of mass M and radius R is

$\text g = \frac{\text{GM}}{\text R^2} = \frac{\text G}{\text R^2} \times \frac{4}{3}\pi \text R^3 \rho = \frac{4}{3}\pi \text{GR}\rho$

that is, g ∝ R ρ. Applying this to the planet and the earth,

$\frac{\text g_p}{\text g} = \frac{\text R_p \rho_p}{\text R_e \rho_e} \quad \Rightarrow \quad \frac{\text R_p}{\text R_e} = \frac{\text g_p}{\text g} \times \frac{\rho_e}{\rho_p}$

Substituting the given values,

$\frac{\text R_p}{\text R_e} = \frac{\sqrt6}{11} \times \frac{3}{2} = \frac{3\sqrt6}{22}$

The escape speed from the surface of a planet is $\text v_e = \sqrt{2\text{gR}}$, therefore

$$\begin{aligned}\frac{\text v_{ep}}{\text v_e} = \sqrt{\frac{\text g_p \text R_p}{\text g\text R_e}} = \sqrt{\frac{\sqrt6}{11} \times \frac{3\sqrt6}{22}} \\[1em] = \sqrt{\frac{3 \times 6}{11 \times 22}} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}\end{aligned}$$

$\text v_{ep} = \frac{3}{11} \times 11 = 3\ \text{km s}^{-1}$

Hence, the escape speed on the surface of the planet is 3 km s^-1.

Given,

• Velocity of projection of the bullet = v
• Acceleration due to gravity at the maximum height $= \frac{1}{4}$ of its value at the surface
• Escape velocity from the planet, $\text v_e = \text v\sqrt{\text N}$

Let R be the radius of the planet and h the maximum height reached by the bullet. The acceleration due to gravity at a height h is

$\text g' = \frac{\text{GM}}{(\text R + \text h)^2} \quad \text{and at the surface} \quad \text g = \frac{\text{GM}}{\text R^2}$

Since $\text g' = \frac{\text g}{4}$,

$\frac{\text R^2}{(\text R + \text h)^2} = \frac{1}{4} \quad \Rightarrow \quad \text R + \text h = 2\text R \quad \Rightarrow \quad \text h = \text R$

Applying the conservation of energy between the surface and the highest point, where the bullet is momentarily at rest,

$\frac{1}{2}\text{mv}^2 - \frac{\text{GMm}}{\text R} = 0 - \frac{\text{GMm}}{\text R + \text h}$

Putting h = R,

$\frac{1}{2}\text{mv}^2 = \frac{\text{GMm}}{\text R} - \frac{\text{GMm}}{2\text R} = \frac{\text{GMm}}{2\text R}$

$\text v^2 = \frac{\text{GM}}{\text R} \quad \Rightarrow \quad \text v = \sqrt{\frac{\text{GM}}{\text R}}$

The escape velocity from the planet is

$\text v_e = \sqrt{\frac{2\text{GM}}{\text R}} = \sqrt2 \times \sqrt{\frac{\text{GM}}{\text R}} = \text v\sqrt2$

Comparing this with the given relation $\text v_e = \text v\sqrt{\text N}$,

$\text N = 2$

Hence, the value of N is 2.

Given,

• Large fixed spherical mass = M
• Two identical point masses, each = m, joined by a rigid rod of length l
• Distance of the nearer point mass from M, r = 3l
• Tension in the rod is zero when $\text m = \text k\left(\frac{\text M}{288}\right)$

Since the rod is rigid, both the point masses move with the same acceleration a. When the tension in the rod is zero, each point mass is acted upon only by the gravitational forces due to M and due to the other point mass.

For the point mass at r = 3l : It is attracted towards M and away from M by the farther point mass, so

$\frac{\text{GMm}}{(3\text l)^2} - \frac{\text{Gm}^2}{\text l^2} = \text{ma} \qquad \text{...(i)}$

For the point mass at r = 4l : It is attracted towards M both by M and by the nearer point mass, so

$\frac{\text{GMm}}{(4\text l)^2} + \frac{\text{Gm}^2}{\text l^2} = \text{ma} \qquad \text{...(ii)}$

Equating the right-hand sides of equations (i) and (ii),

$\frac{\text{GM}}{9\text l^2} - \frac{\text{Gm}}{\text l^2} = \frac{\text{GM}}{16\text l^2} + \frac{\text{Gm}}{\text l^2}$

Cancelling $\frac{\text G}{\text l^2}$ throughout,

$$\begin{aligned}\frac{\text M}{9} - \frac{\text M}{16} = 2\text m \\[1em] \text M\left(\frac{16 - 9}{144}\right) = 2\text m\end{aligned}$$

$\frac{7\text M}{144} = 2\text m \quad \Rightarrow \quad \text m = \frac{7\text M}{288} = 7\left(\frac{\text M}{288}\right)$

Comparing this with $\text m = \text k\left(\frac{\text M}{288}\right)$,

$\text k = 7$

Hence, the value of k is 7.

6.4 × 10¹⁰ J.

Reason — Given,

• Mass of the spaceship, m = 1000 kg
• Acceleration due to gravity, g = 10 m s^-2
• Radius of the earth, R = 6400 km = 6.4 × 10⁶ m

To launch the spaceship from the earth's surface out into free space, energy equal to its binding energy must be supplied. The gravitational potential energy of the spaceship on the earth's surface is

$\text U = -\frac{\text{GM}_e \text m}{\text R} = -\text{mgR} \qquad (\because\ \text{GM}_e = \text{gR}^2)$

In free space the potential energy is zero, so the energy required is

$\text E = 0 - \text U = \text{mgR}$

Substituting the values,

$$\begin{aligned}\text E = 1000 \times 10 \times (6.4 \times 10^6) \\[1em] = 6.4 \times 10^{10}\ \text J\end{aligned}$$

2 R

Reason — Given,

• Acceleration due to gravity at the height h, $\text g' = \frac{\text g}{9}$
• Radius of the earth = R

The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R}\right)^2}$

Putting $\text g' = \frac{\text g}{9}$,

$\frac{\text g}{9} = \frac{\text g}{\left(1 + \frac{\text h}{\text R}\right)^2} \quad \Rightarrow \quad \left(1 + \frac{\text h}{\text R}\right)^2 = 9$

$1 + \frac{\text h}{\text R} = 3 \quad \Rightarrow \quad \frac{\text h}{\text R} = 2$

$\text h = 2\text R$

6.67 × 10^-10 J

Reason — Given,

• Mass of the particle, m = 10 g = 10 × 10^-3 kg
• Mass of the sphere, M = 100 kg
• Radius of the sphere, R = 10 cm = 10 × 10^-2 m
• G = 6.67 × 10^-11 N m² kg^-2

The particle rests on the surface of the sphere, so its distance from the centre is R. The gravitational potential energy of the particle at the surface is

$\text U = -\frac{\text{GMm}}{\text R}$

When the particle is taken far away from the sphere its potential energy becomes zero. Hence the work to be done against the gravitational force is

$\text W = 0 - \text U = \frac{\text{GMm}}{\text R}$

Substituting the values,

$$\begin{aligned}\text W = \frac{(6.67 \times 10^{-11}) \times 100 \times (10 \times 10^{-3})}{10 \times 10^{-2}} \\[1em] = \frac{6.67 \times 10^{-11}}{10 \times 10^{-2}} \times 1\end{aligned}$$

$\text W = 6.67 \times 10^{-10}\ \text J$

Reason — The gravitational pull of the enclosed mass supplies the centripetal force needed by the test mass for its circular motion, so at a distance r,

$\frac{\text{GM}'\text m}{\text r^2} = \frac{\text{mv}^2}{\text r} \quad \Rightarrow \quad \text v = \sqrt{\frac{\text{GM}'}{\text r}}$

where M' is the mass enclosed within the radius r.

For r ≤ R : The mass enclosed within radius r is

$\text M' = \frac{4}{3}\pi \text r^3 \rho_0$

Therefore,

$\text v = \sqrt{\frac{\text G}{\text r} \times \frac{4}{3}\pi \text r^3 \rho_0} = \sqrt{\frac{4}{3}\pi \text G\rho_0}\ \text r \quad \Rightarrow \quad \text v \propto \text r$

Thus the speed increases linearly with r up to r = R.

For r > R : The total mass enclosed within radius r is M, So

$\text v = \sqrt{\frac{\text{GM}}{\text r}} \quad \Rightarrow \quad \text v \propto \frac{1}{\sqrt{\text r}}$

Thus beyond R the speed falls off as $\frac{1}{\sqrt{\text r}}$.

The graph must therefore rise linearly from the centre up to r = R and then decrease gradually, which is the variation shown in option (c).

Reason — The intensity of the gravitational field of the earth at a distance r from its centre is obtained from the mass enclosed within that distance.

Inside the earth (r < R) : By Newton's shell theorem only the inner sphere of radius r attracts the body, and its mass is $\frac{4}{3}\pi \text r^3 \rho$. Hence

$\text E = \frac{\text G}{\text r^2} \times \frac{4}{3}\pi \text r^3 \rho = \frac{4}{3}\pi \text G\rho\text r \quad \Rightarrow \quad \text E \propto \text r$

So E is zero at the centre and increases linearly with r, becoming maximum at the surface.

Outside the earth (r > R) : The whole mass M_e of the earth acts as if concentrated at the centre, so

$\text E = \frac{\text{GM}_e}{\text r^2} \quad \Rightarrow \quad \text E \propto \frac{1}{\text r^2}$

So beyond the surface E decreases sharply as $\frac{1}{\text r^2}$ and tends to zero as r → ∞.

The correct graph is therefore the one that rises linearly up to r = R and then falls off as $\frac{1}{\text r^2}$, which is option (a).

2 h

Reason — Given,

• Radius of the orbit of the geostationary satellite, r = 36000 km
• Period of the geostationary satellite, T = 24 h
• The spy satellite revolves at a few hundred km above the earth's surface, so its orbital radius is r' ≈ 6400 km (radius of the earth)

By Kepler's third law, T² ∝ r³. Therefore for the two satellites,

$\left(\frac{\text T'}{\text T}\right)^2 = \left(\frac{\text r'}{\text r}\right)^3 \quad \Rightarrow \quad \text T' = \text T\left(\frac{\text r'}{\text r}\right)^{3/2}$

Substituting the values,

$$\begin{aligned}\text T' = 24 \times \left(\frac{6400}{36000}\right)^{3/2} = 24 \times (0.178)^{3/2} \\[1em] = 24 \times 0.075 = 1.8\ \text h\end{aligned}$$

This is nearest to 2 h.

$-\frac{9Gm}{r}$

Reason — Given,

• Masses of the two bodies = m and 4m
• Distance between them = r

Let the gravitational field be zero at a point P on the line joining them, at a distance x from the mass m. Then its distance from the mass 4m is (r − x). At P the two field intensities are equal and opposite,

$\frac{\text{Gm}}{\text x^2} = \frac{\text G(4\text m)}{(\text r - \text x)^2} \quad \Rightarrow \quad \left(\frac{\text r - \text x}{\text x}\right)^2 = 4$

$\frac{\text r - \text x}{\text x} = 2 \quad \Rightarrow \quad \text r = 3\text x \quad \Rightarrow \quad \text x = \frac{\text r}{3}$

The distance of P from the mass 4m is therefore

$\text r - \text x = \text r - \frac{\text r}{3} = \frac{2\text r}{3}$

Gravitational potential is a scalar quantity, so the potentials due to the two masses are added algebraically,

$$\begin{aligned}\text V = -\frac{\text{Gm}}{\text r/3} - \frac{\text G(4\text m)}{2\text r/3} \\[1em] = -\frac{3\text{Gm}}{\text r} - \frac{6\text{Gm}}{\text r}\end{aligned}$$

$\text V = -\frac{9\text{Gm}}{\text r}$

$\frac{-GM}{R}$

Reason — Given,

• Mass of the solid sphere = M, radius = R
• Radius of the spherical portion removed $= \frac{\text R}{2}$
• Gravitational potential V = 0 at r = ∞

The potential at the point P is obtained by treating the given body as the complete solid sphere together with a sphere of negative mass filling the cavity.

Potential inside a uniform solid sphere : At a point distant r from the centre, only the mass enclosed within the radius r contributes to the field. Since the density is uniform,

$\text M' = \text M \times \frac{\frac{4}{3}\pi \text r^3}{\frac{4}{3}\pi \text R^3} = \frac{\text{Mr}^3}{\text R^3}$

so the field intensity inside the sphere is

$\text I = \frac{\text{GM}'}{\text r^2} = \frac{\text G}{\text r^2} \times \frac{\text{Mr}^3}{\text R^3} = \frac{\text{GMr}}{\text R^3} \qquad \text{...(i)}$

while outside the sphere the whole mass acts as if concentrated at the centre,

$\text I = \frac{\text{GM}}{\text r^2} \qquad \text{...(ii)}$

Integrating the field from infinity up to the surface, using equation (ii), and taking V(∞) = 0,

$\text V(\text R) = \int_\infty^{\text R} \frac{\text{GM}}{\text x^2}\text{dx} = \text{GM}\left[-\frac{1}{\text x}\right]_\infty^{\text R} = -\frac{\text{GM}}{\text R} \qquad \text{...(iii)}$

Continuing inward from the surface to the point r, now using equation (i),

$$\begin{aligned}\text V(\text r) - \text V(\text R) = \int_{\text R}^{\text r} \frac{\text{GMx}}{\text R^3}\text{dx} = \frac{\text{GM}}{\text R^3}\left[\frac{\text x^2}{2}\right]_{\text R}^{\text r} \\[1em] = \frac{\text{GM}}{2\text R^3}(\text r^2 - \text R^2)\end{aligned}$$

Substituting V(R) from equation (iii) and taking $\frac{\text{GM}}{2\text R^3}$ common,

$\text V(\text r) = -\frac{\text{GM}}{\text R} + \frac{\text{GM}}{2\text R^3}(\text r^2 - \text R^2) = \frac{\text{GM}}{2\text R^3}\left(-2\text R^2 + \text r^2 - \text R^2\right)$

$\text V(\text r) = -\text{GM}\left(\frac{3\text R^2 - \text r^2}{2\text R^3}\right) \qquad \text{...(iv)}$

Potential at P due to the complete solid sphere : The centre of the cavity lies at a distance $\text r = \frac{\text R}{2}$ from the centre of the sphere. Substituting this in equation (iv),

$\text V_1 = -\text{GM}\left[\frac{3\text R^2 - \left(\frac{\text R}{2}\right)^2}{2\text R^3}\right]$

Simplifying the numerator,

$3\text R^2 - \frac{\text R^2}{4} = \frac{12\text R^2 - \text R^2}{4} = \frac{11\text R^2}{4}$

Therefore,

$\text V_1 = -\text{GM} \times \frac{11\text R^2}{4 \times 2\text R^3} = -\frac{11}{8}\frac{\text{GM}}{\text R}$

Potential at P due to the removed sphere : The mass of the removed portion of radius $\frac{\text R}{2}$ is

$\text M'' = \text M \times \frac{\frac{4}{3}\pi\left(\frac{\text R}{2}\right)^3}{\frac{4}{3}\pi \text R^3} = \frac{\text M}{8}$

The point P is at the centre of this removed sphere. Putting r = 0 in equation (iv), the potential at the centre of a uniform sphere of mass M'' and radius $\frac{\text R}{2}$ is

$\text V = -\text{GM}''\left(\frac{3\left(\frac{\text R}{2}\right)^2}{2\left(\frac{\text R}{2}\right)^3}\right) = -\frac{3}{2}\frac{\text{GM}''}{\left(\frac{\text R}{2}\right)}$

Since this mass has been removed, it is taken as negative,

$\text V_2 = -\frac{3}{2} \times \frac{\text G\left(-\frac{\text M}{8}\right)}{\frac{\text R}{2}} = \frac{3}{2} \times \frac{\text{GM}}{8} \times \frac{2}{\text R} = \frac{3}{8}\frac{\text{GM}}{\text R}$

Resultant potential at the centre of the cavity : Potential is a scalar quantity, so the two are added algebraically,

$$\begin{aligned}\text V = \text V_1 + \text V_2 = -\frac{11}{8}\frac{\text{GM}}{\text R} + \frac{3}{8}\frac{\text{GM}}{\text R} \\[1em] = -\frac{8}{8}\frac{\text{GM}}{\text R} = -\frac{\text{GM}}{\text R}\end{aligned}$$

Given,

• Speed of projection, v = 2 km s^-1 = 2 × 10³ m s^-1
• Energy lost to atmospheric resistance = 20% of the initial energy
• Mass of Mars, M = 6.4 × 10²³ kg
• Radius of Mars, R = 3395 km = 3.395 × 10⁶ m
• G = 6.67 × 10^-11 N m² kg^-2

The gravitational potential energies of the rocket (mass m) on the surface of Mars and at a distance r from the centre of Mars are

$\text U_R = -\frac{\text{GMm}}{\text R} \quad \text{and} \quad \text U_r = -\frac{\text{GMm}}{\text r}$

Therefore the increase in the potential energy of the rocket is

$\text U_r - \text U_R = \text{GMm}\left(\frac{1}{\text R} - \frac{1}{\text r}\right)$

Only 80% of the initial kinetic energy is available for this increase, since 20% is lost against atmospheric resistance. Hence

$\text{GMm}\left(\frac{1}{\text R} - \frac{1}{\text r}\right) = 0.8\left(\frac{1}{2}\text{mv}^2\right)$

$\frac{1}{\text R} - \frac{1}{\text r} = \frac{0.4\text v^2}{\text{GM}} \quad \Rightarrow \quad \frac{1}{\text r} = \frac{1}{\text R} - \frac{0.4\text v^2}{\text{GM}}$

Substituting the values,

$$\begin{aligned}\frac{1}{\text r} = \frac{1}{3.395 \times 10^6} - \frac{0.4 \times (2 \times 10^3)^2}{(6.67 \times 10^{-11}) \times (6.4 \times 10^{23})} \\[1em] = 0.2945 \times 10^{-6} - 0.0375 \times 10^{-6}\end{aligned}$$

$\frac{1}{\text r} = 0.257 \times 10^{-6}\ \text m^{-1} \quad \Rightarrow \quad \text r = 3.89 \times 10^6\ \text m = 3890\ \text{km}$

The distance of the rocket above the surface of Mars is therefore

$\text r - \text R = 3890 - 3395 = 495\ \text{km}$

Hence, the rocket goes 495 km above the surface of Mars before returning to it.

The gravitational force between two masses is directly proportional to the product of their masses.

Reason — Newton's universal law of gravitation states that the gravitational force of attraction between any two material particles is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them, acting along the line joining them. Thus

$\text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$

Options 1, 2 and 4 contradict this relation, since the force is not inversely proportional to the product of the masses, not directly proportional to the square of the distance, and certainly not independent of the distance.

four times smaller

Reason — By Newton's law of gravitation $\text F \propto \frac{1}{\text r^2}$. If the distance is doubled, r becomes 2r, so

$\frac{\text F'}{\text F} = \left(\frac{\text r}{2\text r}\right)^2 = \frac{1}{4}$

Hence the force is reduced to one-fourth of its original value, that is, it becomes four times smaller. This is why the force-distance relationship is called the inverse-square law.

None of these

Reason — The gravitational constant G is the constant of proportionality in $\text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$. Its value is the same for all pairs of bodies, at all places, at all distances of separation, and it is also independent of the presence of other bodies or of the properties of the intervening medium. For this reason G is called a 'universal constant', and it does not vary with distance or with mass.

inverse-square law

Reason — Since the gravitational force varies as $\frac{1}{\text r^2}$, doubling the distance between two objects results in a force which is four times smaller. A relationship of this kind is termed the 'inverse-square law'. An inverse-cube law would give $\text F \propto \frac{1}{\text r^3}$, which is not the case for gravitation.

Centre of the Earth

Reason — At a depth h below the earth's surface the acceleration due to gravity varies as

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

At the centre of the earth h = R_e, so that

$\text g' = \text g\left(1 - \frac{\text R_e}{\text R_e}\right) = 0$

Thus the acceleration due to gravity is exactly zero at the centre of the earth, which is the least value it can have anywhere.

Among the three locations on the surface, g is least at the equator (about 9.78 m s^-2) and greatest at the poles (about 9.83 m s^-2), the tropics lying in between. This is because the earth is an oblate spheroid, so that $\text g \propto \frac{1}{\text R_e^2}$ gives a smaller value where the radius is larger, and because the rotation of the earth reduces g by the term R_eω²cos²λ, which is maximum at the equator and zero at the poles. Even the smallest of these surface values, however, is far greater than zero.

Note: The printed answer key marks option 1, considering only locations on the earth's surface. However, since g = 0 at the centre of the earth, option 3 is the correct answer as given.

increase in altitude

Reason — The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

so that g' < g. As we go above the surface of the earth, the acceleration due to gravity goes on decreasing. The value of g in fact increases with increasing latitude, and it does not depend on the core temperature of the earth at all.

is always finite and independent of the presence of any external body

Reason — The gravitational field due to a mass is the region of space around it in which any other mass experiences an attractive gravitational force. Theoretically this field extends up to an infinite distance, but at large distances its strength becomes negligibly small, though never exactly zero. The field of an object exists whether or not another object is nearby; the presence of an external body is required only for the gravitational force to be exerted, not for the field to exist.

They have the same velocities at any instant.

Reason — The acceleration due to gravity at the surface of the moon,

$\text g_m = \frac{\text{GM}_m}{\text R_m^2}$

is free from the mass of the falling body. Hence two bodies of different masses fall with the same acceleration, and since the moon has no atmosphere there is no buoyancy or viscous drag to disturb this. Starting together from rest, both therefore have the same velocity at any instant and reach the surface simultaneously.

The forces on them are not the same, since F = mg_m is proportional to the mass, so the heavier body experiences the greater gravitational force. Their inertia also does not change, inertia being a constant property of a body.

Gravity

Reason — The molecules of the atmosphere are held to the earth by the earth's gravitational attraction. Because the escape velocity at the earth's surface (11.2 km s^-1) is much greater than the average thermal speed of the atmospheric molecules, these molecules are generally unable to reach the speed necessary to escape into space, and the earth is able to retain a thick atmosphere. The moon, whose escape velocity is only about 2.4 km s^-1, cannot retain an atmosphere for this reason.

Mass of the Earth.

Reason — The acceleration due to gravity at the earth's surface is

$\text g = \frac{\text{GM}_e}{\text R_e^2}$

This expression contains the mass M_e and the radius R_e of the earth, but is free from the mass of the body. Hence g depends on the mass of the earth and not upon the mass, volume, shape or size of the body placed on it.

Zero

Reason — By Newton's shell theorem, a spherically symmetric shell of mass exerts no net gravitational force on a particle located anywhere inside the shell. Every small mass element of the shell pulls the particle in one direction, but there is always a corresponding arrangement of mass elements on the opposite side pulling it the other way, and because of the symmetry of the shell these contributions cancel out exactly. Hence the gravitational force on a particle inside a spherical shell is zero, whatever its position inside.

Note: Options 1 and 3 are printed identically as $\frac{\text{GM}}{\text R^2}$ in the textbook. This does not affect the answer, since the correct option is 4.

move to a higher orbit.

Reason — When the speed of the satellite is increased slightly, its kinetic energy increases and the balance between the gravitational force and the centripetal force required for the circular orbit is disturbed. The satellite therefore no longer moves in a circular orbit but enters an elliptical orbit, moving away from the earth to a higher altitude at its apogee while coming closer to the earth at its perigee. In this way the new orbit has a greater average distance from the earth than the original circular orbit.

The satellite will gradually spiral inwards towards the Earth.

Reason — Atmospheric drag is a resistive force which acts opposite to the direction of motion of the satellite and therefore does negative work on it. This causes the satellite to lose energy continuously. As its total energy decreases (becomes more negative), the radius of its orbit becomes smaller, resulting in a gradual decrease in altitude. The satellite therefore spirals inward towards the earth.

They appear stationary relative to a fixed point on Earth.

Reason — A geostationary satellite is placed in an orbit such that its period of revolution is exactly equal to the period of the axial motion of the earth, that is, 24 h, and its direction of rotation is the same as that of the earth, in an orbital plane coplanar with the equatorial plane. Being synchronous with the earth's spin, it appears stationary over a fixed point on the earth's equator. Its orbit is circular, not highly elliptical, and its orbital speed (about 3.1 km s^-1) is much smaller than that of a low earth orbit satellite (about 8 km s^-1).

They can cover the entire Earth's surface.

Reason — A geostationary satellite remains fixed over one point on the equator and therefore provides continuous coverage of the same geographical region only. It has limited effectiveness near the poles, which is why polar regions rely on polar-orbiting satellites. Covering the entire earth's surface is a characteristic of polar satellites, which sweep over the whole globe as the earth rotates beneath their orbit. Orbiting in 24 h, being positioned above the equator, and having a circular orbit are all genuine characteristics of geostationary satellites.

be positioned above the equator

Reason — The conditions for a satellite to be geostationary are that its period of revolution (or angular velocity) should be the same as that of the earth, that its direction of rotation should be the same as that of the earth from west to east, and that its orbital plane should be coplanar with the equatorial plane of the earth. Hence it must be positioned above the equator. Its period must be 24 h and not 12 h, its altitude must be about 35,830 km and not 100 km, and a polar orbit would not keep it stationary over any point.

the centripetal force provided by earth's gravitational pull.

Reason — When a satellite revolves in a circular orbit around the earth, the gravitational force of attraction between the earth and the satellite provides exactly the centripetal force required for its circular motion,

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$

No engine, tether or atmospheric drag is involved; this is why a satellite continues to revolve in its orbit without using any fuel.

free fall of the satellite and its occupants.

Reason — Inside an orbiting satellite the gravitational force on a spaceman supplies exactly the centripetal force needed for his circular motion, so the reactionary force R of the base of the satellite on him becomes zero,

$\frac{\text{GM}_e \text m'}{\text r^2} - \text R = \frac{\text m'\text v_o^2}{\text r} \quad \Rightarrow \quad \text R = 0$

Since weight is felt only through this reaction, he feels his weight to be zero. Gravity is certainly present at that height, and the satellite has a large orbital velocity; the weightlessness is entirely due to the satellite and its occupants being in a state of free fall.

negative

Reason — For a satellite revolving close to the earth's surface, the potential energy and the kinetic energy are

$\text U = -\frac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text K = \frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e}$

so the total energy is

$\text E = \text U + \text K = -\frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e}$

The kinetic energy is less in magnitude than the potential energy, so the total energy is negative. This negative total energy indicates that the satellite is bound to its orbit, and additional energy must be supplied to free it.

The centripetal force is equal to the gravitational force.

Reason — A satellite remains in a stable circular orbit only when the gravitational force of the earth on it supplies precisely the centripetal force needed for that orbit,

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$

If the satellite's speed were equal to the escape velocity it would leave the orbit altogether, and a satellite moving in an orbit certainly does not have zero angular momentum.

The orbit of a planet is an ellipse with the Sun at one focus.

Reason — Kepler's first law is the law of orbits, which states that all planets move around the sun in elliptical orbits having the sun at one focus of the orbit. Option 1 is the statement of Kepler's third law (the law of periods) and option 3 is a consequence of the second law (the law of areas). A perfect circle is only a special case, and is not what the first law asserts.

equal time intervals

Reason — Kepler's second law is the law of areas, according to which a line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant. As a consequence the planet's velocity is maximum when it is nearest the sun and minimum when it is farthest from the sun.

a constant.

Reason — By Kepler's law of periods, the square of the period of revolution of a planet is directly proportional to the cube of its mean distance from the sun,

$\text T^2 \propto \text r^3 \quad \text{or} \quad \text T^2 = \text K\text r^3$

so that $\frac{\text T^2}{\text r^3} = \text K$. The constant K is the same for all the planets and does not change with time, with distance from the sun, or with orbital velocity.

The satellite is in free fall towards Earth.

Reason — The satellite and everything inside it are in a state of free fall towards the earth, since the gravitational force is entirely used up in providing the centripetal force for the circular orbit. The reactionary force of the floor of the satellite on the astronaut therefore becomes zero, and he feels weightless. If he stands on a spring balance, the balance will read zero.

Gravitational forces are present, but there is no normal force.

Reason — The weight of a body is felt due to the reactionary force applied on the body by some other body in contact with it. When this reaction, that is the normal force, becomes zero, we feel as if our weight has also become zero, and this is called the state of weightlessness. Gravitational forces continue to act throughout; it is only the absence of the normal reaction that produces the sensation of weightlessness.

Skydiving before deploying a parachute.

Reason — Before the parachute is deployed, the skydiver is falling freely under gravity with no upward reaction acting on him, and this free fall is the condition closest to weightlessness. Jumping off a tall building would also produce free fall, but only very briefly and it is not a controlled situation. Standing on a high mountain and swimming underwater both involve a normal or buoyant reaction acting on the body, so weight continues to be felt.

the radius of the orbit.

Reason — The period of revolution of a satellite in an orbit of radius r = R_e + h is

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}}$

This expression is free from the mass of the satellite. It shows that the period depends only upon the height of the satellite above the earth's surface, that is, on the radius of its orbit. Two satellites of different masses revolving in the same orbit therefore have the same period.

lower because the moon's gravitational pull is weaker.

Reason — The escape velocity from the surface of a body of mass M and radius R is

$\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$

which depends only on the mass and radius of that body, not on the mass of the escaping object. The moon has a much smaller mass than the earth, so its gravitational pull is weaker and its escape velocity is only about 2.4 km s^-1, against 11.2 km s^-1 for the earth.

$g' = g\left(1 + \frac{h}{R_e}\right)^{-2}$

Reason — If g is the acceleration due to gravity at the earth's surface, then at a height h above the surface

$\text g = \frac{\text{GM}_e}{\text R_e^2} \quad \text{and} \quad \text g' = \frac{\text{GM}_e}{(\text R_e + \text h)^2}$

Dividing the second by the first,

$\frac{\text g'}{\text g} = \frac{\text R_e^2}{(\text R_e + \text h)^2} = \frac{1}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

$\text g' = \text g\left(1 + \frac{\text h}{\text R_e}\right)^{-2}$

4 AU

Reason — By Kepler's third law, T² = k r³. Comparing the planet with the earth, for which T_e = 1 year and r_e = 1 AU,

$\left(\frac{\text T_p}{\text T_e}\right)^2 = \left(\frac{\text r_p}{\text r_e}\right)^3$

Substituting T_p = 8 years,

$\left(\frac{8}{1}\right)^2 = \left(\frac{\text r_p}{1}\right)^3 \quad \Rightarrow \quad \text r_p^3 = 64$

$\text r_p = 4\ \text{AU}$

decrease

Reason — The acceleration due to gravity at the surface of a planet is

$\text g = \frac{\text{GM}}{\text r^2}$

Here G and the mass M of Saturn are constants, so $\text g \propto \frac{1}{\text r^2}$. As the radius r increases, the acceleration due to gravity at its surface falls off with the square of the radius, and therefore decreases.

Spherical shell

Reason — The graph shows that the force of gravity is zero up to a distance R and then falls off as $\frac{1}{\text r^2}$ beyond R.

By Newton's shell theorem, a spherically symmetric shell of mass exerts no net gravitational force on a particle located anywhere inside it, so F = 0 for r < R. For a point outside the shell, the shell behaves like a point mass located at its centre, so

$\text F = \frac{\text{GMm}}{\text r^2} \quad \Rightarrow \quad \text F \propto \frac{1}{\text r^2}$

This is exactly the variation shown, so the graph is that of a spherical shell.

Solid sphere

Reason — The graph shows that the force of gravity increases linearly with r up to r = R and then decreases as $\frac{1}{\text r^2}$.

For a point inside a solid sphere of uniform density, only the mass enclosed within the radius r contributes to the force, and

$\text F = \frac{\text{GMm}}{\text R^3}\text r \quad \Rightarrow \quad \text F \propto \text r$

so the force is zero at the centre and rises linearly to a maximum at the surface. For a point outside, the whole mass acts as if concentrated at the centre and

$\text F = \frac{\text{GMm}}{\text r^2} \quad \Rightarrow \quad \text F \propto \frac{1}{\text r^2}$

This is the variation shown, so the graph is that of a solid sphere.

The satellite will enter an elliptical orbit with apogee at the point where velocity was increased.

Reason — Increasing the satellite's velocity slightly increases its kinetic energy and hence its total energy, disturbing the balance required for a circular orbit. The satellite therefore changes its path to an elliptical orbit. The point at which the velocity was increased becomes the apogee of the new orbit, since the satellite begins to move away from the earth from that point. The increase being small, the satellite does not attain the escape velocity and so it does not escape earth's gravity.

[LT^-2]

Reason — The gravitational field intensity is the force experienced by a unit mass placed at a point in the field,

$\text I = \frac{\text{GM}}{\text r^2} = \frac{\text F}{\text m}$

Its dimensional formula is therefore

$[\text I] = \frac{[\text{MLT}^{-2}]}{[\text M]} = [\text{LT}^{-2}]$

which is the same as that of acceleration, as it should be, since the gravitational field intensity of the earth is numerically equal to g.

About 84 min

Reason — For a satellite orbiting very close to the earth's surface (h << R_e), the period of revolution is

$\text T = 2\pi\sqrt{\frac{\text R_e}{\text g}}$

Substituting R_e = 6.37 × 10⁶ m and g = 9.8 m s^-2,

$\text T = 2 \times 3.14 \times \sqrt{\frac{6.37 \times 10^6}{9.8}} = 5063\ \text s \approx 84\ \text{min}$

$E = -\frac{GM_em}{2R_e}$

Reason — For a satellite of mass m revolving close to the earth in an orbit of radius R_e, the potential energy and the kinetic energy are

$\text U = -\frac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text K = \frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e}$

Hence the total energy of the satellite is

$\text E = \text U + \text K = -\frac{\text{GM}_e \text m}{\text R_e} + \frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e} = -\frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e}$

Note: Options 1 and 4 are printed identically as $-\frac{\text{GM}_e \text m}{2\text R_e}$ in the textbook, and the answer key marks option 1.

1 : $\sqrt2$

Reason — The orbital speed of a satellite in an orbit of radius r is

$\text v_o = \sqrt{\frac{\text{GM}_e}{\text r}} \quad \Rightarrow \quad \text v_o \propto \frac{1}{\sqrt{\text r}}$

which is independent of the mass of the satellite. Given that r_x = 2r_y,

$\frac{\text v_x}{\text v_y} = \sqrt{\frac{\text r_y}{\text r_x}} = \sqrt{\frac{\text r_y}{2\text r_y}} = \frac{1}{\sqrt2}$

$\text v_x : \text v_y = 1 : \sqrt2$

$(rp)^{\frac{1}{2}}$

Reason — The escape velocity from the surface of a planet of radius R on which the acceleration due to gravity is g is

$\text v_e = \sqrt{2\text{gR}}$

For the two planets A and B,

$\frac{\text v_{e(A)}}{\text v_{e(B)}} = \sqrt{\frac{2\text g_A \text R_A}{2\text g_B \text R_B}} = \sqrt{\frac{\text g_A}{\text g_B}} \times \sqrt{\frac{\text R_A}{\text R_B}}$

Given $\frac{\text R_A}{\text R_B} = \text r$ and $\frac{\text g_A}{\text g_B} = \text p$,

$\frac{\text v_{e(A)}}{\text v_{e(B)}} = \sqrt{\text r \times \text p} = (\text{rp})^{1/2}$

increase by about 5%

Reason — The escape velocity from the earth's surface is

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} \quad \Rightarrow \quad \text v_e \propto \sqrt{\text G}$

If G is increased by 10%, the new value is G' = 1.10 G, so the new escape velocity is

$\text v_e' = \text v_e \times \sqrt{1.10}$

The percentage increase in the escape velocity is therefore

$$\begin{aligned}\frac{\text v_e' - \text v_e}{\text v_e} \times 100 = (\sqrt{1.10} - 1) \times 100 \\[1em] = (1.0488 - 1) \times 100 = 4.88% \approx 5%\end{aligned}$$

increase by approximately 2%

Reason — The acceleration due to gravity at the earth's surface is

$\text g = \frac{\text{GM}_e}{\text R_e^2} \quad \Rightarrow \quad \text g \propto \frac{1}{\text R_e^2}$

Taking the fractional change on both sides, since the mass is kept constant,

$\frac{\Delta \text g}{\text g} = -2\frac{\Delta \text R_e}{\text R_e}$

The radius shrinks by 1%, that is $\frac{\Delta \text R_e}{\text R_e} = -1%$, so

$\frac{\Delta \text g}{\text g} = -2 \times (-1%) = +2%$

Hence g increases by approximately 2%.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By Newton's universal law of gravitation, the force of attraction between two material particles is inversely proportional to the square of the distance between them, $\text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$.

Reason (R) is also correct: This dependence on $\frac{1}{\text r^2}$ is precisely what is termed the inverse-square law, and it is a part of the statement of Newton's law of universal gravitation.

Since the inverse-square dependence stated in the Reason is the very relation asserted in (A), the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Gravitational force is a central force, that is, it acts along the line joining the centres of two bodies, and central forces are conservative in nature. Hence gravitational force is an example of a conservative force.

Reason (R) is also correct: A conservative force is one for which the work done in moving a body from one position to another depends only on the initial and the final positions and not on the path followed, that is, the work done is path-independent.

Path-independence of the work done is the defining property of a conservative force, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The gravitational forces between two bodies are completely independent of the presence of other bodies and of the properties of the intervening medium. If the force of attraction between two bodies in air is F, it remains the same when they are placed in water, the separation being unchanged.

Reason (R) is also correct: By Newton's law of gravitation, $\text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$, the force depends only on the masses of the two bodies and on the distance between them.

Since the expression for the force contains only the masses and the separation, and no term for the medium, the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: Gravitational force exists even in a vacuum and does not require any medium to act. Two particles placed in vacuum attract each other with the force $\text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$, which is certainly not zero.

Reason (R) is correct as a statement of fact: Gravitational force is a central force, that is, it acts along the line joining the centres of the two masses.

Therefore, both assertion and reason are false.

Note: The answer key is incorrect. The assertion is false, but the reason is true. Therefore, none of the given options is correct.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: As we go above the surface of the earth, the acceleration due to gravity goes on decreasing, since $\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$, which is less than g.

Reason (R) is also correct: At a height h the distance from the earth's centre becomes (R_e + h), and $\text g' = \frac{\text{GM}_e}{(\text R_e + \text h)^2}$, so that $\text g' \propto \frac{1}{\text r^2}$.

As the height increases the distance from the earth's centre increases, and by the inverse-square dependence the value of g must fall. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: The variation of g over the earth's surface arises from two separate causes, the rotation of the earth and its oblate shape. Stopping the rotation removes only the term R_eω²cos²λ. The earth would still be an oblate spheroid, and since $\text g = \frac{\text{GM}_e}{\text R_e^2}$, that is $\text g \propto \frac{1}{\text R_e^2}$, the value of g would still be greater at the poles, where the radius is smaller, than at the equator. Hence g would not be the same at all points on the surface.

Reason (R) is false: The earth is not a perfect sphere. It is an oblate spheroid, slightly flattened at the poles and bulging at the equator, its equatorial diameter being about 21 km greater than the polar diameter.

Therefore, both assertion and reason are false.

Note: The textbook’s explanation shows that variations in g are caused by both the earth’s rotation and its flattened shape. Since some variation would remain even if the earth stopped rotating, the assertion is false. Therefore, option 4 is correct as the options stand.

If both assertion and reason are false.

Explanation

Assertion (A) is false: The value of g actually increases as we move from the equator to the poles, both because the distance from the earth's centre decreases and because the centrifugal effect of rotation, $\text R_e\omega^2\cos^2\lambda$, becomes smaller as the latitude increases.

Reason (R) is also false: The radius of the earth is greater at the equator than at the poles, the equatorial radius being about 6378 km against the polar radius of about 6357 km.

Therefore, both assertion and reason are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: At a depth h below the surface, $\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$, so that g' < g. As we go below the surface of the earth the acceleration due to gravity goes on decreasing, and it becomes zero at the centre.

Reason (R) is also correct: At a depth h, by Newton's shell theorem the outer spherical shell exerts no force, and only the inner solid sphere of radius (R_e − h) attracts the body. Hence the mass contributing to the gravitational pull decreases as we go deeper, and the force varies linearly with the distance from the centre.

Because the effective attracting mass decreases linearly with the distance from the centre, the value of g must also decrease as we move inward. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: The value of g at a height h above the earth is less than that at the earth's surface, since $\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$ is smaller than g.

Reason (R) is also false: The gravitational force is inversely proportional to the square of the distance from the earth's centre, and is not directly proportional to the height above the earth's surface.

Therefore, both assertion and reason are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The value of g is about 9.83 m s^-2 at the poles and about 9.78 m s^-2 at the equator, so g is slightly higher at the poles.

Reason (R) is also correct: The earth is an oblate spheroid, so its polar radius is smaller than its equatorial radius.

Since $\text g = \frac{\text{GM}_e}{\text R_e^2}$, that is $\text g \propto \frac{1}{\text R_e^2}$, a smaller radius at the poles gives a larger value of g there. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: The orbital velocity of a satellite, $\text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$, is free from the mass of the satellite. Hence it does not depend on the satellite's mass.

Reason (R) is also false: Although the orbital velocity is obtained by balancing the gravitational force against the required centripetal force, the mass m of the satellite appears on both sides of $\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$ and cancels out. The result therefore depends only on the mass of the central body and the radius of the orbit.

Therefore, both assertion and reason are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The orbital velocity $\text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$ varies as $\frac{1}{\sqrt{\text r}}$, so a satellite in an orbit closer to the earth (smaller r) has a higher orbital velocity.

Reason (R) is also correct: The gravitational force $\frac{\text{GM}_e \text m}{\text r^2}$ is stronger at lower altitudes, since the distance from the earth's centre is smaller.

A stronger gravitational pull at a lower altitude demands a larger centripetal force, and hence a higher orbital velocity, to maintain the orbit. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Since $\text v_o = \sqrt{\frac{\text{GM}_e}{\text R_e + \text h}}$, the orbital velocity decreases as the height h of the orbit increases.

Reason (R) is also correct: The gravitational force varies inversely as the square of the distance, so it decreases as the distance from the earth increases.

At higher altitudes the weaker gravitational pull requires a smaller centripetal force, so a lower orbital velocity is sufficient to maintain the orbit. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: Geostationary satellites are in high orbits, at a height of about 35,830 km, and have a comparatively low orbital velocity of about 3.1 km s^-1, against about 8 km s^-1 for satellites in low earth orbits.

Reason (R) is also false: Geostationary satellites do not orbit close to the earth's surface; they are placed far above it in the parking orbit above the equator.

Therefore, both assertion and reason are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The mass m of the satellite cancels out in $\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$, giving $\text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$, which is independent of the satellite's mass.

Reason (R) is also correct: The resulting expression contains only the mass M_e of the central body around which the satellite revolves and the radius r of the orbit.

Because the orbital velocity is determined entirely by the central mass and the orbital radius, it cannot depend on the mass of the satellite. This is why two satellites of different masses revolving in the same orbit have the same speed. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Escape velocity is the minimum velocity that an object needs in order to escape the gravitational pull of a celestial body without any further propulsion.

Reason (R) is false: The escape velocity, $\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}}$, is independent of the angle, that is the direction, of projection. It depends only on the mass and radius of the earth.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a body near the earth's surface, $\text v_e = \sqrt{2\text{gR}_e}$ and $\text v_o = \sqrt{\text{gR}_e}$, so that $\text v_e = \sqrt2\text v_o$. The escape velocity is therefore greater than the orbital velocity.

Reason (R) is also correct: Escape velocity is the minimum speed required to leave the gravitational field of the planet without further propulsion, whereas the orbital velocity is only the speed needed to remain bound in a circular orbit.

Since escaping the field altogether demands more energy than merely remaining in a bound orbit, the escape velocity must exceed the orbital velocity. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: By Newton's law of gravitation the force of attraction is directly proportional to the mass of the attracting body, so a more massive planet exerts a stronger gravitational pull.

Reason (R) is false: Since $\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$, the escape velocity increases with the mass of the planet, because a stronger gravitational pull requires a higher speed to escape.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The escape velocity $\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}}$ varies as $\frac{1}{\sqrt{\text R_e}}$, so it is higher at the poles, where the radius is smaller.

Reason (R) is also correct: The earth is an oblate spheroid, so its radius is smaller at the poles than at the equator.

A smaller radius at the poles gives a larger value of $\frac{2\text{GM}_e}{\text R_e}$ and hence a higher escape velocity there. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The escape velocity from the moon's surface is about 2.4 km s^-1, which is much less than the 11.2 km s^-1 required at the earth's surface.

Reason (R) is also correct: The moon has a considerably smaller mass and a smaller radius than the earth.

With a much smaller mass, the moon's gravitational pull is weaker, and $\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$ works out smaller for the moon in spite of its smaller radius. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A geosynchronous satellite is synchronous with the earth's spin, so it appears stationary with respect to an observer on the earth.

Reason (R) is also correct: These satellites have a period of revolution exactly equal to the period of the axial motion of the earth, that is, 24 h.

Because the satellite completes one orbit in exactly the time the earth takes to complete one rotation, it stays above the same point and appears stationary. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Using $\text h = \left(\frac{\text T^2 \text{gR}_e^2}{4\pi^2}\right)^{1/3} - \text R_e$ with T = 24 h, the height of a geostationary satellite works out to about 35,830 km above the earth's surface.

Reason (R) is also correct: It is at exactly this altitude that the period of revolution of the satellite becomes equal to the rotational period of the earth.

The height is determined by the requirement that the satellite's period should match the earth's rotational period, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: One of the conditions for a satellite to be geostationary is that its orbital plane should be coplanar with the equatorial plane of the earth. Hence geostationary satellites orbit above the equator.

Reason (R) is false: Precisely because they are positioned over the equator, geostationary satellites have limited effectiveness near the poles. Polar regions therefore rely on polar-orbiting satellites for communication and weather monitoring.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: A satellite is called geostationary if it appears stationary over a fixed point on the earth's equator, that is, it remains fixed relative to a point on the earth's surface.

Reason (R) is false: All geostationary satellites are geosynchronous, but the converse is not true. A geosynchronous satellite need not have an equatorial orbit, and only those with an equatorial orbital plane are geostationary.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Since a geostationary satellite moves exactly in step with the earth's rotation, its position relative to a point on the earth's surface does not change, so its velocity relative to that surface is constant.

Reason (R) is also correct: The satellite's angular velocity, and hence its orbital motion, matches the rotational speed of the earth.

Because the satellite's orbital motion is synchronised with the earth's rotation, it maintains a constant position, and hence a constant velocity, relative to a point on the surface. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Kepler's first law, the law of orbits, states that all planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

Reason (R) is false: Kepler's first law asserts elliptical orbits with the sun at one focus, not circular orbits with the sun at the centre. A circle is only a special limiting case of an ellipse.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Kepler's second law, the law of areas, states that a line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant.

Reason (R) is also correct: When the planet is nearest the sun its velocity is maximum, and when it is farthest from the sun its velocity is minimum.

For equal areas to be swept in equal times, the planet must move faster when the radius vector is short, that is near the sun, and slower when it is long. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Kepler's third law, the law of periods, states that the square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun, T² = K r³.

Reason (R) is also correct: The constant K is the same for all the planets, so the law relates the orbital period of every planet to its distance from the sun.

Since the relation T² = K r³ holds with the same constant for all planets, it is exactly the statement made in the Assertion. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The gravitational potential energy is $\text U = -\frac{\text{GM}_e \text m}{\text r}$. As r increases, the magnitude of U decreases, that is, U becomes less negative. Hence the potential energy increases as the object moves away from the earth.

Reason (R) is false: It is the magnitude of the potential energy that is inversely proportional to the distance from the earth's centre. Since U itself is negative, an inverse relation of the kind stated would wrongly imply that U decreases as r increases.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The period of revolution of a satellite is $\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}}$, which increases with the height of the orbit. Hence a satellite in a higher orbit has a longer period of revolution.

Reason (R) is false: By Kepler's third law the square of the orbital period is proportional to the cube of the radius of the orbit, T² ∝ r³, and not the period to the square of the radius.

Therefore, assertion is true but reason is false.

The electrostatic force between two electrons is about 10⁴³ times the gravitational force between them.

Gravitational forces are the weakest forces existing in nature. This is why the gravitational attraction between ordinary bodies is never felt in daily life, while their electrostatic interaction is easily observed.

Both will be equal.

By Newton's law of gravitation the gravitational forces between two bodies form an action-reaction pair, that is, the forces exerted by two bodies on each other are equal in magnitude but oppositely directed. Hence the force of the earth on the moon and that of the moon on the earth are equal in magnitude, however different their masses may be.

It will remain 6.67 × 10^-11 N m² kg^-2.

The gravitational constant G is a universal constant. Its value is the same for all pairs of bodies at all places and at all distances of separation, and gravitational forces are completely independent of the properties of the intervening medium. Hence the value of G is unaffected by the presence of a dense medium.

The value of g decreases below the earth's surface.

At a depth h below the surface only the inner solid sphere of radius (R_e − h) attracts the body, the outer spherical shell exerting no force. Hence

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

so that g' < g, and g becomes zero at the centre of the earth, where h = R_e.

(i) Weight : The weight of the body will become 1/6th of that on the earth, since W = mg and the value of g on the moon is 1/6th of that on the earth.

(ii) Inertial mass : There will be no change. Inertial mass is a measure of the inertia of the body and is a constant property of the body itself.

(iii) Gravitational mass : There will be no change. The gravitational mass is also a constant property of the body and does not depend on the place where the body is taken.

The time-period of a simple pendulum is $\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$, that is, $\text T \propto \frac{1}{\sqrt{\text g}}$.

On a mountain the value of g is less than that on the plane, because the acceleration due to gravity decreases with altitude. Hence the time-period of the pendulum increases, and the pendulum clock becomes slow.

A wrist-watch, on the other hand, is controlled by a spring, and its time-period depends upon the elastic properties of the spring and not upon the value of g. Hence it remains unaffected by the variation of g.

If the earth stops rotating about its axis, the value of g will increase at all places except at the poles.

The apparent value of the acceleration due to gravity is

$\text g' = \text g - \text R_e\omega^2\cos^2\lambda$

On stopping the rotation, ω becomes zero and the term R_eω²cos²λ vanishes.

No, the effect will not be the same at all places. The increase depends upon the latitude λ. It will be maximum at the equator (λ = 0°) and zero at the poles (λ = 90°), since there cos λ = 0 and the rotation makes no contribution.

The earth is flat at the poles, that is, it is an oblate spheroid, due to the rotation of the earth about its polar axis.

Because of this rotation a centrifugal effect acts outwards from the axis of rotation. This effect is greatest at the equator and zero at the poles, so over a long time the earth has bulged at the equator and become flattened at the poles. The equatorial diameter is about 21 km greater than the polar diameter.

The weight of a body is W = mg, and the value of g is smaller at the equator than at the poles.

This is due to two causes. Firstly, the earth is an oblate spheroid, so its equatorial radius is greater than its polar radius, and since $\text g \propto \frac{1}{\text R_e^2}$, the value of g is smaller where the radius is larger. Secondly, the rotation of the earth reduces g by the term R_eω²cos²λ, which is maximum at the equator and zero at the poles.

Hence the same body weighs less at the equator than at the poles.

A satellite gets the necessary centripetal force from the gravitational force of attraction exerted on it by its planet.

When the satellite revolves in a circular orbit of radius r, this gravitational force supplies exactly the centripetal force required,

$\frac{\text{GMm}}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$

This is why a satellite continues to revolve in its orbit without using any fuel.

The gravitational force due to the earth on the satellite is F, and the net force on the satellite is also F.

The gravitational force of the earth on the satellite is the only force acting on it, and this force itself provides the required centripetal force. Hence both the gravitational force and the net force on the satellite are equal to F, directed towards the centre of the earth.

The earth does not fall towards the sun because it is revolving around the sun in a stable orbit.

The gravitational force of attraction of the sun on the earth is exactly the centripetal force needed to keep the earth moving in its orbit. This force acts along the line joining the earth and the sun, that is, perpendicular to the direction of the earth's velocity at every instant. It therefore only changes the direction of the earth's motion and does not pull the earth inward to the sun.

The orbital velocity of a satellite revolving in an orbit of radius r is

$\text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$

In deriving this, the mass m of the satellite appears on both sides of $\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$ and cancels out. Hence the expression is free from the mass of the satellite and depends only on the mass of the earth and the radius of the orbit.

Therefore two artificial satellites revolving in the same orbit must have the same speed, irrespective of the difference between their masses.

No, its period of revolution would not change.

The period of revolution of a satellite in an orbit of radius r is

$\text T = 2\pi\sqrt{\frac{\text r^3}{\text{GM}_e}}$

This expression contains the mass M_e of the earth and the radius of the orbit, but is free from the mass of the moon. Hence doubling the mass of the moon, the orbit remaining the same, would leave its period of revolution unchanged.

A geostationary satellite completes one revolution in 24 h, the same as the period of the axial motion of the earth. Hence its angular velocity is

$\omega = \frac{2\pi}{\text T} = \frac{2\pi}{24} = \frac{\pi}{12}\ \text{radian per hour}$

Hence, the angular velocity of a geostationary satellite is $\frac{\pi}{12}$ radian per hour.

No, the packet will not reach the earth. It will also continue to revolve around the earth along with the satellite.

When the packet is released, it already possesses the same orbital velocity as the satellite at that height. Since the orbital velocity is independent of the mass of the body, the packet continues to move in the same orbit with the same speed, and it therefore behaves as a satellite of the earth itself.

No, the weight of a body cannot be determined inside an artificial satellite.

Inside the satellite every body is in a state of weightlessness, since the artificial satellite is like a freely-falling body. The gravitational force is entirely used up in providing the centripetal force, so the reactionary force on the body becomes zero. If the body is placed on a spring balance inside the satellite, the balance will read zero.

Given,

• Gravitational potential energy of the body on Venus, U = − 7.5 × 10⁶ J

To escape from the planet, the body must be taken to infinity, where its gravitational potential energy is zero. Hence the energy required is

$\text E = \text U_\infty - \text U = 0 - (-7.5 \times 10^6)$

$\text E = + 7.5 \times 10^6\ \text J$

Hence, the energy required for the body to escape from Venus is + 7.5 × 10⁶ J.

Escape velocity is the minimum velocity that an object needs in order to escape the gravitational pull of a celestial body, without any further propulsion.

If a body is launched from the earth's surface with this speed, it will continue moving away from the earth, overcoming the earth's gravitational pull, and will not fall back or enter an orbit. Its value at the earth's surface is

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} = \sqrt{2\text{gR}_e} = 11.2\ \text{km s}^{-1}$

It will remain 11.2 km s^-1.

The escape velocity from the earth is

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}}$

This expression is free from the mass of the body. Hence the escape velocity does not depend upon the mass of the piece, and a piece of 10 g escapes with exactly the same velocity as a piece of 1 g.

The velocity required to send a body from the earth's surface into space, never to return, is the escape velocity v_e, and the orbital speed of a satellite close to the earth is v_o. Their values are

$\text v_e = \sqrt{2\text{gR}_e} \quad \text{and} \quad \text v_o = \sqrt{\text{gR}_e}$

Therefore

$\frac{\text v_o}{\text v_e} = \frac{\sqrt{\text{gR}_e}}{\sqrt{2\text{gR}_e}} = \frac{1}{\sqrt2} \quad \Rightarrow \quad \text v_e = \sqrt2\ \text v_o$

Hence, the escape velocity is $\sqrt2$ times the orbital speed.

Yes, the satellite will escape.

If the kinetic energy of the satellite is doubled,

$\frac{1}{2}\text{mv}^2 = 2 \times \frac{1}{2}\text{mv}_o^2 \quad \Rightarrow \quad \text v = \sqrt2\ \text v_o$

But for a satellite revolving close to the earth, the escape velocity is exactly $\text v_e = \sqrt2\text v_o$. Hence its velocity becomes equal to the escape velocity, and the satellite will leave its orbit and escape.

Given,

• Distance of the planet from the sun = 40 times the distance of the earth from the sun
• Masses of the earth and the planet are equal

By Newton's law of gravitation the force of the sun on a body of mass m at a distance r is

$\text F = \frac{\text{GM}_s \text m}{\text r^2} \quad \Rightarrow \quad \text F \propto \frac{1}{\text r^2}$

since M_s and m are the same in both cases. Therefore

$\frac{\text F_{earth}}{\text F_{planet}} = \left(\frac{\text r_{planet}}{\text r_{earth}}\right)^2 = (40)^2 = 1600$

Hence, the gravitational forces of the sun on the earth and on the planet are in the ratio 1600 : 1.

Given,

• Acceleration due to gravity on the satellite, g_s = 1.96 m s^-2
• Acceleration due to gravity on the earth, g_e = 9.80 m s^-2
• Safe height of jumping on the earth, h_e = 5 m

While jumping down from a height h, the potential energy mgh is converted into kinetic energy on landing. For the jump to be equally safe, this energy must be the same in both cases,

$\text{mg}_e \text h_e = \text{mg}_s \text h_s \quad \Rightarrow \quad \text h_s = \text h_e \times \frac{\text g_e}{\text g_s}$

Substituting the values,

$\text h_s = 5 \times \frac{9.80}{1.96} = 5 \times 5 = 25\ \text m$

Hence, jumping from a height of 25 m will be safe on the satellite.

Given,

• Time taken to strike the earth, t_e = 1 s
• Acceleration due to gravity at the moon, $\text g_m = \frac{\text g_e}{6}$
• The stone is dropped from the same height h in both cases

For a body dropped from rest, $\text h = \frac{1}{2}\text{gt}^2$, so that for the same height

$\text t = \sqrt{\frac{2\text h}{\text g}} \quad \Rightarrow \quad \text t \propto \frac{1}{\sqrt{\text g}}$

Therefore,

$\frac{\text t_m}{\text t_e} = \sqrt{\frac{\text g_e}{\text g_m}} = \sqrt{\frac{\text g_e}{\text g_e/6}} = \sqrt6$

$\text t_m = 1 \times \sqrt6 = \sqrt6\ \text s$

Hence, the stone will take $\sqrt6$ s to strike the surface of the moon.

Given,

• Distance between the bodies A and B = r
• Acceleration of A under the inverse square law = a

Under the inverse square law the force on A is $\text F = \frac{\text{Gm}_A \text m_B}{\text r^2}$, so its acceleration is

$\text a = \frac{\text F}{\text m_A} = \frac{\text{Gm}_B}{\text r^2}$

If the gravitational force follows an inverse fourth power law, the force becomes $\text F' = \frac{\text{Gm}_A \text m_B}{\text r^4}$, and the acceleration of A is

$\text a' = \frac{\text F'}{\text m_A} = \frac{\text{Gm}_B}{\text r^4} = \frac{1}{\text r^2}\left(\frac{\text{Gm}_B}{\text r^2}\right)$

$\text a' = \frac{\text a}{\text r^2}$

Hence, the acceleration of the body A would be $\frac{\text a}{\text r^2}$.

The weight will reduce to one-fourth of its present value.

The acceleration due to gravity at the earth's surface is

$\text g = \frac{\text{GM}_e}{\text R_e^2}$

The mass M_e remains unchanged while the diameter, and hence the radius, becomes twice, that is R_e' = 2R_e. Therefore

$\frac{\text g'}{\text g} = \left(\frac{\text R_e}{2\text R_e}\right)^2 = \frac{1}{4} \quad \Rightarrow \quad \text g' = \frac{\text g}{4}$

Since W = mg, the weight of the object also becomes one-fourth.

The weight will be halved.

Since the average density ρ remains unchanged, the acceleration due to gravity is

$\text g = \frac{\text G}{\text R_e^2} \times \frac{4}{3}\pi \text R_e^3 \rho = \frac{4}{3}\pi \text{GR}_e\rho \quad \Rightarrow \quad \text g \propto \text R_e$

The diameter, and hence the radius, becomes half, so

$\text g' = \frac{\text g}{2}$

Since W = mg, the weight of the object on the surface is also halved.

Given,

• Mass of the planet, M_p = 2M_e
• Diameter of the planet = twice that of the earth, so R_p = 2R_e
• Acceleration due to gravity on the earth = g

The acceleration due to gravity at the surface of a planet is

$\text g = \frac{\text{GM}}{\text R^2}$

Therefore,

$\frac{\text g_p}{\text g} = \frac{\text M_p}{\text M_e} \times \left(\frac{\text R_e}{\text R_p}\right)^2 = 2 \times \left(\frac{1}{2}\right)^2 = \frac{1}{2}$

$\text g_p = \frac{\text g}{2}$

Hence, the acceleration due to gravity at the planet is $\frac{\text g}{2}$.

Given,

• Weight of the person on the earth, W_e = 600 N
• Gravitational field of the moon $= \frac{1}{6}$ of that of the earth
• Height of jump on the earth = 2 m
• g on earth = 10 N kg^-1

(a) Weight on the moon : Since W = mg and the gravitational field of the moon is 1/6th that of the earth,

$\text W_m = \frac{\text W_e}{6} = \frac{600}{6} = 100\ \text N$

(b) Height of jump on the moon : While jumping, the same muscular energy mgh is expended, so

$\text{mg}_e \text h_e = \text{mg}_m \text h_m \quad \Rightarrow \quad \text h_m = \text h_e \times \frac{\text g_e}{\text g_m} = 2 \times 6 = 12\ \text m$

(c) Mass of the person : The mass of a body is a constant property and does not change with place,

$\text m = \frac{\text W_e}{\text g_e} = \frac{600}{10} = 60\ \text{kg}$

Hence, the weight on the moon is 100 N, the height of the jump on the moon is 12 m, and the mass is 60 kg both on the earth and on the moon.

At a height equal to the radius of the earth : The acceleration due to gravity at a height h is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Putting h = R_e,

$\text g' = \frac{\text g}{(1 + 1)^2} = \frac{\text g}{4}$

Hence his weight will be one-fourth of that on the earth's surface.

At a depth equal to the radius of the earth : The acceleration due to gravity at a depth h is

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

Putting h = R_e, we get g' = 0. This point is the centre of the earth, so his weight will be zero.

Given,

• Acceleration due to gravity at the required point $= \frac{\text g}{4}$

(i) Below the surface of the earth : At a depth h,

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

Putting $\text g' = \frac{\text g}{4}$,

$\frac{1}{4} = 1 - \frac{\text h}{\text R_e} \quad \Rightarrow \quad \frac{\text h}{\text R_e} = \frac{3}{4}$

$\text h = \frac{3}{4}\text R_e$

(ii) Above the surface of the earth : At a height h,

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Putting $\text g' = \frac{\text g}{4}$,

$\left(1 + \frac{\text h}{\text R_e}\right)^2 = 4 \quad \Rightarrow \quad 1 + \frac{\text h}{\text R_e} = 2$

$\text h = \text R_e$

Hence, the distances are $\frac{3}{4}\text R_e$ below the surface and R_e above the surface.

Given,

• Weight at the depth $= \frac{1}{4}$ of the weight at the surface
• Radius of the earth, R_e = 6400 km

Since W = mg, the acceleration due to gravity at that depth must be $\text g' = \frac{\text g}{4}$. At a depth h,

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

Substituting $\text g' = \frac{\text g}{4}$,

$\frac{1}{4} = 1 - \frac{\text h}{\text R_e} \quad \Rightarrow \quad \frac{\text h}{\text R_e} = \frac{3}{4}$

$\text h = \frac{3}{4} \times 6400 = 4800\ \text{km}$

Hence, at a depth of 4800 km the weight of the person will be one-fourth of his weight at the surface.

Given,

• Distance of the planet from the sun, r_p = 4 r_e
• Period of revolution of the earth, T_e = 1 year

By Kepler's law of periods, T² ∝ r³, so that $\text T \propto \text r^{3/2}$. Therefore

$\frac{\text T_p}{\text T_e} = \left(\frac{\text r_p}{\text r_e}\right)^{3/2} = (4)^{3/2} = 8$

$\text T_p = 8 \times 1 = 8\ \text{years}$

Hence, the planet will complete one revolution around the sun in 8 years.

The statement is not correct. No work is done on the earth.

The gravitational force of the sun on the earth is a centripetal force, directed towards the sun along the radius of the orbit. The earth's velocity at every instant is along the tangent to its circular path, that is, perpendicular to this force. Hence the angle between the force and the displacement is 90°, and

$\text W = \text F\text s\cos 90^\circ = 0$

Hence, the work done by the gravitational force of the sun on the earth is zero.

For a satellite revolving at a height h above the earth's surface, in an orbit of radius r = R_e + h, the period of revolution may be written in the following equivalent forms :

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}}$

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{gR}_e^2}} \qquad (\because\ \text{GM}_e = \text{gR}_e^2)$

$\text T = \sqrt{\frac{3\pi(\text R_e + \text h)^3}{\text G\rho \text R_e^3}}$

where ρ is the mean density of the earth.

For a satellite revolving very close to the earth's surface (h << R_e), the orbital speed and the period of revolution are

$\text v_o = \sqrt{\text{gR}_e} = \sqrt{9.8 \times (6.37 \times 10^6)} \approx 8\ \text{km s}^{-1}$

$\text T = 2\pi\sqrt{\frac{\text R_e}{\text g}} = 2 \times 3.14 \times \sqrt{\frac{6.37 \times 10^6}{9.8}} \approx 84\ \text{min}$

Hence, the speed is nearly 8 km s^-1 and the period of revolution is about 84 min.

The periodic time of revolution of the communication satellite INSAT II B is 24 h.

INSAT II B is a geostationary satellite. For a satellite to appear stationary over a point on the earth's equator, its period of revolution must be exactly equal to the period of the axial motion of the earth, that is, 24 h.

The maximum height attained by a projectile thrown upward from the earth's surface with a velocity v is

$\text h = \frac{\text v^2 \text R_e}{2\text{gR}_e - \text v^2}$

where R_e is the radius of the earth and g is the acceleration due to gravity at its surface. This formula is used when the height h is not negligible in comparison with R_e.

The escape velocity of a body from the surface of the earth is

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}}$

where M_e is the mass and R_e is the radius of the earth. Since GM_e = gR_e², this may also be written as

$\text v_e = \sqrt{2\text{gR}_e}$

Given,

• Increase in the orbital speed = 41.4%

The orbital speed of a satellite close to the earth is v_o. On increasing it by 41.4%, the new speed is

$\text v = \text v_o + \frac{41.4}{100}\text v_o = \frac{141.4}{100}\text v_o = 1.414\ \text v_o$

$\text v = \sqrt2\ \text v_o$

But for a satellite revolving close to the earth, the escape velocity is related to the orbital speed by

$\text v_e = \sqrt2\ \text v_o$

Hence the new speed of the satellite is exactly equal to the escape velocity.

Therefore the satellite will leave its orbit and escape to infinity.

Moon travellers tie heavy weights at their back because the value of g on the moon is small, being about 1/6th of its value on the earth.

Because of this small value of g, their weight W = mg on the moon is only one-sixth of their weight on the earth. The reduced weight means a reduced normal reaction and therefore reduced friction between their feet and the lunar surface, so they would tend to bounce off the surface and would find it difficult to walk steadily. Tying heavy weights increases their mass, and hence their weight on the moon, allowing them to walk normally.

Kepler gave the following three laws of planetary motion :

(i) Law of Orbits : All planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

(ii) Law of Areas : A line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant. Consequently the velocity of a planet is maximum when it is nearest the sun and minimum when it is farthest from the sun.

(iii) Law of Periods : The square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun,

$\text T^2 \propto \text r^3$

Kepler's third law, the law of periods, states that if the period of a planet around the sun is T and the mean radius of its orbit is r, then

$\text T^2 \propto \text r^3 \quad \text{or} \quad \text T^2 = \text K\text r^3$

where K is a constant which is the same for all the planets and has the value 2.97 × 10^-19 s² m^-3. If the orbit is an ellipse, then r may be replaced by the semi-major axis a of the ellipse.

Thus, larger the distance of a planet from the sun, larger will be its period of revolution around the sun.

Newton's law of gravitation : The gravitational force of attraction between any two material particles is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. It acts along the line joining the two particles.

If two particles of masses m₁ and m₂ are situated a distance r apart and F is the force of attraction between them, then

$\text F \propto \frac{\text m_1 \text m_2}{\text r^2} \quad \text{or} \quad \text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$

where G is the gravitational constant, whose value is the same for all pairs of particles in the universe, and is therefore called a universal constant.

Above the earth's surface : At a height h above the surface, the distance from the centre becomes (R_e + h), so

$\text g' = \frac{\text{GM}_e}{(\text R_e + \text h)^2} \quad \Rightarrow \quad \text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

that is, g' < g. If h is negligible compared with R_e, this may be written by the binomial theorem as

$\text g' = \text g\left(1 - \frac{2\text h}{\text R_e}\right)$

Below the earth's surface : At a depth h, only the inner solid sphere of radius (R_e − h) attracts the body, and

$\text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

that is, g' < g. At the centre of the earth h = R_e and g' becomes zero.

Hence, the value of g decreases both on going above and on going below the surface of the earth.

Kepler's second law is based on the conservation of angular momentum of the planet about the sun.

The gravitational force on a planet is a central force directed towards the sun, so it exerts no torque about the sun. The angular momentum $\vec{\text L} \space$of the planet therefore remains constant, and since the areal velocity is

$\frac{\text d\vec{\text A}}{\text{dt}} = \frac{1}{2\text m}\vec{\text L}$

a constant $\vec{\text L} \space$means a constant areal velocity, which is exactly Kepler's second law.

Newton's law of gravitation is $\text F = \text G\frac{\text m_1 \text m_2}{\text r^2}$.

If we put m₁ = m₂ = 1 and r = 1, then G = F.

Thus, the universal gravitational constant G is numerically equal to the force with which two particles, each of unit mass and placed a unit distance apart, attract each other.

Its S.I. unit is N m² kg^-2 and its value is 6.67 × 10^-11 N m² kg^-2. It is called a universal constant because its value is the same for all pairs of bodies, at all places, at all distances of separation, and it is independent of the intervening medium.

The observations (a), (b), (c), (e) and (f) point to the equivalence of inertial and gravitational masses.

The equivalence of inertial and gravitational masses is established by measuring the accelerations produced in different bodies falling freely under the earth's gravitational field. In the observations (a), (b), (c), (e) and (f) the bodies are in motion and they all point to the equivalence of inertial and gravitational masses.

For a body falling freely, the gravitational force is m_Gg and by Newton's second law this equals m_Ia. Since all bodies are found to fall with the same acceleration, it follows that m_G = m_I.

The observations (d) and (g) do not point to this equivalence. Observation (d) is a consequence of Newton's shell theorem, and observation (g) is simply a statement of Newton's third law.

Up to ordinary heights the change in the distance of a projectile from the centre of the earth is negligible compared with the radius of the earth. Hence the projectile moves under a nearly uniform gravitational force which is constant in both magnitude and direction, and under such a force the path of the projectile is parabolic.

For projectiles going to a very great height, however, the gravitational force decreases in inverse proportion to the square of the distance of the projectile from the centre of the earth, and its direction also changes as the projectile moves. Under such a variable force the path of the projectile becomes elliptical.

At the centre of the earth the weight of the body will be zero, since g = 0 there.

As the body moves from the centre to the earth's surface, the value of g increases according to $\text g' = \text g\left(\frac{\text r}{\text R_e}\right)$, so the weight increases and becomes maximum at the surface.

On moving above the earth's surface towards the moon the value of g decreases, so the weight decreases. At one place between the earth and the moon, the gravitational forces of the earth and the moon on the body are equal and opposite, so the weight becomes zero there.

Beyond this point, on moving further towards the moon, the gravitational force of the moon on the body increases, and hence the weight goes on increasing again.

An artificial satellite revolves at a great height where the air is very much rare, so the frictional force due to air on it is nearly negligible. The gravitational force of the earth on the satellite provides exactly the centripetal force required for its circular motion, and this force does no work on the satellite since it is always perpendicular to the satellite's velocity. Hence the satellite continues to revolve in its orbit without using any fuel.

An aeroplane, on the other hand, flies at a comparatively low height where the air is dense. It therefore requires fuel continuously to do work against the frictional force due to air, and also to obtain the lift needed to stay at that height.

No, it is not possible.

For a satellite to remain always visible over a particular place, it must be a geostationary satellite, that is, it must have a period of revolution of 24 h and it must revolve in a circular orbit lying in the equatorial plane of the earth.

Lucknow (or New Delhi) is not situated on the equator, so a satellite in the equatorial plane cannot remain fixed vertically above it. Hence it is not possible to place an artificial satellite in an orbit such that it is always visible over Lucknow or New Delhi.

No, the missile will not hit the satellite.

A satellite appears stationary over a point on the earth's equator only when it is at such a height that its period of revolution about the earth is 24 h, the same as the period of the spin of the earth about its own axis. The satellite is therefore not actually at rest; it is moving along its orbit at the same angular rate as the earth.

The missile, after being fired towards the satellite, takes some time in covering the distance between the earth and the satellite. In this time the satellite moves along its orbit and changes its position. Hence the missile will miss the satellite.

The gravitational potential energy of a body on the earth's surface is

$\text U = -\frac{\text{GM}_e \text m}{\text R_e}$

which is negative because the potential energy at infinity is taken to be zero and work is obtained in bringing the body from infinity to the earth's surface.

The statement therefore means that 6.4 × 10⁶ J of energy must be supplied to the body on the surface of the earth in order to send it away, out of the gravitational field of the earth, that is, to take it to infinity where its potential energy becomes zero.

(a) Negative.

The gravitational potential energy of a system of two mass-points is $\text U = -\frac{\text{Gm}_1 \text m_2}{\text r}$, taking the potential energy at infinite separation to be zero. Since gravitational force is always attractive, every such pair in a galaxy contributes a negative term. Hence the total gravitational potential energy of a galaxy is negative.

(b) Gravitational; electromagnetic; strong nuclear.

On the large scale the universe is shaped by gravitational forces, because these are long-range forces and act between all masses, and on this scale bodies are electrically neutral. On the atomic scale the electromagnetic forces between the charged nuclei and electrons dominate. On the nuclear scale the strong nuclear force, which is the strongest of all and acts over distances of the order of nuclear dimensions, is dominant.

Nuclear forces are short-range forces which act only over distances of the order of nuclear dimensions (about 10^-15 m), so they cannot act between bodies separated by large distances.

Electrical forces are long-range forces, but they act only between charged bodies. Bodies on the terrestrial, astronomical and cosmological scale, such as planets and stars, are massive but electrically neutral, so these forces play no role in their motion.

Gravitational force, though the weakest of all, is a long-range force and acts between all bodies having mass, and it can never be cancelled since it is always attractive. Hence, even though it is the weakest, only the gravitational force plays a dominant role in the motion of bodies on the earth as well as in the motion of massive bodies such as planets and stars in outer space.

Given,

• Mass of the planet, M_p = 2M_e
• Diameter of the planet = twice that of the earth, so R_p = 2R_e
• The pendulum is a second's pendulum on the earth, that is, T_e = 2 s

The acceleration due to gravity at the surface of a planet is $\text g = \frac{\text{GM}}{\text R^2}$, so

$\frac{\text g_p}{\text g_e} = \frac{\text M_p}{\text M_e} \times \left(\frac{\text R_e}{\text R_p}\right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$

Thus g is reduced to half on the planet. The time-period of a simple pendulum is

$\text T = 2\pi\sqrt{\frac{\text l}{\text g}} \quad \Rightarrow \quad \text T \propto \frac{1}{\sqrt{\text g}}$

Therefore,

$\frac{\text T_p}{\text T_e} = \sqrt{\frac{\text g_e}{\text g_p}} = \sqrt2$

$\text T_p = \sqrt2 \times 2 = 2\sqrt2\ \text s$

Hence, the time-period of the pendulum on this planet will be $2\sqrt2$ second.

Given,

• Radius of the earth = R_e
• Kinetic energy of projection $= \frac{1}{2}$ of the minimum escape energy

The escape energy of a particle of mass m on the earth's surface is $\frac{\text{GM}_e \text m}{\text R_e}$. Hence the kinetic energy of projection is

$\text K = \frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e}$

The gravitational potential energies of the particle at the surface and at a height h are

$\text U = -\frac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text U_h = -\frac{\text{GM}_e \text m}{\text R_e + \text h}$

The particle rises until its whole kinetic energy is converted into the increase in potential energy,

$$\begin{aligned}\text U_h - \text U = \text K \\[1em] \text{GM}_e \text m\left(\frac{1}{\text R_e} - \frac{1}{\text R_e + \text h}\right) = \frac{1}{2}\frac{\text{GM}_e \text m}{\text R_e}\end{aligned}$$

$\frac{\text h}{\text R_e(\text R_e + \text h)} = \frac{1}{2\text R_e} \quad \Rightarrow \quad \frac{\text h}{\text R_e + \text h} = \frac{1}{2}$

$2\text h = \text R_e + \text h \quad \Rightarrow \quad \text h = \text R_e$

Hence, the particle will gain a height equal to the radius of the earth, R_e.

Given,

• Three equal masses, each m, at the vertices A, B and C of an equilateral triangle
• Mass placed at the centroid G = 2m

Let the distance of each vertex from the centroid be GA = GB = GC = l.

The gravitational force on the mass 2m at G due to each of the masses m at A, B and C is

$\text F = \frac{\text G(2\text m)(\text m)}{\text l^2}$

The three forces are therefore equal in magnitude, and they are directed from G towards A, towards B and towards C respectively. Since ABC is an equilateral triangle and G is its centroid, these three directions are equally inclined to one another at 120°.

Three equal forces acting at a point at 120° to one another form a closed triangle, so their vector sum is zero.

Hence, by symmetry, the resultant gravitational force acting on the mass 2m placed at the centroid G is zero.

Let two spheres of the same material, each of radius r and density ρ, be touching each other. Their masses may be assumed to be concentrated at their centres.

The mass of each sphere is

$\text M = \text{volume} \times \text{density} = \frac{4}{3}\pi \text r^3 \rho$

Since the spheres are touching each other, the distance between their centres is

$\text d = \text r + \text r = 2\text r$

By Newton's law of gravitation, the force of attraction between them is

$\text F = \frac{\text{GMM}}{\text d^2} = \frac{\text G\left(\frac{4}{3}\pi \text r^3 \rho\right) \times \left(\frac{4}{3}\pi \text r^3 \rho\right)}{(2\text r)^2}$

$\text F = \frac{\text G \times \frac{16}{9}\pi^2 \text r^6 \rho^2}{4\text r^2} = \frac{4}{9}\text G\pi^2 \rho^2 \text r^4$

Since G, π and ρ are all constants,

$\text F \propto \text r^4$

Hence, the gravitational force between the two touching spheres of the same material is directly proportional to r⁴.

Given,

• Gravitational force on the rocket at a height h $= \frac{1}{3}$ of the force at sea level

The gravitational force on a body of mass m at sea level is

$\text F = \frac{\text{GM}_e \text m}{\text R_e^2}$

and at a height h above the surface it is

$\text F' = \frac{\text{GM}_e \text m}{(\text R_e + \text h)^2}$

According to the question, $\text F' = \frac{1}{3}\text F$, so

$\frac{\text{GM}_e \text m}{(\text R_e + \text h)^2} = \frac{1}{3} \times \frac{\text{GM}_e \text m}{\text R_e^2}$

$(\text R_e + \text h)^2 = 3\text R_e^2 \quad \Rightarrow \quad \text R_e + \text h = \sqrt3\text R_e$

$\text h = (\sqrt3 - 1)\text R_e = (1.732 - 1)\text R_e = 0.732\ \text R_e$

Hence, the relation between h and R_e is h = 0.732 R_e.

Given,

• Radii of the two planets = R₁ and R₂
• Mean densities of the two planets = ρ₁ and ρ₂

Assuming a planet to be spherical, its mass may be taken to be concentrated at its centre. If M is the mass, R the radius and ρ the density of a planet, then

$\text M = \frac{4}{3}\pi \text R^3 \rho$

The gravitational force experienced by a body of mass m placed on its surface is

$\text F = \text G\frac{\text{Mm}}{\text R^2} = \frac{\text{Gm}}{\text R^2}\left(\frac{4}{3}\pi \text R^3 \rho\right) = \frac{4}{3}\pi \text{GmR}\rho$

By Newton's second law of motion, F = mg, where g is the acceleration due to gravity on the planet. Therefore

$\text{mg} = \frac{4}{3}\pi \text{GmR}\rho \quad \Rightarrow \quad \text g = \frac{4}{3}\pi \text{GR}\rho \quad \text{or} \quad \text g \propto \text R\rho$

Thus, the ratio of the accelerations due to gravity on the two given planets is

$\frac{\text g_1}{\text g_2} = \frac{\text R_1 \rho_1}{\text R_2 \rho_2}$

Hence, the accelerations due to gravity on the two planets are in the ratio R₁ρ₁ : R₂ρ₂.

Given,

• Period of revolution of the satellite just above the earth = T
• Density of the earth = ρ

The period of revolution of a satellite revolving just above the earth's surface, in an orbit of radius R_e, is

$\text T = 2\pi\sqrt{\frac{\text R_e^3}{\text{GM}_e}}$

If the earth be a sphere of mean density ρ, its mass is

$\text M_e = \text{volume} \times \text{density} = \frac{4}{3}\pi \text R_e^3 \rho$

Making this substitution and squaring both sides,

$\text T^2 = 4\pi^2 \times \frac{\text R_e^3}{\text G} \times \frac{1}{\frac{4}{3}\pi \text R_e^3 \rho} = \frac{4\pi^2 \times 3}{4\pi \text G\rho}$

$\text T^2 = \frac{3\pi}{\text G\rho} \quad \Rightarrow \quad \rho\text T^2 = \frac{3\pi}{\text G}$

Since π and G are both constants, the quantity $\frac{3\pi}{\text G}$ is a constant.

Hence, ρT² is a universal constant.

The gravitational attraction of the sun on the earth is greater than that of the moon on the earth. This is because of the very large mass of the sun, although the distance between the sun and the earth is much larger than that between the moon and the earth, and the gravitational force varies inversely as the square of the distance.

The tidal effect, however, is produced by the variation of the gravitational force across the diameter of the earth, and this tidal force varies inversely as the cube of the distance.

Therefore, although the mass of the moon is much smaller compared with the mass of the sun, its tidal effect is greater than that of the sun because of its much smaller distance from the earth.

No, the weight of a body will be the same at mid-night and at noon.

The earth is itself a satellite of the sun. A body placed on the earth is also a satellite of the sun, and both the body and the earth have the same acceleration towards the sun, since the acceleration produced by the sun's gravitational field does not depend upon the mass of the body.

Hence there will be no relative gravitational acceleration between the body and the earth, that is, a body placed on the earth will experience no gravitational effect due to the sun. It will experience a gravitational force only due to the earth, and this will be the weight of the body measured on the earth, which will remain the same for all the 24 h.

No, the statement is not correct.

The unit of gravitational potential is J kg^-1, while the unit of gravitational potential energy is J. Two quantities of different units cannot be equated, so the given statement is dimensionally incorrect.

The correct statement will be : If the gravitational potential at a place is 1 J kg^-1, then the gravitational potential energy of 1 kg mass at that place will be 1 J.

Given,

• Mass of the star = M
• Radius of the circular orbit of the planet = R
• Period of revolution = T
• Gravitational force between the planet and the star is proportional to R^-5/2

Let m be the mass of the light planet. Its period of revolution in a circular orbit of radius R is

$\text T = \frac{2\pi \text R}{\text v}$

The gravitational force between the planet and the star provides the necessary centripetal force for the circular motion. Since the force is proportional to R^-5/2,

$\frac{\text{mv}^2}{\text R} = \frac{\text{GMm}}{\text R^{5/2}}$

$\frac{\text v^2}{\text R^2} = \frac{\text{GM}}{\text R^{7/2}}$

From the expression for T,

$\text T^2 = \frac{4\pi^2 \text R^2}{\text v^2}$

Substituting the value of $\frac{\text v^2}{\text R^2}$ obtained above,

$\text T^2 = \frac{4\pi^2}{\left(\frac{\text v^2}{\text R^2}\right)} = \frac{4\pi^2}{\frac{\text{GM}}{\text R^{7/2}}}$

$\text T^2 = \frac{4\pi^2}{\text{GM}}\text R^{7/2}$

Hence proved.

(i) becomes one-fourth

By Newton's law of universal gravitation $\text F \propto \frac{1}{\text r^2}$. If the distance is doubled, r becomes 2r, so

$\frac{\text F'}{\text F} = \left(\frac{\text r}{2\text r}\right)^2 = \frac{1}{4}$

Hence the force is reduced to one-fourth of its original value.

(ii) triples

The gravitational force is directly proportional to the product of the masses, $\text F \propto \text m_1 \text m_2$. If m₁ is tripled and both m₂ and r remain the same, the force also becomes three times its original value.

(iii) a universal constant.

The value of G is the same for all pairs of bodies, at all places, at all distances of separation, and it is independent of the presence of other bodies or of the properties of the intervening medium. Hence G is regarded as a universal constant.

(iv) It applies universally to all objects with mass.

Newton stated that his formula applies not only between the sun and the planets, but between any two bodies of the universe. The law is therefore universal and applies to all objects with mass, regardless of their location or distance.

(v) remains same

Gravitational forces are completely independent of the properties of the intervening medium. Hence, when the bodies are immersed in a liquid with their separation unchanged, the gravitational force of attraction between them remains F.

(i) It increases

As we move from the equator to the poles, the distance from the earth's centre decreases because the earth is an oblate spheroid, and since $\text g \propto \frac{1}{\text R_e^2}$, the value of g increases. Further, the centrifugal effect of the earth's rotation, given by the term R_eω²cos²λ, decreases as the latitude λ increases, which also increases g.

(ii) It decreases

At an altitude h the distance from the earth's centre increases, and

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

so that g' < g. Hence the value of g decreases with increasing altitude.

(iii) The mass of the object

The acceleration due to gravity is $\text g = \frac{\text{GM}_e}{\text R_e^2}$, an expression which is free from the mass of the body. Altitude, latitude and the distribution of the earth's mass all affect g, but the mass of the object placed on the earth does not.

(iv) Graph (a)

Starting from the centre of the earth, the variation with depth is governed by $\text g' = \left(\frac{\text g}{\text R_e}\right)\text r$, so g' ∝ r. Hence g' is zero at the centre (r = 0) and increases linearly to a maximum value g at the surface (r = R_e).

Beyond the surface the variation is governed by $\text g' = \frac{\text{gR}_e^2}{\text r^2}$, so $\text g' \propto \frac{1}{\text r^2}$. Hence g' decreases sharply as a hyperbola and tends to zero as r → ∞, the X-axis being an asymptote to the curve.

Only graph (a) shows this linear rise up to r = R_e followed by the hyperbolic fall.

(v) It would be the same everywhere

If the earth were a perfect sphere, the radius R_e would be the same at every point on the surface, so $\text g = \frac{\text{GM}_e}{\text R_e^2}$ would not vary with position. With no rotation, the term R_eω²cos²λ also vanishes, so g' = g everywhere.

Note: The passage prints the variation of g with height as $\text g' = \frac{\text{rR}_e^2}{\text r^2}$. The leading r is a misprint; the correct relation is $\text g' = \frac{\text{gR}_e^2}{\text r^2}$, as used above.

(i) decreases

The orbital velocity of a satellite in an orbit of radius r is

$\text v_o = \sqrt{\frac{\text{GM}_e}{\text r}} \quad \Rightarrow \quad \text v_o \propto \frac{1}{\sqrt{\text r}}$

Hence, as the radius of the orbit increases, the orbital velocity of the satellite decreases.

(ii) 11.2 kms^-1

The escape velocity from the earth's surface is

$$\begin{aligned}\text v_e = \sqrt{2\text{gR}_e} = \sqrt{2 \times 9.8 \times (6.37 \times 10^6)} \\[1em] = 11.2 \times 10^3\ \text{m s}^{-1} = 11.2\ \text{km s}^{-1}\end{aligned}$$

(iii) less than the escape velocity

For a satellite to remain in a stable circular orbit, the gravitational force must supply exactly the centripetal force required. Its velocity must therefore be the orbital velocity $\text v_o = \frac{\text v_e}{\sqrt2}$, which is less than the escape velocity. If the velocity became equal to or greater than v_e, the satellite would leave its orbit and escape.

(iv) The satellite will gain enough speed to escape Earth's gravitational pull and drift into space.

On increasing the orbital velocity by 41.4%, the new velocity is

$\text v = \text v_o + \frac{41.4}{100}\text v_o = 1.414\ \text v_o = \sqrt2\ \text v_o = \text v_e$

Since the new velocity is exactly the escape velocity, the satellite will leave its orbit and escape into space.

(v) Mars has a smaller gravitational pull than Earth

The escape velocity is $\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$, which depends on the mass and radius of the planet. A smaller escape velocity for Mars means a weaker gravitational pull at its surface, which is why Mars cannot retain a thick atmosphere as the earth does.

Note: The passage prints both the orbital velocity and the escape velocity as $\text v_0=\sqrt{\frac{2\text{GM}_e}{\text R_e^2}}$, but The correct relations are $\text v_o = \sqrt{\frac{\text{GM}_e}{\text R_e}} = \sqrt{\text{gR}_e}$ and $\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} = \sqrt{2\text{gR}_e}$.

(i) calculating the correct orbital insertion velocity for the moon.

The spacecraft must first escape the earth's gravity and then enter the moon's gravitational influence. The critical requirement in this transition is that its velocity must be precisely calculated so that it is captured into a stable lunar orbit, neither crashing into the moon nor passing by it.

(ii) Potential energy decreases, and kinetic energy increases.

As the spacecraft descends towards the moon its distance from the moon's centre decreases, so its gravitational potential energy $\text U = -\frac{\text{GMm}}{\text r}$ becomes more negative, that is, it decreases. By the conservation of mechanical energy this lost potential energy appears as kinetic energy, so the kinetic energy increases and the spacecraft accelerates.

(iii) To prevent it from bouncing off the surface on landing.

The acceleration due to gravity on the moon is only about 1/6th of that on the earth, so the downward force holding the lander on the surface is small. Unless the lander is designed for this reduced gravitational force, it would rebound on touchdown instead of settling, and a stable, controlled landing would not be possible.

(iv) The roughness and composition of the lunar terrain.

The rover must navigate a surface consisting of dust, rocks and uneven ground, with minimal energy expenditure. There is no atmosphere on the moon, and its gravitational force is low rather than high, so the terrain is the significant factor affecting the rover's mobility.

(v) provides insights into the moon's internal structure.

The lander carries instruments to detect "moonquakes". By studying how the seismic waves generated by these quakes propagate through the moon, information can be obtained about the different layers within the moon, that is, about its internal structure.

(i) Elliptical with the Sun at one focus

This is Kepler's first law, the law of orbits, according to which all planets move around the sun in elliptical orbits having the sun at one focus of the orbit, and not at the centre.

(ii) Faster when it is closer to the Sun

By Kepler's second law the line joining a planet and the sun sweeps out equal areas in equal times. When the planet is nearer the sun the radius vector is shorter, so the planet must move faster for the area swept to remain the same. Hence its velocity is maximum when it is nearest the sun and minimum when it is farthest.

(iii) The orbital period to the distance from the Sun

Kepler's third law, the law of periods, states that

$\text T^2 \propto \text r^3$

that is, it relates the square of the orbital period of a planet to the cube of its mean distance from the sun.

(iv) angular momentum.

The gravitational force on a planet is a central force, directed always towards the sun. It therefore exerts no torque about the sun, and hence the angular momentum of the planet about the sun remains constant with time. The velocity, the centripetal force and the linear momentum all change continuously along the elliptical orbit.

(v) All planets in the solar system follow the same elliptical path

Kepler's laws do imply that the orbital speed of a planet is not constant, that planets closer to the sun have shorter orbital periods, and that the force on a planet is directed towards the sun. However, each planet has its own elliptical orbit with its own size and eccentricity, so all planets certainly do not follow the same elliptical path.

(i) They appear stationary relative to a specific point on Earth.

A geostationary satellite has a period of revolution of 24 h, exactly equal to the period of the axial motion of the earth, and it revolves in the equatorial plane in the same direction as the earth. Being synchronous with the earth's spin, it appears stationary over a fixed point on the earth's surface. It orbits once in 24 h and not in 12 h, it does not pass over the poles, and it is used mainly for communication, broadcasting and weather monitoring.

(ii) coverage of the entire earth's surface over time.

Polar satellites move in a north-south direction, passing close to the poles on each revolution. As the earth rotates beneath their orbital path, they gradually scan the entire surface of the globe in successive passes. A geostationary satellite, in contrast, provides continuous coverage of only one specific region.

(iii) 600 – 800 km

Polar satellites operate in low earth orbit, typically at altitudes ranging from about 500 km to 800 km above the earth's surface, completing one revolution in about 100 minutes.

(iv) The orbital period would decrease.

By Kepler's third law, T² ∝ r³. If the satellite were placed at a lower altitude, the radius of its orbit would be smaller and hence its period of revolution would be shorter. It would then move faster than the earth's rotation and would no longer appear stationary over a point on the earth.

(v) Monitoring polar ice caps and global weather patterns.

Because polar satellites repeatedly scan every part of the earth's surface and operate at low altitudes, they provide comprehensive, high-resolution and frequent coverage. They are therefore used for weather forecasting, environmental monitoring such as tracking glacier retreat and snow cover, and climate studies. Continuous television coverage and real-time communication over a fixed region are the functions of geostationary satellites.

Kepler's laws of planetary motion :

(i) Law of Orbits : All planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

(ii) Law of Areas : A line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant.

(iii) Law of Periods : The square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun, T² = K r³.

Proof that the force is inversely proportional to the square of the distance :

Newton found that the orbits of most of the planets are nearly circular. Let a planet of mass m move on a circular path of radius r with a linear speed v. Since it is moving on a circular path, it is acted upon by a centripetal force directed towards the sun,

$\text F = \frac{\text{mv}^2}{\text r}$

If T be the period of revolution of the planet, then

$\text v = \frac{\text{linear distance travelled in one revolution}}{\text{period of revolution}} = \frac{2\pi \text r}{\text T}$

Substituting this value of v,

$\text F = \frac{\text m}{\text r}\left(\frac{2\pi \text r}{\text T}\right)^2 = \frac{4\pi^2 \text{mr}}{\text T^2}$

But according to Kepler's third law, T² = K r³. Therefore

$\text F = \frac{4\pi^2 \text{mr}}{\text K\text r^3} = \frac{4\pi^2}{\text K}\left(\frac{\text m}{\text r^2}\right)$

Since 4π² and K are constants,

$\text F \propto \frac{\text m}{\text r^2} \quad \text{or} \quad \text F \propto \frac{1}{\text r^2}$

Hence, the force acting on a planet is inversely proportional to the square of its distance from the sun.

Gravitational potential : If we bring a body from outside into a gravitational field, then the field itself will work on the body. The work done in bringing a unit mass from infinity to a point in the gravitational field is called the 'gravitational potential' at that point. This work is obtained, and not done by the agent bringing the mass.

If W joule be the work obtained in bringing a body of mass m kg from infinity to a point in the gravitational field, then the gravitational potential at that point is

$\text V = \frac{\text W}{\text m}\ \text{J/kg}$

Since the work W is obtained, that is, it is negative, the gravitational potential is always negative. It is a scalar quantity and its dimensions are [L²T^-2].

Derivation of gravitational potential energy of a body on the surface of earth :

Let M_e be the mass and R_e the radius of the earth. Suppose a body of mass m is placed at a point B at a height x from the centre O of the earth. The gravitational force of attraction on the body is

$\text F = \text G\frac{\text M_e \text m}{\text x^2}$

The work done in taking the body through an infinitesimally small distance dx from C to B is

$\text{dW} = \text F\text{dx} = \text G\frac{\text M_e \text m}{\text x^2}\text{dx}$

Therefore the work done in taking the body from infinity to the point A on the earth's surface is obtained by integrating between the limits x = ∞ and x = R_e,

$\text W = \int_\infty^{\text R_e} \frac{\text{GM}_e \text m}{\text x^2}\text{dx} = \text{GM}_e \text m\left[-\frac{1}{\text x}\right]_\infty^{\text R_e}$

$\text W = \text{GM}_e \text m\left[-\frac{1}{\text R_e} + \frac{1}{\infty}\right] = -\frac{\text{GM}_e \text m}{\text R_e}$

The gravitational potential energy of the body on the earth's surface is equal to this work obtained,

$\text U = \text W = -\frac{\text{GM}_e \text m}{\text R_e}$

Since $\text g = \frac{\text{GM}_e}{\text R_e^2}$, that is GM_e = gR_e², this may also be written as

$\text U = -\frac{\text{gR}_e^2 \text m}{\text R_e} = -\text{gR}_e \text m$

Escape velocity : Escape velocity is the minimum velocity that an object needs in order to escape the gravitational pull of a celestial body, without any further propulsion. If an object is launched from the earth's surface with this speed, it will continue moving away from the earth, overcoming the earth's gravitational pull, and will not fall back or enter an orbit.

Derivation of the formula :

The work done in moving a body of mass m from the earth's surface (R_e) to infinity, against the gravitational pull of the earth, is

$\text W = \frac{\text{GM}_e \text m}{\text R_e}$

This work is supplied by the kinetic energy given to the body. If v_e is the velocity with which the body just breaks free from the gravitational pull of the earth, then

$\frac{1}{2}\text{mv}_e^2 = \frac{\text{GM}_e \text m}{\text R_e} \quad \Rightarrow \quad \frac{1}{2}\text v_e^2 = \frac{\text{GM}_e}{\text R_e}$

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} \qquad \text{...(i)}$

Since GM_e = gR_e², equation (i) may also be written as

$\text v_e = \sqrt{2\text{gR}_e} \qquad \text{...(ii)}$

The expression is free from the mass m of the body, so the escape velocity is independent of the mass of the body.

Relation with the orbital velocity : The orbital velocity of a satellite revolving close to the earth's surface is

$\text v_o = \sqrt{\text{gR}_e}$

Therefore

$\frac{\text v_o}{\text v_e} = \frac{\sqrt{\text{gR}_e}}{\sqrt{2\text{gR}_e}} = \frac{1}{\sqrt2}$

$\text v_e = \sqrt2\ \text v_o$

Hence proved. If the orbital velocity of a satellite revolving close to the earth happens to increase to $\sqrt2$ times, the satellite would escape.

Gravitational potential energy of a body on earth's surface : The work obtained in bringing a body from infinity to a point in a gravitational field is called the gravitational potential energy of the body at that point. For a body of mass m on the surface of the earth of mass M_e and radius R_e,

$\text U = -\frac{\text{GM}_e \text m}{\text R_e}$

The gravitational potential energy at infinity is assumed to be zero, and since work is obtained (not done) in bringing the body from infinity into the gravitational field, the gravitational potential energy is always negative.

Derivation of the escape velocity :

The total energy of the body on the earth's surface, when it is projected with a velocity v_e, is the sum of its kinetic energy and its gravitational potential energy,

$\text E = \frac{1}{2}\text{mv}_e^2 - \frac{\text{GM}_e \text m}{\text R_e}$

For the body just to escape, it must reach infinity, where both its kinetic energy and its potential energy become zero, so that the total energy there is zero. By the conservation of energy,

$\frac{1}{2}\text{mv}_e^2 - \frac{\text{GM}_e \text m}{\text R_e} = 0$

$\frac{1}{2}\text{mv}_e^2 = \frac{\text{GM}_e \text m}{\text R_e} \quad \Rightarrow \quad \text v_e^2 = \frac{2\text{GM}_e}{\text R_e}$

$\text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} = \sqrt{2\text{gR}_e}$

Substituting g = 9.8 m s^-2 and R_e = 6.37 × 10⁶ m,

$\text v_e = \sqrt{2 \times 9.8 \times (6.37 \times 10^6)} = 11.2 \times 10^3\ \text{m/s} = 11.2\ \text{km/s}$

Hence, the escape velocity of a body from the earth is 11.2 km s^-1.

Kepler's laws of planetary motion :

(i) Law of Orbits : All planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

(ii) Law of Areas : A line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant. Hence the velocity of a planet is maximum when it is nearest the sun and minimum when it is farthest.

(iii) Law of Periods : The square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun, T² = K r³.

Newton's conclusions from Kepler's laws :

Newton found that the orbits of most of the planets are nearly circular, and that from the second law the areal speed of a planet remains constant, which for a circular orbit means that its linear speed is constant. Since the planet moves on a circular path, it must be acted upon by a centripetal force directed towards the sun. On combining this with the third law he obtained $\text F = \frac{4\pi^2}{\text K}\left(\frac{\text m}{\text r^2}\right)$, and drew the following conclusions :

(i) A planet is acted upon by a centripetal force which is directed towards the sun.

(ii) This force is inversely proportional to the square of the distance between the planet and the sun, that is, $\text F \propto \frac{1}{\text r^2}$.

(iii) This force is directly proportional to the mass of the planet ($\text F \propto \text m$). Since the force between the planet and the sun is mutual, the force F is also proportional to the mass M of the sun ($\text F \propto \text M$).

Since the mass of the sun is constant, the constant $\frac{4\pi^2}{\text K}$ was replaced by GM, giving

$\text F = \text G\frac{\text{Mm}}{\text r^2}$

Newton then stated that this formula applies not only between the sun and the planets, but between any two bodies of the universe. This is Newton's law of gravitation.

The acceleration due to gravity at the earth's surface is related to the gravitational constant by

$\text g = \frac{\text{GM}_e}{\text R_e^2}$

(i) Mass of the earth : Rearranging the above relation, the mass of the earth is

$\text M_e = \frac{\text{gR}_e^2}{\text G}$

Thus, by measuring g, the radius R_e of the earth and the gravitational constant G, the mass of the earth can be computed. Substituting the known values g = 9.80 m s^-2, R_e = 6.37 × 10⁶ m and G = 6.67 × 10^-11 N m² kg^-2,

$\text M_e = \frac{9.80 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = 6.0 \times 10^{24}\ \text{kg (approx.)}$

(ii) Mean density of the earth : If ρ be the average density of the earth, then its mass is

$\text M_e = \text{volume} \times \text{density} = \frac{4}{3}\pi \text R_e^3 \rho$

Substituting this value of M_e in the expression above,

$\frac{4}{3}\pi \text R_e^3 \rho = \frac{\text{gR}_e^2}{\text G} \quad \Rightarrow \quad \rho = \frac{3\text g}{4\pi \text R_e \text G}$

Substituting the known values,

$$\begin{aligned}\rho = \frac{3 \times 9.80}{4 \times 3.14 \times (6.37 \times 10^6) \times (6.67 \times 10^{-11})} \\[1em] = 5.5 \times 10^3\ \text{kg m}^{-3}\ \text{(approx.)}\end{aligned}$$

Hence, a knowledge of g enables us to determine the mass of the earth as 6.0 × 10²⁴ kg and its mean density as 5.5 × 10³ kg m^-3.

Newton found that the orbits of most of the planets are nearly circular. According to Kepler's second law the areal speed of a planet remains constant, which means that in a circular orbit the linear speed of the planet is constant.

Let a planet of mass m move on a circular path of radius r with a linear speed v. Since it is moving on a circular path, it is being acted upon by a centripetal force directed towards the centre, that is, towards the sun,

$\text F = \frac{\text{mv}^2}{\text r}$

If T be the period of revolution of the planet, then

$\text v = \frac{2\pi \text r}{\text T}$

Substituting this value of v,

$\text F = \frac{\text m}{\text r}\left(\frac{2\pi \text r}{\text T}\right)^2 = \frac{4\pi^2 \text{mr}}{\text T^2} \qquad \text{...(i)}$

But according to Kepler's third law, T² = K r³. Substituting this in equation (i),

$\text F = \frac{4\pi^2 \text{mr}}{\text K\text r^3} = \frac{4\pi^2}{\text K}\left(\frac{\text m}{\text r^2}\right) \quad \text{or} \quad \text F \propto \frac{\text m}{\text r^2}$

From this Newton concluded that the force is inversely proportional to the square of the distance between the planet and the sun and directly proportional to the mass m of the planet. Since the force between the planet and the sun is mutual, F is also proportional to the mass M of the sun. Hence

$\text F \propto \frac{\text{Mm}}{\text r^2} \quad \Rightarrow \quad \text F = \text G\frac{\text{Mm}}{\text r^2}$

where the constant $\frac{4\pi^2}{\text K}$ has been replaced by GM, the mass of the sun being constant.

Hence, Newton's gravitational inverse square law is established from Kepler's laws of planetary motion.

Let M_e be the mass of the earth and R_e its radius. A satellite of mass m revolves in a circular orbit of radius r at a height h above the earth's surface, so that

$\text r = \text R_e + \text h$

Orbital velocity : The centripetal force required for the circular motion of the satellite is $\frac{\text{mv}_o^2}{\text r}$, and this is provided by the gravitational force exerted by the earth on the satellite. Therefore

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r} \quad \Rightarrow \quad \text v_o^2 = \frac{\text{GM}_e}{\text r}$

$\text v_o = \sqrt{\frac{\text{GM}_e}{\text R_e + \text h}} \qquad \text{...(i)}$

If g is the acceleration due to gravity on the earth's surface, then $\text g = \frac{\text{GM}_e}{\text R_e^2}$, that is GM_e = gR_e². Substituting this value in equation (i),

$\text v_o = \sqrt{\frac{\text{gR}_e^2}{\text R_e + \text h}} = \text R_e\sqrt{\frac{\text g}{\text R_e + \text h}}$

Hence proved.

Time-period of revolution : The satellite covers the circumference 2πr of its orbit in one revolution, so

$\text T = \frac{2\pi \text r}{\text v_o} = \frac{2\pi(\text R_e + \text h)}{\text v_o}$

Substituting the value of v_o from equation (i),

$\text T = \frac{2\pi(\text R_e + \text h)}{\left[\text{GM}_e / (\text R_e + \text h)\right]^{1/2}} \quad \Rightarrow \quad \text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}}$

Since GM_e = gR_e², this may also be written as

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{gR}_e^2}}$

The period of revolution of a satellite revolving at a height h above the earth's surface, in an orbit of radius r = R_e + h, is

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}} = 2\pi\sqrt{\frac{\text r^3}{\text{GM}_e}}$

Squaring both sides,

$\text T^2 = 4\pi^2 \times \frac{\text r^3}{\text{GM}_e}$

Rearranging so that the orbital radius and the period appear on the same side,

$\frac{\text r^3}{\text T^2} = \frac{\text{GM}_e}{4\pi^2}$

In this expression G is the universal gravitational constant, M_e is the mass of the earth and 4π² is a number. All three are constants, and therefore

$\frac{\text r^3}{\text T^2} = \text{constant}$

Hence, the ratio of the cube of the radius of the satellite's orbit to the square of its period of revolution is constant, which is exactly Kepler's third law. This shows that an artificial satellite of the earth obeys the same law that governs the motion of the planets around the sun.

Distinction between g and G :

Acceleration due to gravity (g)	Gravitational constant (G)
It is the acceleration produced in a body falling freely towards the earth.	It is numerically equal to the force between two particles, each of unit mass, placed a unit distance apart.
It is a vector quantity.	It is a scalar quantity.
Its S.I. unit is m s-2 or N kg-1.	Its S.I. unit is N m2 kg-2.
Its dimensions are [LT-2].	Its dimensions are [M-1L3T-2].
It is not a universal constant. Its value varies with altitude, depth, latitude and the shape of the earth.	It is a universal constant. Its value is the same for all pairs of bodies at all places and at all times.
Its value at the earth's surface is about 9.8 m s-2.	Its value is 6.67 × 10-11 N m2 kg-2.

Expression for g in terms of G :

Let M_e be the mass and R_e the radius of the earth. Assuming the earth to be spherical, its total mass may be imagined as being concentrated at a single point at its centre O.

Let a body of mass m be situated at the earth's surface. By Newton's law of gravitation, the force of attraction between the body and the earth is

$\text F = \frac{\text{GM}_e \text m}{\text R_e^2} \qquad \text{...(i)}$

The gravitational force on the body is its weight, F = mg. Its reaction is the equal and opposite gravitational force exerted by the body on Earth.

$\text F = \text{mg} \qquad \text{...(ii)}$

From equations (i) and (ii),

$\text{mg} = \frac{\text{GM}_e \text m}{\text R_e^2} \quad \Rightarrow \quad \text g = \frac{\text{GM}_e}{\text R_e^2}$

This expression is free from the mass m of the body, which shows that the value of g does not depend upon the mass of the body.

(i) Variation of g above the surface of the earth :

Let O be the centre of the earth, R_e its radius and M_e its mass. The acceleration due to gravity at the surface is

$\text g = \frac{\text{GM}_e}{\text R_e^2} \qquad \text{...(i)}$

If the body is raised to a height h above the surface, its distance from the centre becomes (R_e + h), and the acceleration due to gravity there is

$\text g' = \frac{\text{GM}_e}{(\text R_e + \text h)^2} \qquad \text{...(ii)}$

Dividing equation (ii) by equation (i),

$\frac{\text g'}{\text g} = \frac{\text R_e^2}{(\text R_e + \text h)^2} = \frac{1}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2} \quad \Rightarrow \quad \text g' \lt \text g$

Thus, as we go above the surface of the earth, the acceleration due to gravity goes on decreasing. For example, at a height equal to the radius of the earth, g' = g/4.

(ii) Variation of g below the surface of the earth :

Let a body P of mass m be situated at a depth h below the earth's surface, so that its distance from the centre is (R_e − h). By Newton's shell theorem the gravitational force on a body inside a spherical shell is always zero, so the body experiences attraction only due to the inner solid sphere of radius (R_e − h), whose mass is

$\text M' = \frac{4}{3}\pi(\text R_e - \text h)^3 \rho$

The force of attraction on the body is therefore

$\text{mg}' = \text G\frac{\text M'\text m}{(\text R_e - \text h)^2} = \frac{4}{3}\pi \text G(\text R_e - \text h)\rho\text m \qquad \text{...(iii)}$

Similarly at the surface, where h = 0,

$\text{mg} = \frac{4}{3}\pi \text{GR}_e \rho\text m \qquad \text{...(iv)}$

Dividing equation (iii) by equation (iv),

$\frac{\text g'}{\text g} = \frac{\text R_e - \text h}{\text R_e} \quad \Rightarrow \quad \text g' = \text g\left(1 - \frac{\text h}{\text R_e}\right)$

so that g' < g. Thus, as we go below the surface of the earth, the acceleration due to gravity goes on decreasing and becomes zero at the centre of the earth (where h = R_e).

Relation between g and G : The two are related by

$\text g = \frac{\text{GM}_e}{\text R_e^2}$

where G is a universal constant, while g is not constant and varies from place to place.

Factors on which g depends : The acceleration due to gravity at the earth's surface is

$\text g = \frac{\text{GM}_e}{\text R_e^2}$

so it depends upon the mass M_e of the earth and the radius R_e of the earth, but not upon the mass, shape or size of the body. Its value further changes with the following four factors :

(i) Altitude : g decreases as we go above the earth's surface.

(ii) Depth : g decreases as we go below the earth's surface and becomes zero at the centre.

(iii) Shape of the earth : Since the earth is an oblate spheroid, its equatorial radius is greater than its polar radius, so g is about 9.83 m s^-2 at the poles and about 9.78 m s^-2 at the equator.

(iv) Rotation of the earth : Due to the rotation, the apparent value becomes g' = g − R_eω²cos²λ, which is minimum at the equator and maximum at the poles.

Value of g at a height equal to the radius of the earth :

The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Putting h = R_e,

$\text g' = \frac{\text g}{\left(1 + \frac{\text R_e}{\text R_e}\right)^2} = \frac{\text g}{(1 + 1)^2} = \frac{\text g}{4}$

Hence, at a height equal to the radius of the earth from the surface, the value of g becomes one-fourth of its value at the surface.

Given, the mass of the earth is M_e and its mean radius is R_e. A satellite of mass m revolves in a circular orbit at a height h above the earth's surface, so that the radius of its orbit is

$\text r = \text R_e + \text h$

When the satellite revolves in a circular orbit around the earth, a centripetal force acts between them. This force is provided by the gravitational force of attraction exerted by the earth on the satellite.

The centripetal force required for the motion of the satellite is $\frac{\text{mv}_o^2}{\text r}$, and the gravitational force exerted by the earth on the satellite is $\frac{\text{GM}_e \text m}{\text r^2}$. Since the gravitational force provides the required centripetal force,

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r}$

Cancelling m and one power of r from both sides,

$\text v_o^2 = \frac{\text{GM}_e}{\text r} \quad \Rightarrow \quad \text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$

Substituting r = R_e + h,

$\text v_o = \sqrt{\frac{\text{GM}_e}{\text R_e + \text h}} \qquad \text{...(i)}$

If g is the acceleration due to gravity on the earth's surface, then GM_e = gR_e², and equation (i) becomes

$\text v_o = \text R_e\sqrt{\frac{\text g}{\text R_e + \text h}} \qquad \text{...(ii)}$

Factors on which the orbital speed depends : Equations (i) and (ii) are free from the mass m of the satellite. Hence the orbital speed depends only upon the mass of the earth, the radius of the earth and the height h of the satellite above the earth's surface. Greater the height of the satellite, smaller is its orbital speed. Since the speed does not depend upon the mass of the satellite, two satellites of different masses revolving in the same orbit around the earth have the same speed.

Orbital speed of a satellite : When a satellite of mass m revolves in a circular orbit of radius r around the earth, the gravitational force of the earth provides the necessary centripetal force,

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r} \quad \Rightarrow \quad \text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$

If h be the height of the satellite above the earth's surface, then r = R_e + h, so that

$\text v_o = \sqrt{\frac{\text{GM}_e}{\text R_e + \text h}} = \text R_e\sqrt{\frac{\text g}{\text R_e + \text h}}$

Period of revolution : The satellite covers the circumference of its orbit in one revolution, so

$\text T = \frac{2\pi \text r}{\text v_o} = \frac{2\pi(\text R_e + \text h)}{\text v_o} = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}}$

Geostationary satellite : The period of revolution of a satellite depends upon its height above the earth's surface. If the height of an artificial satellite above the earth's surface is such that its period of revolution is exactly equal to the period of the axial motion of the earth (24 h), then the satellite would appear stationary over a point on the earth's equator. Such a satellite is known as a geostationary satellite or a geosynchronous satellite, and its orbit is called the parking orbit.

The conditions for a satellite to be geostationary are :

(i) The angular velocity, or the time period, of the satellite should be the same as that of the earth.

(ii) The direction of rotation of the satellite should be the same as that of the earth, that is, anticlockwise from west to east.

(iii) The orbital plane of the satellite should be coplanar with the equatorial plane of the earth.

Putting T = 24 h in $\text h = \left(\frac{\text T^2 \text{gR}_e^2}{4\pi^2}\right)^{1/3} - \text R_e$, the height comes out to be about 35,830 km above the equator.

Usefulness : Since a geostationary satellite remains fixed relative to a specific point on the earth's surface, it provides continuous coverage of the same geographical area. It is therefore ideal for communication, broadcasting services such as television, radio and internet, and weather monitoring. This is also why a satellite dish need not be constantly realigned. Such satellites are further crucial in disaster management, since they provide real-time monitoring of weather patterns, enabling early warnings for cyclones and other natural disasters.

Derivation of the periodic-time : A satellite of mass m revolves around the earth in a circular orbit of radius r = R_e + h. The gravitational force of the earth provides the necessary centripetal force,

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}_o^2}{\text r} \quad \Rightarrow \quad \text v_o = \sqrt{\frac{\text{GM}_e}{\text r}}$

The satellite covers the circumference 2πr of its orbit in one revolution, so its period of revolution is

$\text T = \frac{2\pi \text r}{\text v_o} = \frac{2\pi \text r}{\sqrt{\text{GM}_e / \text r}}$

$\text T = 2\pi\sqrt{\frac{\text r^3}{\text{GM}_e}} = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{GM}_e}} \qquad \text{...(i)}$

Dependence on the density of the earth : If the earth be supposed to be a sphere of mean density ρ, then its mass is

$\text M_e = \text{volume} \times \text{density} = \frac{4}{3}\pi \text R_e^3 \rho$

Substituting this value of M_e in equation (i),

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text G \times \frac{4}{3}\pi \text R_e^3 \rho}} = \sqrt{\frac{4\pi^2 \times 3(\text R_e + \text h)^3}{4\pi \text G\rho \text R_e^3}}$

$\text T = \sqrt{\frac{3\pi(\text R_e + \text h)^3}{\text G\rho \text R_e^3}}$

For a satellite revolving very close to the earth's surface (h << R_e), this reduces to

$\text T = \sqrt{\frac{3\pi}{\text G\rho}}$

Hence, the period of revolution of a satellite revolving close to the earth's surface varies inversely as the square root of the mean density of the earth, and is independent of the radius of the earth and of the mass of the satellite.

Escape energy : The gravitational potential energy of a body of mass m on the earth's surface is $-\frac{\text{GM}_e \text m}{\text R_e}$, and at infinity it is zero. The escape energy of a body is the minimum energy that must be supplied to it in order to take it from the surface of the earth to infinity, that is, to free it completely from the gravitational field of the earth. Its value is

$\text{Escape energy} = 0 - \left(-\frac{\text{GM}_e \text m}{\text R_e}\right) = +\frac{\text{GM}_e \text m}{\text R_e}$

Proof that escape velocity is independent of mass :

This escape energy is supplied to the body in the form of kinetic energy. If v_e is the escape velocity, then

$\frac{1}{2}\text{mv}_e^2 = \frac{\text{GM}_e \text m}{\text R_e}$

The mass m of the body appears on both sides of this equation and therefore cancels out,

$\frac{1}{2}\text v_e^2 = \frac{\text{GM}_e}{\text R_e} \quad \Rightarrow \quad \text v_e = \sqrt{\frac{2\text{GM}_e}{\text R_e}} = \sqrt{2\text{gR}_e}$

The final expression contains only the mass M_e and the radius R_e of the earth, and no term involving the mass of the escaping body.

Hence, the escape velocity of a body is independent of its mass. This is why the escape velocity from the earth is 11.2 km s^-1 for a body of 1 g as well as for a body of 1000 kg.

Escape velocity : Escape velocity is the minimum velocity that an object needs in order to escape the gravitational pull of a celestial body, without any further propulsion. An object launched with this speed will continue moving away, overcoming the gravitational pull, and will not fall back or enter an orbit.

Derivation for a planet of mass M and radius R :

Let a body of mass m be projected from the surface of the planet with a velocity v_e. The gravitational potential energy of the body on the surface of the planet is

$\text U = -\frac{\text{GMm}}{\text R}$

and its initial kinetic energy is $\frac{1}{2}\text{mv}_e^2$. Hence the total energy of the body at the surface is

$\text E = \frac{1}{2}\text{mv}_e^2 - \frac{\text{GMm}}{\text R}$

For the body just to escape, it must reach infinity, where its potential energy is zero and its kinetic energy is also just zero. Hence its total energy at infinity is zero.

By the conservation of energy,

$\frac{1}{2}\text{mv}_e^2 - \frac{\text{GMm}}{\text R} = 0$

$\frac{1}{2}\text{mv}_e^2 = \frac{\text{GMm}}{\text R} \quad \Rightarrow \quad \text v_e^2 = \frac{2\text{GM}}{\text R}$

$\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$

If g is the acceleration due to gravity at the surface of the planet, then GM = gR², and

$\text v_e = \sqrt{2\text{gR}}$

The expression is free from the mass m of the body, so the escape velocity depends only upon the mass and the radius of the planet, and is different for different planets.

Gravitational potential energy : The work obtained in bringing a body from infinity to a point in a gravitational field is called the 'gravitational potential energy' of the body at that point. The gravitational potential energy at infinity is assumed to be zero. Because work is obtained, and not done, in bringing the body from infinity into a gravitational field, the gravitational potential energy is always negative.

(i) Potential energy at a distance r (r > R_e) from the centre :

If F be the gravitational force acting on the body of mass m when it is at a distance x from the centre of the earth, then

$\text F = \frac{\text{GM}_e \text m}{\text x^2}$

The work obtained in taking the body through an infinitesimally small distance dx is dW = F dx. Hence the total work obtained in bringing the body from infinity to a distance r is

$\text U = \int_\infty^{\text r} \frac{\text{GM}_e \text m}{\text x^2}\text{dx} = \text{GM}_e \text m\left[-\frac{1}{\text x}\right]_\infty^{\text r}$

$\text U = \text{GM}_e \text m\left[-\frac{1}{\text r} + \frac{1}{\infty}\right] = -\frac{\text{GM}_e \text m}{\text r}$

If the body is at a height h above the earth's surface, then r = R_e + h, so

$\text U = -\frac{\text{GM}_e \text m}{\text R_e + \text h} = -\frac{\text{gR}_e^2 \text m}{\text R_e + \text h}$

(ii) Potential energy on the surface of the earth :

Putting r = R_e in the above expression, the gravitational potential energy of the body on the earth's surface is

$\text U = -\frac{\text{GM}_e \text m}{\text R_e}$

Since GM_e = gR_e², this may also be written as

$\text U = -\frac{\text{gR}_e^2 \text m}{\text R_e} = -\text{gR}_e \text m$

Given,

• Velocity of projection = v
• Maximum height attained = h, not negligible compared with R_e

Derivation of the expression for h :

The gravitational potential energies of the body of mass m on the earth's surface and at a height h are

$\text U = -\frac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text U_h = -\frac{\text{GM}_e \text m}{\text R_e + \text h}$

Therefore the increase in gravitational potential energy is

$\text U_h - \text U = \text{GM}_e \text m\left(\frac{1}{\text R_e} - \frac{1}{\text R_e + \text h}\right) = \frac{\text{GM}_e \text{mh}}{\text R_e(\text R_e + \text h)}$

Using GM_e = gR_e², this becomes

$\text U_h - \text U = \frac{\text{mgh}}{1 + (\text h / \text R_e)}$

As the body goes up, its whole initial kinetic energy is converted into this increase in potential energy at the maximum height. Hence

$\frac{1}{2}\text{mv}^2 = \frac{\text{mgh}}{1 + (\text h / \text R_e)} \quad \Rightarrow \quad \text v^2 = \frac{2\text{gh}}{1 + (\text h / \text R_e)} \qquad \text{...(i)}$

Solving equation (i) for h,

$\text v^2\left(1 + \frac{\text h}{\text R_e}\right) = 2\text{gh} \quad \Rightarrow \quad \text v^2 + \frac{\text v^2 \text h}{\text R_e} = 2\text{gh}$

$\text v^2 \text R_e = \text h(2\text{gR}_e - \text v^2) \quad \Rightarrow \quad \text h = \frac{\text v^2 \text R_e}{2\text{gR}_e - \text v^2}$

(i) When h = R_e : Substituting h = R_e in equation (i),

$\text v^2 = \frac{2\text{gR}_e}{1 + \frac{\text R_e}{\text R_e}} = \frac{2\text{gR}_e}{2} = \text{gR}_e \quad \Rightarrow \quad \text v = \sqrt{\text{gR}_e}$

(ii) When h = ∞ : Substituting h → ∞ in equation (i), the term $\frac{\text h}{\text R_e}$ dominates, so

$\text v^2 = \frac{2\text{gh}}{\text h / \text R_e} = 2\text{gR}_e \quad \Rightarrow \quad \text v = \sqrt{2\text{gR}_e}$

Hence, $\text h = \frac{\text v^2 \text R_e}{2\text{gR}_e - \text v^2}$, and the velocities are $\sqrt{\text{gR}_e}$ for h = R_e and $\sqrt{2\text{gR}_e}$ for h = ∞. The second value is the escape velocity, as it should be.

There cannot be an atmosphere on the moon because the escape velocity on the moon is very small, being only about 2.4 km s^-1, against 11.2 km s^-1 on the earth.

The escape velocity from the surface of a body of mass M and radius R is

$\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$

The moon has a much smaller mass than the earth, so its gravitational pull is weak and its escape velocity is correspondingly small.

The molecules of a gas are in continuous random motion with a wide range of speeds, and their average speed increases with temperature. On the moon, because the escape velocity is so low, the molecules of any atmosphere would easily attain speeds greater than the escape velocity of the moon. They would therefore escape from the moon's gravitational field into outer space.

In this way any gaseous particles that might once have been present on the moon have long since escaped, leaving a barren surface with no air, no weather and extreme temperature variations between day and night.

On the earth, in contrast, the high escape velocity ensures that most atmospheric molecules, despite their thermal motion, remain bound to the earth. Only the lightest gases, such as hydrogen and helium, have velocities close to the escape velocity and can slowly escape over time.

Hence, there cannot be an atmosphere on the moon.

Given,

• Mass of the electron, m_e = 9.1 × 10^-31 kg
• Mass of the proton, m_p = 1.7 × 10^-27 kg
• Charge, e = 1.6 × 10^-19 C
• G = 6.67 × 10^-11 N m² kg^-2
• $\frac{1}{4\pi\varepsilon_0}$ = 9 × 10⁹ N m² C^-2

If r be the distance between the electron and the proton, the gravitational force between them is

$\text F_g = \frac{\text{Gm}_e \text m_p}{\text r^2}$

and the electrostatic force between them is

$\text F_e = \frac{1}{4\pi\varepsilon_0}\frac{\text e^2}{\text r^2}$

Taking the ratio, the distance r cancels out,

$\frac{\text F_g}{\text F_e} = \frac{\text{Gm}_e \text m_p}{\left(\frac{1}{4\pi\varepsilon_0}\right)\text e^2}$

Substituting the values,

$$\begin{aligned}\frac{\text F_g}{\text F_e} = \frac{(6.67 \times 10^{-11}) \times (9.1 \times 10^{-31}) \times (1.7 \times 10^{-27})}{(9 \times 10^9) \times (1.6 \times 10^{-19})^2} \\[1em] = \frac{1.032 \times 10^{-67}}{2.304 \times 10^{-28}}\end{aligned}$$

$\frac{\text F_g}{\text F_e} = 4.5 \times 10^{-40}$

Hence, the ratio of the gravitational force to the electrostatic force is 4.5 × 10^-40. This shows that gravitational forces are the weakest forces existing in nature.

Given,

• Masses of the two bodies, M₁ = 100 kg and M₂ = 1000 kg
• Distance between them = 1.00 m
• G = 6.67 × 10^-11 N m² kg^-2

The distance of the middle-point from either body is

$\text r = \frac{1.00}{2} = 0.5\ \text m$

Gravitational field intensity : The intensities due to the two bodies are directed opposite to each other at the middle-point, since the point lies between them. Hence the resultant intensity is

$\text I = \frac{\text{GM}_2}{\text r^2} - \frac{\text{GM}_1}{\text r^2} = \frac{\text G(\text M_2 - \text M_1)}{\text r^2}$

Substituting the values,

$$\begin{aligned}\text I = \frac{(6.67 \times 10^{-11}) \times (1000 - 100)}{(0.5)^2} = \frac{(6.67 \times 10^{-11}) \times 900}{0.25} \\[1em] = 2.40 \times 10^{-7}\ \text{N kg}^{-1}\end{aligned}$$

directed towards the 1000 kg body.

Gravitational potential : Potential is a scalar quantity, so the potentials due to the two bodies are added algebraically,

$\text V = -\frac{\text{GM}_1}{\text r} - \frac{\text{GM}_2}{\text r} = -\frac{\text G(\text M_1 + \text M_2)}{\text r}$

$\text V = -\frac{(6.67 \times 10^{-11}) \times (100 + 1000)}{0.5} = -1.47 \times 10^{-7}\ \text{J kg}^{-1}$

Hence, the gravitational field intensity is 2.40 × 10^-7 N kg^-1 and the potential is −1.47 × 10^-7 J kg^-1.

Given,

• Masses of the two bodies, M₁ = 100 kg and M₂ = 10000 kg
• Distance between them = 1 m

Let the gravitational field-intensity be zero at a point P on the line joining the two bodies, at a distance x from the smaller body of mass 100 kg. Its distance from the larger body is then (1 − x).

At this point the two field intensities are equal and opposite,

$\frac{\text{GM}_1}{\text x^2} = \frac{\text{GM}_2}{(1 - \text x)^2}$

$\left(\frac{1 - \text x}{\text x}\right)^2 = \frac{\text M_2}{\text M_1} = \frac{10000}{100} = 100$

Taking the square root,

$\frac{1 - \text x}{\text x} = 10 \quad \Rightarrow \quad 1 - \text x = 10\text x$

$11\text x = 1 \quad \Rightarrow \quad \text x = \frac{1}{11}\ \text m$

Hence, the gravitational field-intensity is zero at a point in between the two bodies, at a distance of $\frac{1}{11}$ m from the smaller body.

Given,

• Period of the moon around the earth, T_m = 27 days, radius of orbit r_m = 3.8 × 10⁵ km
• Period of the earth around the sun, T_e = 365 days, radius of orbit r_e = 1.5 × 10⁸ km

For a body revolving in a circular orbit of radius r with period T around a central mass M, the gravitational force provides the centripetal force, giving

$\text M = \frac{4\pi^2 \text r^3}{\text{GT}^2}$

Applying this to the earth (from the moon's motion) and to the sun (from the earth's motion),

$\text M_e = \frac{4\pi^2 \text r_m^3}{\text{GT}_m^2} \quad \text{and} \quad \text M_s = \frac{4\pi^2 \text r_e^3}{\text{GT}_e^2}$

Taking the ratio,

$\frac{\text M_s}{\text M_e} = \frac{\text r_e^3}{\text T_e^2} \times \frac{\text T_m^2}{\text r_m^3} = \left(\frac{\text r_e}{\text r_m}\right)^3 \times \left(\frac{\text T_m}{\text T_e}\right)^2$

Substituting the values,

$$\begin{aligned}\frac{\text M_s}{\text M_e} = \left(\frac{1.5 \times 10^8}{3.8 \times 10^5}\right)^3 \times \left(\frac{27}{365}\right)^2 \\[1em] = (394.7)^3 \times (0.07397)^2\end{aligned}$$

$\frac{\text M_s}{\text M_e} = (6.15 \times 10^7) \times (5.47 \times 10^{-3}) = 3.366 \times 10^5$

Hence, the mass of the sun is 3.366 × 10⁵ times the mass of the earth, that is, M_s : M_e = 3.366 × 10⁵ : 1.

Given,

• Height to be reached, h = 4R_e
• Radius of the earth, R_e = 6400 km = 6.4 × 10⁶ m
• g = 9.8 m s^-2

The velocity required to reach a height h, when h is not negligible compared with R_e, is given by

$\text v^2 = \frac{2\text{gh}}{1 + \frac{\text h}{\text R_e}}$

Substituting h = 4R_e,

$\text v^2 = \frac{2\text g(4\text R_e)}{1 + \frac{4\text R_e}{\text R_e}} = \frac{8\text{gR}_e}{5}$

Substituting the values,

$\text v^2 = \frac{8 \times 9.8 \times (6.4 \times 10^6)}{5} = \frac{5.018 \times 10^8}{5} = 1.0035 \times 10^8$

$\text v = \sqrt{1.0035 \times 10^8} = 1.0 \times 10^4\ \text{m s}^{-1} = 10\ \text{km s}^{-1}$

Hence, the body must be thrown with a velocity of 10 km s^-1.

Given,

• Velocity of projection, v = 10 km s^-1 = 1.0 × 10⁴ m s^-1
• Radius of the earth, R_e = 6400 km = 6.4 × 10⁶ m
• g = 10 m s^-2

The maximum height attained by a body projected upward with a velocity v, when the height is not negligible compared with R_e, is

$\text h = \frac{\text v^2 \text R_e}{2\text{gR}_e - \text v^2}$

Substituting the values,

$$\begin{aligned}\text h = \frac{(1.0 \times 10^4)^2 \times (6.4 \times 10^6)}{2 \times 10 \times (6.4 \times 10^6) - (1.0 \times 10^4)^2} \\[1em] = \frac{(1.0 \times 10^8) \times (6.4 \times 10^6)}{(1.28 \times 10^8) - (1.0 \times 10^8)}\end{aligned}$$

$\text h = \frac{6.4 \times 10^{14}}{2.8 \times 10^7} = 2.28 \times 10^7\ \text m = 2.28 \times 10^4\ \text{km}$

Hence, the body will go up to a height of 2.28 × 10⁴ km.

Given,

• Escape velocity, v_e = 11.2 km s^-1
• Velocity of projection, v_i = 2v_e

By the conservation of energy, if v_f is the speed of the body at infinity where the potential energy is zero,

$\frac{1}{2}\text{mv}_i^2 - \frac{\text{GM}_e \text m}{\text R_e} = \frac{1}{2}\text{mv}_f^2$

Since $\frac{1}{2}\text{mv}_e^2 = \frac{\text{GM}_e \text m}{\text R_e}$, this becomes

$\text v_i^2 - \text v_e^2 = \text v_f^2 \quad \Rightarrow \quad \text v_f = \sqrt{\text v_i^2 - \text v_e^2}$

Substituting v_i = 2v_e,

$$\begin{aligned}\text v_f = \sqrt{4\text v_e^2 - \text v_e^2} = \sqrt3\ \text v_e \\[1em] = 1.732 \times 11.2\end{aligned}$$

$\text v_f = 19.4\ \text{km s}^{-1}$

Hence, the speed of the body at infinity is 19.4 km s^-1.

Given,

• Orbital speed of the satellite, v_o = 8 km s^-1

For a satellite orbiting very close to the earth, the escape velocity and the orbital speed are related by

$\text v_e = \sqrt2\ \text v_o$

Substituting the given value,

$\text v_e = \sqrt2 \times 8 = 1.414 \times 8 = 11.3\ \text{km s}^{-1}$

Hence, the speed of the satellite should be increased to $\sqrt2 \times 8$ = 11.3 km s^-1 so that it may escape leaving its orbit.

Given,

• Mass of the planet, M_p = 8M_e
• Radius of the planet, R_p = 2R_e
• Acceleration due to gravity on the earth, g_e = 10 m s^-2

The acceleration due to gravity at the surface of a planet is

$\text g = \frac{\text{GM}}{\text R^2}$

Taking the ratio for the planet and the earth,

$\frac{\text g_p}{\text g_e} = \frac{\text M_p}{\text M_e} \times \left(\frac{\text R_e}{\text R_p}\right)^2$

Substituting the values,

$\frac{\text g_p}{\text g_e} = 8 \times \left(\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2$

$\text g_p = 2 \times 10 = 20\ \text{m s}^{-2}$

Hence, the acceleration due to gravity at the planet is 20 m s^-2.

Given,

• Distance between the centres, r = 1.0 m
• Gravitational force between the spheres, F = 1.0 N
• G = 6.67 × 10^-11 N m² kg^-2

The spheres are identical, so let the mass of each be m. By Newton's law of gravitation,

$\text F = \frac{\text{Gmm}}{\text r^2} = \frac{\text{Gm}^2}{\text r^2}$

Rearranging for m,

$\text m^2 = \frac{\text F\text r^2}{\text G} \quad \Rightarrow \quad \text m = \text r\sqrt{\frac{\text F}{\text G}}$

Substituting the values,

$\text m = 1.0 \times \sqrt{\frac{1.0}{6.67 \times 10^{-11}}} = \sqrt{1.499 \times 10^{10}}$

$\text m = 1.225 \times 10^5\ \text{kg}$

Hence, the mass of each sphere is 1.225 × 10⁵ kg.

Given,

• Masses of the two bodies, m₁ = 40 kg and m₂ = 80 kg
• Distance between them, r = 0.15 m
• Force of gravitation, F = 1.0 mg-wt
• g = 10 m s^-2

First the force is expressed in newton. Since 1 mg = 10^-6 kg,

$\text F = (1.0 \times 10^{-6}\ \text{kg}) \times 10\ \text{m s}^{-2} = 1.0 \times 10^{-5}\ \text N$

By Newton's law of gravitation,

$\text F = \text G\frac{\text m_1 \text m_2}{\text r^2} \quad \Rightarrow \quad \text G = \frac{\text F\text r^2}{\text m_1 \text m_2}$

Substituting the values,

$$\begin{aligned}\text G = \frac{(1.0 \times 10^{-5}) \times (0.15)^2}{40 \times 80} = \frac{(1.0 \times 10^{-5}) \times 0.0225}{3200} \\[1em] = \frac{2.25 \times 10^{-7}}{3200}\end{aligned}$$

$\text G = 7.0 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$

Hence, the constant of gravitation is 7.0 × 10^-11 N m² kg^-2.

Given,

• Mass of the moon, M = 7.34 × 10²² kg
• Radius of the moon, R = 1.75 × 10⁶ m
• G = 6.67 × 10^-11 N m² kg^-2

The acceleration due to gravity at the surface of the moon is

$\text g = \frac{\text{GM}}{\text R^2}$

Substituting the values,

$$\begin{aligned}\text g = \frac{(6.67 \times 10^{-11}) \times (7.34 \times 10^{22})}{(1.75 \times 10^6)^2} \\[1em] = \frac{4.896 \times 10^{12}}{3.0625 \times 10^{12}}\end{aligned}$$

$\text g = 1.6\ \text{m s}^{-2}$

Hence, the acceleration due to gravity at the moon is 1.6 m s^-2.

Given,

• Radius of the earth, R_e = 6.37 × 10⁶ m
• Acceleration due to gravity, g = 9.81 m s^-2
• G = 6.67 × 10^-11 N m² kg^-2

Mass of the earth : From the relation $\text g = \frac{\text{GM}_e}{\text R_e^2}$,

$\text M_e = \frac{\text{gR}_e^2}{\text G}$

Substituting the values,

$$\begin{aligned}\text M_e = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = \frac{9.81 \times (4.058 \times 10^{13})}{6.67 \times 10^{-11}} \\[1em] = \frac{3.981 \times 10^{14}}{6.67 \times 10^{-11}} = 6.0 \times 10^{24}\ \text{kg}\end{aligned}$$

Density of the earth : If ρ be the average density, then $\text M_e = \frac{4}{3}\pi \text R_e^3 \rho$. Substituting this in the relation above,

$\frac{4}{3}\pi \text R_e^3 \rho = \frac{\text{gR}_e^2}{\text G} \quad \Rightarrow \quad \rho = \frac{3\text g}{4\pi \text R_e \text G}$

Substituting the values,

$$\begin{aligned}\rho = \frac{3 \times 9.81}{4 \times 3.14 \times (6.37 \times 10^6) \times (6.67 \times 10^{-11})} \\[1em] = \frac{29.43}{5.337 \times 10^{-3}}\end{aligned}$$

$\rho = 5.5 \times 10^3\ \text{kg m}^{-3}$

Hence, the mass of the earth is 6.0 × 10²⁴ kg and its density is 5.5 × 10³ kg m^-3.

Given,

• Acceleration due to gravity on the earth, g_e = 10 m s^-2
• Mass of Mars, $\text M_M = \frac{\text M_e}{10}$
• Radius of Mars, $\text R_M = \frac{\text R_e}{2}$

The acceleration due to gravity at the surface of a planet is $\text g = \frac{\text{GM}}{\text R^2}$. Taking the ratio for Mars and the earth,

$\frac{\text g_M}{\text g_e} = \frac{\text M_M}{\text M_e} \times \left(\frac{\text R_e}{\text R_M}\right)^2$

Substituting the values,

$\frac{\text g_M}{\text g_e} = \frac{1}{10} \times (2)^2 = \frac{4}{10} = 0.4$

$\text g_M = 0.4 \times 10 = 4\ \text{m s}^{-2}$

Hence, the gravitational acceleration of a body on the surface of Mars is 4 m s^-2.

Given,

• Weight of the body on the earth, W_e = 100 kg-weight
• Mass of Mars, $\text M_M = \frac{\text M_e}{10}$
• Radius of Mars, $\text R_M = \frac{\text R_e}{2}$

The acceleration due to gravity at the surface of a planet is $\text g = \frac{\text{GM}}{\text R^2}$, so

$\frac{\text g_M}{\text g_e} = \frac{\text M_M}{\text M_e} \times \left(\frac{\text R_e}{\text R_M}\right)^2 = \frac{1}{10} \times (2)^2 = 0.4$

The mass of the body remains the same everywhere, and the weight is W = mg. Therefore

$\frac{\text W_M}{\text W_e} = \frac{\text g_M}{\text g_e} = 0.4$

$\text W_M = 0.4 \times 100 = 40\ \text{kg-weight}$

Hence, the weight of the body on Mars is 40 kg-weight.

Given,

• Height above the earth's surface, h = 3200 km
• Acceleration due to gravity at the earth's surface, g = 9.8 m s^-2
• Radius of the earth, R_e = 6400 km

The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Substituting the values,

$\text g' = \frac{9.8}{\left(1 + \frac{3200}{6400}\right)^2} = \frac{9.8}{\left(1 + \frac{1}{2}\right)^2} = \frac{9.8}{\left(\frac{3}{2}\right)^2}$

$\text g' = \frac{9.8}{2.25} = 4.35\ \text{m s}^{-2}$

Hence, the value of g at a height of 3200 km is 4.35 m s^-2.

Given,

• Radius of the earth, R_e = 6400 km

The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

(i) When $\text g' = \frac{\text g}{2}$ :

$\left(1 + \frac{\text h}{\text R_e}\right)^2 = 2 \quad \Rightarrow \quad 1 + \frac{\text h}{\text R_e} = \sqrt2 = 1.414$

$\text h = 0.414\ \text R_e = 0.414 \times 6400 = 2650\ \text{km}$

(ii) When $\text g' = \frac{\text g}{4}$ :

$\left(1 + \frac{\text h}{\text R_e}\right)^2 = 4 \quad \Rightarrow \quad 1 + \frac{\text h}{\text R_e} = 2$

$\text h = \text R_e = 6400\ \text{km}$

Hence, the acceleration due to gravity becomes half at a height of 2650 km and one-fourth at a height of 6400 km.

Given,

• Weight of the man on the earth's surface = W
• Height above the earth's surface, h = 3R_e

The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \frac{\text g}{\left(1 + \frac{\text h}{\text R_e}\right)^2}$

Substituting h = 3R_e,

$\text g' = \frac{\text g}{\left(1 + \frac{3\text R_e}{\text R_e}\right)^2} = \frac{\text g}{(1 + 3)^2} = \frac{\text g}{16}$

The mass m of the man remains unchanged, and his weight is W = mg. Hence his weight at that height is

$\text W' = \text{mg}' = \frac{\text{mg}}{16} = \frac{\text W}{16}$

Hence, his weight at that place will be $\frac{\text W}{16}$.

Given,

• Force acting on the body, F = 74 N
• Mass of the body, m = 3.7 kg

The intensity of the gravitational field at a point is the force experienced by a unit mass placed at that point,

$\text I = \frac{\text F}{\text m}$

Substituting the values,

$\text I = \frac{74}{3.7} = 20\ \text{N kg}^{-1}$

Hence, the intensity of the gravitational field is 20 N kg^-1.

Given,

• Distance from the centre of the earth, r = 10000 km = 1.0 × 10⁷ m
• Gravitational field intensity, I = 4.8 N kg^-1

The gravitational field intensity and the gravitational potential at a distance r from the centre of the earth are

$\text I = \frac{\text{GM}_e}{\text r^2} \quad \text{and} \quad \text V = -\frac{\text{GM}_e}{\text r}$

Dividing the second by the first,

$\frac{\text V}{\text I} = -\frac{\text{GM}_e}{\text r} \times \frac{\text r^2}{\text{GM}_e} = -\text r \quad \Rightarrow \quad \text V = -\text I\text r$

Substituting the values,

$\text V = -4.8 \times (1.0 \times 10^7)$

$\text V = -4.8 \times 10^7\ \text{J kg}^{-1}$

Hence, the gravitational potential at that point is −4.8 × 10⁷ J kg^-1.

Given,

• Radius of the earth, R_e = 6.4 × 10⁶ m
• Mean density of the earth, ρ = 5.5 × 10³ kg m^-3
• G = 6.67 × 10^-11 N m² kg^-2

The gravitational potential at the earth's surface is

$\text V = -\frac{\text{GM}_e}{\text R_e}$

The mass of the earth in terms of its density is

$\text M_e = \frac{4}{3}\pi \text R_e^3 \rho$

Substituting this value of M_e,

$\text V = -\frac{\text G}{\text R_e} \times \frac{4}{3}\pi \text R_e^3 \rho = -\frac{4}{3}\pi \text{GR}_e^2 \rho$

Substituting the values,

$$\begin{aligned}\text V = -\frac{4}{3} \times 3.14 \times (6.67 \times 10^{-11}) \times (6.4 \times 10^6)^2 \times (5.5 \times 10^3) \\[1em] = -4.187 \times (6.67 \times 10^{-11}) \times (4.096 \times 10^{13}) \times (5.5 \times 10^3)\end{aligned}$$

$\text V = -6.3 \times 10^7\ \text{J kg}^{-1}$

Hence, the gravitational potential at the earth's surface is −6.3 × 10⁷ J kg^-1.

Given,

• Mass of the earth, M_e = 6.0 × 10²⁴ kg
• Mass of the body, m = 33.5 kg
• Distance from the centre of the earth, r = 3.35 × 10¹⁰ m
• G = 6.7 × 10^-11 SI units

(ii) Gravitational potential : The gravitational potential at a distance r from the earth's centre is

$\text V = -\frac{\text{GM}_e}{\text r}$

Substituting the values,

$$\begin{aligned}\text V = -\frac{(6.7 \times 10^{-11}) \times (6.0 \times 10^{24})}{3.35 \times 10^{10}} = -\frac{4.02 \times 10^{14}}{3.35 \times 10^{10}} \\[1em] = -12 \times 10^3\ \text{J kg}^{-1}\end{aligned}$$

(i) Potential energy of the body : The gravitational potential energy of a body of mass m at that point is

$\text U = \text V \times \text m = (-12 \times 10^3) \times 33.5$

$\text U = -4.02 \times 10^5\ \text J$

Hence, the potential energy of the body is −4.02 × 10⁵ J and the gravitational potential at that distance is −12 × 10³ J kg^-1.

Given,

• Mass of the earth, M_e = 6.0 × 10²⁴ kg
• Radius of the earth, R_e = 6400 km = 6.4 × 10⁶ m
• Mass of the body, m = 200 kg
• Height above the earth's surface, h = 600 km = 0.6 × 10⁶ m
• G = 6.7 × 10^-11 SI units

The gravitational potential energy of a body of mass m at a height h above the earth's surface is

$\text U = -\frac{\text{GM}_e \text m}{\text R_e + \text h}$

Here,

$\text R_e + \text h = (6.4 \times 10^6) + (0.6 \times 10^6) = 7.0 \times 10^6\ \text m$

Substituting the values,

$$\begin{aligned}\text U = -\frac{(6.7 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 200}{7.0 \times 10^6} \\[1em] = -\frac{8.04 \times 10^{16}}{7.0 \times 10^6}\end{aligned}$$

$\text U = -1.15 \times 10^{10}\ \text J$

Hence, the potential energy of the body is −1.15 × 10¹⁰ J.

Given,

• Distance of the moon from the earth, r = 3.8 × 10⁵ km = 3.8 × 10⁸ m
• Mass of the earth, M_e = 6.1 × 10²⁴ kg
• G = 6.7 × 10^-11 N m² kg^-2

The moon revolves around the earth in a circular orbit, and the gravitational force of the earth provides the necessary centripetal force,

$\frac{\text{GM}_e \text m}{\text r^2} = \frac{\text{mv}^2}{\text r} \quad \Rightarrow \quad \text v = \sqrt{\frac{\text{GM}_e}{\text r}}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{\frac{(6.7 \times 10^{-11}) \times (6.1 \times 10^{24})}{3.8 \times 10^8}} = \sqrt{\frac{4.087 \times 10^{14}}{3.8 \times 10^8}} \\[1em] = \sqrt{1.0755 \times 10^6}\end{aligned}$$

$\text v = 1.03 \times 10^3\ \text{m s}^{-1} = 1.03\ \text{km s}^{-1}$

Hence, the speed of the moon revolving around the earth is 1.03 km s^-1.

Given,

• Period of revolution, T = 90 min = 90 × 60 = 5400 s
• Radius of the earth, R_e = 6370 km = 6.37 × 10⁶ m
• g = 9.8 m s^-2

The period of revolution of a satellite at a height h above the earth's surface is

$\text T = 2\pi\sqrt{\frac{(\text R_e + \text h)^3}{\text{gR}_e^2}}$

Squaring both sides and solving for (R_e + h),

$\text T^2 = \frac{4\pi^2(\text R_e + \text h)^3}{\text{gR}_e^2} \quad \Rightarrow \quad \text R_e + \text h = \left[\frac{\text T^2 \text{gR}_e^2}{4\pi^2}\right]^{1/3}$

Substituting the values,

$$\begin{aligned}\text R_e + \text h = \left[\frac{(5400)^2 \times 9.8 \times (6.37 \times 10^6)^2}{4 \times (3.14)^2}\right]^{1/3} \\[1em] = \left[\frac{(2.916 \times 10^7) \times (3.977 \times 10^{14})}{39.48}\right]^{1/3}\end{aligned}$$

$\text R_e + \text h = \left[2.937 \times 10^{20}\right]^{1/3} = 6.65 \times 10^6\ \text m$

Therefore,

$\text h = (6.65 \times 10^6) - (6.37 \times 10^6) = 0.28 \times 10^6\ \text m = 280\ \text{km}$

Hence, the value of h is 280 km.

Given,

• Mass of the satellite, m = 500 kg
• Height above the earth's surface, h = 10³ km = 1.0 × 10⁶ m
• Radius of the earth, R_e = 6.4 × 10⁶ m
• g = 9.80 m s^-2

The radius of the orbit is

$\text r = \text R_e + \text h = (6.4 \times 10^6) + (1.0 \times 10^6) = 7.4 \times 10^6\ \text m$

(i) Speed of the satellite : The orbital speed is

$$\begin{aligned}\text v_o = \text R_e\sqrt{\frac{\text g}{\text R_e + \text h}} = (6.4 \times 10^6) \times \sqrt{\frac{9.80}{7.4 \times 10^6}} \\[1em] = (6.4 \times 10^6) \times (1.151 \times 10^{-3})\end{aligned}$$

$\text v_o = 7.36 \times 10^3\ \text{m s}^{-1}$

(ii) Angular velocity of the satellite :

$\omega = \frac{\text v_o}{\text r} = \frac{7.36 \times 10^3}{7.4 \times 10^6}$

$\omega = 9.95 \times 10^{-4}\ \text{rad s}^{-1}$

(iii) Gravitational force on the satellite : This force provides the necessary centripetal force,

$\text F = \frac{\text{mv}_o^2}{\text r} = \frac{500 \times (7.36 \times 10^3)^2}{7.4 \times 10^6} = \frac{500 \times (5.417 \times 10^7)}{7.4 \times 10^6}$

$\text F = 3.66 \times 10^3\ \text N$

Hence, the speed of the satellite is 7.36 × 10³ m s^-1, its angular velocity is 9.95 × 10^-4 rad s^-1 and the gravitational force on it is 3.66 × 10³ N.

Given,

• Radius of the circular orbit, r = 8000 km = 8.0 × 10⁶ m
• Radius of the earth, R_e = 6400 km = 6.4 × 10⁶ m
• g = 9.8 m s^-2

Orbital speed : The speed with which the satellite must be projected in the orbit is

$\text v_o = \text R_e\sqrt{\frac{\text g}{\text r}}$

Substituting the values,

$$\begin{aligned}\text v_o = (6.4 \times 10^6) \times \sqrt{\frac{9.8}{8.0 \times 10^6}} = (6.4 \times 10^6) \times (1.107 \times 10^{-3}) \\[1em] = 7.08 \times 10^3\ \text{m s}^{-1} = 7.08\ \text{km s}^{-1}\end{aligned}$$

Period of revolution : The satellite covers the circumference of its orbit in one revolution,

$$\begin{aligned}\text T = \frac{2\pi \text r}{\text v_o} = \frac{2 \times 3.14 \times (8.0 \times 10^6)}{7.08 \times 10^3} \\[1em] = \frac{5.024 \times 10^7}{7.08 \times 10^3} = 7096\ \text s\end{aligned}$$

$\text T = \frac{7096}{60} = 118\ \text{min}$

Hence, the satellite should be projected with a speed of 7.08 km s^-1 and its period of revolution will be 118 min.

Given,

• Height of the satellite, h = 1800 km = 1.8 × 10⁶ m
• Radius of the earth, R_e = 6300 km = 6.3 × 10⁶ m
• g = 10 m s^-2

The radius of the orbit is

$\text r = \text R_e + \text h = (6.3 \times 10^6) + (1.8 \times 10^6) = 8.1 \times 10^6\ \text m$

(i) Orbital velocity :

$$\begin{aligned}\text v_o = \text R_e\sqrt{\frac{\text g}{\text R_e + \text h}} = (6.3 \times 10^6) \times \sqrt{\frac{10}{8.1 \times 10^6}} \\[1em] = (6.3 \times 10^6) \times (1.111 \times 10^{-3})\end{aligned}$$

$\text v_o = 7.0 \times 10^3\ \text{m s}^{-1} = 7\ \text{km s}^{-1}$

(ii) Radial acceleration : This is the centripetal acceleration of the satellite,

$\text a = \frac{\text v_o^2}{\text r} = \frac{(7.0 \times 10^3)^2}{8.1 \times 10^6} = \frac{4.9 \times 10^7}{8.1 \times 10^6}$

$\text a = 6\ \text{m s}^{-2}$

(iii) Period of revolution :

$\text T = \frac{2\pi \text r}{\text v_o} = \frac{2 \times 3.14 \times (8.1 \times 10^6)}{7.0 \times 10^3} = \frac{5.087 \times 10^7}{7.0 \times 10^3}$

$\text T = 7267\ \text s$

Hence, the orbital velocity is 7 km s^-1, the radial acceleration is 6 m s^-2 and the period of revolution is 7267 s.

Given,

• Distance between the centres of the earth and the moon, r = 60 R_e
• Acceleration due to gravity on the earth's surface, g = 10 m s^-2

The moon revolves around the earth, and the gravitational force of the earth on it provides the necessary centripetal force. The centripetal acceleration of the moon is therefore equal to the gravitational field intensity of the earth at the moon's distance,

$\text a = \frac{\text{GM}_e}{\text r^2}$

At the earth's surface,

$\text g = \frac{\text{GM}_e}{\text R_e^2}$

Dividing the first by the second,

$\frac{\text a}{\text g} = \left(\frac{\text R_e}{\text r}\right)^2 = \left(\frac{\text R_e}{60\text R_e}\right)^2 = \frac{1}{3600}$

$\text a = \frac{\text g}{3600} = \frac{10}{3600} = \frac{1}{360}\ \text{m s}^{-2}$

Hence, the centripetal acceleration of the moon is $\frac{1}{360}$ m s^-2.

Given,

• Height of the satellite, h = 500 km = 0.5 × 10⁶ m
• Period of revolution, T = 98 min = 98 × 60 = 5880 s
• Radius of the earth, R_e = 6.37 × 10⁶ m
• G = 6.67 × 10^-11 N m² kg^-2

The radius of the orbit is

$\text r = \text R_e + \text h = (6.37 \times 10^6) + (0.5 \times 10^6) = 6.87 \times 10^6\ \text m$

The period of revolution of a satellite in an orbit of radius r is

$\text T = 2\pi\sqrt{\frac{\text r^3}{\text{GM}_e}} \quad \Rightarrow \quad \text M_e = \frac{4\pi^2 \text r^3}{\text{GT}^2}$

Substituting the values,

$$\begin{aligned}\text M_e = \frac{4 \times (3.14)^2 \times (6.87 \times 10^6)^3}{(6.67 \times 10^{-11}) \times (5880)^2} \\[1em] = \frac{39.44 \times (3.243 \times 10^{20})}{(6.67 \times 10^{-11}) \times (3.457 \times 10^7)}\end{aligned}$$

$\text M_e = \frac{1.279 \times 10^{22}}{2.306 \times 10^{-3}} = 5.54 \times 10^{24}\ \text{kg}$

Hence, the mass of the earth is 5.54 × 10²⁴ kg.

Given,

• Period of revolution of the moon, T = 30 days = 30 × 24 × 60 × 60 = 2.592 × 10⁶ s
• Mass of the earth, M_e = 6 × 10²⁴ kg
• G = 6.67 × 10^-11 N m² kg^-2

The moon revolves around the earth in an orbit of radius r, so its period of revolution is

$\text T = 2\pi\sqrt{\frac{\text r^3}{\text{GM}_e}}$

Squaring both sides and solving for r,

$\text T^2 = \frac{4\pi^2 \text r^3}{\text{GM}_e} \quad \Rightarrow \quad \text r = \left[\frac{\text{GM}_e \text T^2}{4\pi^2}\right]^{1/3}$

Substituting the values,

$$\begin{aligned}\text r = \left[\frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times (2.592 \times 10^6)^2}{4 \times (3.14)^2}\right]^{1/3} \\[1em] = \left[\frac{(4.002 \times 10^{14}) \times (6.718 \times 10^{12})}{39.44}\right]^{1/3}\end{aligned}$$

$\text r = \left[\frac{2.688 \times 10^{27}}{39.44}\right]^{1/3} = \left[6.816 \times 10^{25}\right]^{1/3}$

$\text r = 4.08 \times 10^8\ \text m = 4.08 \times 10^5\ \text{km}$

Hence, the distance of the moon from the earth is 4.08 × 10⁵ km.

Given,

• Density of the planet, ρ = 8 × 10³ kg m^-3
• G = 6.67 × 10^-11 N m² kg^-2

For a satellite revolving near the surface of a planet, the height h is negligible compared with the radius R of the planet. The period of revolution is then

$\text T = 2\pi\sqrt{\frac{\text R^3}{\text{GM}}}$

The mass of the planet in terms of its density is

$\text M = \frac{4}{3}\pi \text R^3 \rho$

Substituting this value of M,

$\text T = 2\pi\sqrt{\frac{\text R^3}{\text G \times \frac{4}{3}\pi \text R^3 \rho}} = \sqrt{\frac{4\pi^2 \times 3}{4\pi \text G\rho}} = \sqrt{\frac{3\pi}{\text G\rho}}$

The radius of the planet cancels out. Substituting the values,

$$\begin{aligned}\text T = \sqrt{\frac{3 \times 3.14}{(6.67 \times 10^{-11}) \times (8 \times 10^3)}} = \sqrt{\frac{9.42}{5.336 \times 10^{-7}}} \\[1em] = \sqrt{1.765 \times 10^7}\end{aligned}$$

$\text T = 4202\ \text s$

Hence, the period of revolution of the satellite is 4202 s.

Given,

• Mass of the body, m = 100 kg
• g = 10 m s^-2
• Radius of the earth, R_e = 6.4 × 10⁶ m

The gravitational potential energy of the body on the earth's surface is

$\text U = -\frac{\text{GM}_e \text m}{\text R_e}$

Since GM_e = gR_e², this becomes

$\text U = -\frac{\text{gR}_e^2 \text m}{\text R_e} = -\text{mgR}_e$

To escape from the earth, the body must be taken to infinity, where its potential energy is zero. Hence the energy needed is

$\text E = 0 - \text U = \text{mgR}_e$

Substituting the values,

$\text E = 100 \times 10 \times (6.4 \times 10^6)$

$\text E = 6.4 \times 10^9\ \text J$

Hence, 6.4 × 10⁹ J of energy would be needed for the body to escape from the earth.

Given,

• Radius of the earth, R_e = 6.38 × 10⁶ m
• Acceleration due to gravity, g = 9.8 m s^-2

The escape velocity of a body from the earth's surface is

$\text v_e = \sqrt{2\text{gR}_e}$

Substituting the values,

$\text v_e = \sqrt{2 \times 9.8 \times (6.38 \times 10^6)} = \sqrt{1.250 \times 10^8}$

$\text v_e = 1.12 \times 10^4\ \text{m s}^{-1} = 11.2\ \text{km s}^{-1}$

Hence, the escape velocity of a body from the earth's surface is 11.2 km s^-1.

Given,

• Mass of the moon, M = 7.34 × 10²² kg
• Mean radius of the moon, R = 1.74 × 10⁶ m
• G = 6.67 × 10^-11 N m² kg^-2

The escape velocity from the surface of the moon is

$\text v_e = \sqrt{\frac{2\text{GM}}{\text R}}$

Substituting the values,

$$\begin{aligned}\text v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (7.34 \times 10^{22})}{1.74 \times 10^6}} \\[1em] = \sqrt{\frac{9.792 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.628 \times 10^6}\end{aligned}$$

$\text v_e = 2.37 \times 10^3\ \text{m s}^{-1} = 2.37\ \text{km s}^{-1}$

Hence, the escape velocity on the moon's surface is 2.37 km s^-1. This value is much smaller than that on the earth, which is why the moon cannot retain an atmosphere.