Class 11 Physics Ch 7 Gravitation NCERT Solutions – CBSE Answers

Q.8.1 Answer the following:

(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?

(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull

(You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?

Ans.8.1

(a) No. Unlike electrical forces, gravitational force is independent of the status of the objects.

(b) Yes, the size of the space station is large enough and the astronaut will detect the change in Earth’s gravity.

(c) Tidal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

Q.8.2 Choose the correct alternative:

(a) Acceleration due to gravity increases/decreases with increasing altitude

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density)

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body

(d) The formula –G Mm(1/r₂ – 1/r₁) is more/less accurate than the formula mg(r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth

Ans.8.2

(a) Decreases - Acceleration due to gravity at depth h is given by $g_h$ = (1 –  $\frac{2h}{R_e}$ )g, where  $R_e=Radiusoftheearth,$ g = acceleration due to gravity on the surface of the Earth. From this equation, it is clear that acceleration due to gravity decreases with increase in height

(b) Decreases – Acceleration due to gravity at depth d is given by $g_d$ = (1-  $\frac{d}{R_e}$ )g. So the acceleration due to gravity decreases with increase in depth.

(c) Mass of the body – Acceleration due to gravity of body mass m is given by the relation g = $\frac{GM}{R^2}$ , where G = Universal gravitation constant, M = mass of the Earth and R = radius of the Earth. Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More – Gravitational potential energy of two points $r_1$ and  $r_2$ distance away from the centre of the Earth is respectively given by:

V(  $r_1$ ) =  $-$ -  $\frac{GmM}{r_1}$ and V(  $r_2$ ) =  $-$ -  $\frac{GmM}{r_2}$

Difference in potential energy, V = V(  $r_2$ )  $-$ V(  $r_1$ ) =  $-GmM(\frac{1}{r_2}$  $-\frac{1}{r_1}$ )

Hence this formula is more accurate than mg(  $r_2$ -  $r_1$ )

Ans.8.3

Time taken by the Earth to complete one revolution around the Sun,  $T_e$ = 1 year

Orbital radius of the Earth in its orbit,  $R_e$ = 1 AU

Time taken by the planet to complete one revolution around the Sun,  $T_p$ =  $\frac{1}{2}{T}_e$ =  $\frac{1}{2}$ year

Orbital radius of the planet =  $R_p$

From Kepler’s 3rd law of planetary motion, we can write:

(  ${\frac{R_p}{R_e})}^3$ = (  ${\frac{T_p}{T_e})}^2$

$\frac{R_p}{R_e}$ = (  ${\frac{T_p}{T_e})}^{2/3}$ =(  ${\frac{1}{2})}^{2/3}$ = 0.63

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth

Ans.8.4

The rotation period of the satellite Io,  $T_{ju}$ = 1.769 days = 1.769 x 24 x 60 x 60s

Radius of the Orbit is given by the relation,  $R_{ju}$ = 4.22  $\times {10}^8$ m

Mass is given by the relation:  $M_j$ =  $\frac{4{\pi }^2{R}_{ju}^3}{G{T}_{ju}^2}$ ……(i)

Where  $M_j$ = mass of Jupiter and G = Universal gravitational constant

Orbital period of the Earth,  $T_e$ = 365.25 days = 365.25 x 24 x 60 x 60 s

Orbital radius of the Earth,  $R_e$ = 1 AU = 1.496 x  ${10}^{11}$ m

Mass of Sun is given as :  $M_s$ =  $\frac{4{\pi }^2{R}_e^3}{G{T}_e^2}$ …(ii)

$\frac{M_s}{M_j}$ = $\frac{4{\pi }^2{R}_e^3}{G{T}_e^2}$  $\times$  $\frac{G{T}_{ju}^2}{4{\pi }^2{R}_{ju}^3}$ = $\frac{R_e^3}{T_e^2}$  $\times$  $\frac{T_{ju}^2}{R_{ju}^3}$= ( ${\frac{1.496\mathrm{}\mathrm{x}\mathrm{}{10}^{11}}{4.22\mathrm{}\times {10}^8\mathrm{}})}^3$  $\times ({\frac{1.769\mathrm{}\mathrm{x}\mathrm{}24\mathrm{}\mathrm{x}\mathrm{}60\mathrm{}\mathrm{x}\mathrm{}60}{365.25\mathrm{}\mathrm{x}\mathrm{}24\mathrm{}\mathrm{x}\mathrm{}60\mathrm{}\mathrm{x}\mathrm{}60\mathrm{}})}^2$ = 1045.04

Hence it can be inferred that the mass of Jupiter is about one – thousandth that of the Sun