Two planets A and B of equal mass are having their period of revolutions T A and T B such that T A = 2T B . These planets are revolving in the circular orbits or radii r A and r B respectively. Which out of the following would be the correct relationship of their orbits?

T A 2 − T B 2 = π 2 G M ( r B 3 − 4 r A 3 )

Class 11 Physics Ch 7 Gravitation NCERT Solutions – CBSE Answers

Q: 8.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.

Velocity of the rocket, v = 5 km/s = 5 ×103 m/s Mass of the Earth, Me = 6.0 × 1024 kg Radius of the Earth, Re = 6.4 × 106 m Height reached by rocket mass, m = h At the Earth’s surface Total energy of the rocket = Kinetic energy + Potential energy = 12 m v2 + ( -GmMeRe ) At highest point h, v = 0, and potential energy = -GmMeRe+h Total energy of the rocket at height h = -GmMeRe+h From the law of conservation of energy, we have, Total energy of the rocket at Earth surface = Total energy at height h 12 m v2 + ( -GmMeRe ) = -GmMeRe+h or 12v2 = GMe(1Re-1Re+h) 12v2 = GMe(1Re-1Re+h) = GMehRe(Re+h) = GMeRe2×Reh(Re+h) = gReh(Re+h) where g = GMeRe2 = 9.8 ms2 v2(Re+h) = 2 gReh v2Re+v2h = 2 gReh h(2 gRe-v2) = v2Re h = v2Re(2gRe-v2) = 1.59 x 106 m

Q: 8.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

8.12 Mass of the Sun, Ms = 2×1030 kg, Mass of the Earth, Me = 6×1024 kg Orbital radius, r = 1.5×1011 m Let the mass of the rocket be, m Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero. From Newton’s law of gravitation, we can equate gravitational forces acting on the satellite P under the influence of the Sun and the Earth as: GmMs(r-x)2 = GmMex2 or( r-xx)2 = MsMe( r-xx)2 = 2×10306×1024 , r-xx = 577.35 , r = 578.35x , x = 1.5×1011578.35 = 2.59 ×108 m

Q: 8.14 A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 × 108 km away from the Sun?

Distance of the Earth from the Sun, re = 1.50 × 108 km = 1.5 × 1011 m Time period of the Earth = Te , Time period of the Saturn Ts = 29.5Te Distance of Saturn from the Sun = rs From Kepler’s 3rd law of planetary motion, we have T =( 4π2r3GM)1/2 For Saturn and Sun, we can write, rs3re3 = Ts2Te3 rs = re(29.5TeTe)2/3 = 1.5 × 1011 ×(29.51)2/3 = 1.4 X 1012 m

Q: 8.7 Does the escape speed of a body from the earth depend on (a) The mass of the body (b) The location from where it is projected (c) The direction of projection (d) The height of the location from where the body is launched?

(a) Escape velocity of a body from the Earth is given by the relation: vesc = 2gR …. (1) Where g = acceleration due to gravity and R = radius of the Earth So escape velocity is independent of the mass of the body. (b) It does not depend on the location from where it is projected. (c) Does not depend on the direction of projection (d) Depends on the height of the location from where the body is launched.

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

(a) Decreases - Acceleration due to gravity at depth h is given by gh = (1 – 2hRe )g, where Re=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 gd = (1- dRe )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 = GMR2 , 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 r1 and r2 distance away from the centre of the Earth is respectively given by: V( r1 ) = - - GmMr1 and V( r2 ) = - - GmMr2 Difference in potential energy, V = V( r2 ) - V( r1 ) = -GmM(1r2 -1r1 ) Hence this formula is more accurate than mg( r2 - r1 )

Updated on Jul 11, 2025 17:45 IST

By Pallavi Pathak, Assistant Manager Content

Ch 7 Gravitation Physics NCERT Solutions introduce the universal force that governs the motion of celestial and terrestrial bodies. These NCERT solutions are created by the experts at Shiksha. It provides detailed and accurate answers to all the NCERT textbook questions. NCERT Solutions for Class 11 Physics Chapter 7 covers the following key topics:

Gravitational potential energy
Newton’s law of gravitation
Acceleration due to gravity
Satellite motion

The solutions are according to the CBSE curriculum and hence provide exam-oriented preparation material. By practicing these solutions, the students can improve their problem-solving skills and score high in the CBSE Board exam and other entrance exams such as NEET and JEE Main.

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation: Key Concepts, Weightage

Class 11 Physics Chapter 7 Gravitation is an important chapter for various exams, including CBSE Board, NEET, JEE Main, and other entrance exams. The following are the topics covered in this chapter:

Exercise	Topics Covered
7.1	Introduction
7.2	Kepler's Laws
7.3	Universal Law of Gravitation
7.4	The Gravitational Constant
7.5	Acceleration Due to Gravity of the Earth
7.6	Acceleration Due to Gravity Below and Above the Surface of Earth
7.7	Gravitational Potential Energy
7.8	Escape Speed
7.9	Earth Satellites
7.10	Energy of an Orbiting Satellite

Gravitation Weightage for NEET, JEE Main Exams

Exam	Number of Questions	Weightage
NEET	2-3 questions	3-5%
JEE Main	1 question	3.33%

NCERT Physics Class11 th Solution PDF For Gravitation

Students must download the NCERT Class 11 Physics Solutions PDF for Chapter 7 – Gravitation Free PDF from here to get accurate and step-by-step answers to all textbook questions. The PDF helps students to deepen their understanding of key concepts like satellite motion, gravitational force, escape speed, and Kepler’s laws.

To get the chapters-wise NCERT notes, important topics, solved examples, and weightage information, the students must read here - Class 11 Physics Notes.

Download Here: NCERT Solution for Class XI Physics Chapter Gravitation PDF

NCERT Physics Class11th Gravitation Solutions

Find here the NCERT solutions of the Class 11 Physics Chapter 7 Gravitation

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 – $\frac{2 h}{R_{e}}$ )g, where $R_{e} = R a d i u s o f t h e e a r t h,$ 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 = $\frac{G M}{R^{2}}$ , 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( $r_{1}$ ) = $-$ - $\frac{G m M}{r_{1}}$ and V( $r_{2}$ ) = $-$ - $\frac{G m M}{r_{2}}$

Difference in potential energy, V = V( $r_{2}$ ) $-$ V( $r_{1}$ ) = $- G m M (\frac{1}{r_{2}}$ $- \frac{1}{r_{1}}$ )

Hence this formula is more accurate than mg( $r_{2}$ - $r_{1}$ )

Q.8.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

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

Q.8.4 IO, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Ans.8.4

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

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

Mass is given by the relation: $M_{j}$ = $\frac{4 π^{2} R_{j u}^{3}}{G T_{j u}^{2}}$ ……(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 : $M_{s}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ …(ii)

$\frac{M_{s}}{M_{j}}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ $\times$ $\frac{G T_{j u}^{2}}{4 π^{2} R_{j u}^{3}}$ = $\frac{R_{e}^{3}}{T_{e}^{2}}$ $\times$ $\frac{T_{j u}^{2}}{R_{j u}^{3}}$ = ( ${\frac{1.496 x 10^{11}}{4.22 \times 10^{8}})}^{3}$ $\times ({\frac{1.769 x 24 x 60 x 60}{365.25 x 24 x 60 x 60})}^{2}$ = 1045.04

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

Commonly asked questions

8.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^–11 N m2 kg^–².

Velocity of the rocket, v = 5 km/s = 5 $\times 10^{3}$ m/s

Mass of the Earth, $M_{e}$ = 6.0 × 10²⁴kg

Radius of the Earth, $R_{e}$ = 6.4 × 10⁶ m

Height reached by rocket mass, m = h

At the Earth’s surface

Total energy of the rocket = Kinetic energy + Potential energy = $\frac{1}{2}$ m $v^{2}$ + ( $\frac{- G {m M}_{e}}{R_{e}}$ )

At highest point h, v = 0, and potential energy = $\frac{- G {m M}_{e}}{R_{e} + h}$

Total energy of the rocket at height h = $\frac{- G {m M}_{e}}{R_{e} + h}$

From the law of conservation of energy, we have,

Total energy of the rocket at Earth surface = Total energy at height h

$\frac{1}{2}$ m $v^{2}$ + ( $\frac{- G {m M}_{e}}{R_{e}}$ ) = $\frac{- G {m M}_{e}}{R_{e} + h}$ or $\frac{1}{2} v^{2}$ = $G M_{e} (\frac{1}{R_{e}} - \frac{1}{R_{e} + h})$

$\frac{1}{2} v^{2}$ = $G M_{e} (\frac{1}{R_{e}} - \frac{1}{R_{e} + h})$ = $\frac{G M_{e} h}{R_{e} (R_{e} + h)}$ = $\frac{G M_{e}}{R_{e}^{2}} \times \frac{R_{e} h}{(R_{e} + h)}$ = $\frac{{g R}_{e} h}{(R_{e} + h)}$ where g = $\frac{G M_{e}}{R_{e}^{2}}$ = 9.8 $\frac{m}{s^{2}}$

$v^{2} (R_{e} + h)$ = 2 ${g R}_{e} h$ $v^{2} R_{e} + v^{2} h$ = 2 ${g R}_{e} h$ h(2 ${g R}_{e} - v^{2})$ = $v^{2} R_{e}$

h = $\frac{v^{2} R_{e}}{(2 {g R}_{e} - v^{2})}$ = 1.59 x $10^{6}$ m

8.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2×10³⁰ kg, mass of the earth = 6×10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 10¹¹ m).

8.12

Mass of the Sun, $M_{s}$ = $2 \times 10^{30}$ kg, Mass of the Earth, $M_{e}$ = $6 \times 10^{24}$ kg

Orbital radius, r = $1.5 \times 10^{11}$ m

Let the mass of the rocket be, m

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

From Newton’s law of gravitation, we can equate gravitational forces acting on the satellite P under the influence of the Sun and the Earth as:

$\frac{G_{m} M_{s}}{({r - x)}^{2}}$ = $G_{m} \frac{M_{e}}{x^{2}}$ or( ${\frac{r - x}{x})}^{2}$

= $\frac{M_{s}}{M_{e}}$ ( ${\frac{r - x}{x})}^{2}$ = $\frac{{2 \times 10}^{30}}{{6 \times 10}^{24}}$ ,

$\frac{r - x}{x}$ = 577.35 , r = 578.35x ,

x = $\frac{1.5 \times 10^{11}}{578.35}$ = 2.59 $\times 10^{8}$ m

8.14 A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 × 10⁸ km away from the Sun?

Distance of the Earth from the Sun, $r_{e}$ = 1.50 × 10⁸ km = 1.5 × 10¹¹m

Time period of the Earth = $T_{e}$ , Time period of the Saturn $T_{s}$ = ${29.5 T}_{e}$

Distance of Saturn from the Sun = $r_{s}$

From Kepler’s 3rd law of planetary motion, we have T =( ${\frac{4 π^{2} r^{3}}{G M})}^{1 / 2}$

For Saturn and Sun, we can write, $\frac{r_{s}^{3}}{r_{e}^{3}}$ = $\frac{T_{s}^{2}}{T_{e}^{3}}$

$r_{s}$ = $r_{e} ({\frac{29.5 T_{e}}{T_{e}})}^{2 / 3}$ = 1.5 × 10¹¹ $\times ({\frac{29.5}{1})}^{2 / 3}$ = 1.4 X $10^{12}$ m

8.7 Does the escape speed of a body from the earth depend on

(a) The mass of the body

(b) The location from where it is projected

(c) The direction of projection

(d) The height of the location from where the body is launched?

(a) Escape velocity of a body from the Earth is given by the relation:

$v_{e s c}$ = $\sqrt{2 g R}$ …. (1)

Where g = acceleration due to gravity and R = radius of the Earth

So escape velocity is independent of the mass of the body.

(b) It does not depend on the location from where it is projected.

(c) Does not depend on the direction of projection

(d) Depends on the height of the location from where the body is launched.

8.2 Choose the correct alternative: