Class 11 Physics NCERT Solutions

~6 min read

Gravitation Class 11 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 7, Gravitation — 21 questions from 7.1 to 7.21, each worked through step by step in the CBSE marking pattern. Newton's law of gravitation, gravitational potential, escape speed and Kepler's laws.

Class:11Subject:PhysicsChapter:7

3 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Physics Chapter 7?

Chapter 7 carries 21 exercise questions, numbered 7.1 to 7.21. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents22 sections

Revision & Practice

Chapter Overview

Gravitation applies one universal law to falling apples and orbiting moons: Newton's law of universal gravitation, gravitational field and potential, elliptical orbits with Kepler's laws, satellites, and escape speed. Exams love to test why tides come from the nearer moon, where the net field is null between earth and sun, and the energy bookkeeping that carries a satellite to infinity. Every question below is from the NCERT Class 11 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

For orbit problems equate GMm/r² = mv²/r and express energy as E = −GMm/2r for a bound satellite. Use Kepler's third law as (T₂/T₁)² = (r₂/r₁)³. State why g decreases with altitude, with depth and is independent of the body's mass, and always justify which formula you choose for potential-energy differences.

$Ex 7.1 — Shielding gravity, detecting it in orbit, lunar tides$
$Ex 7.2 — g with altitude, depth, mass; formula accuracy$
$Ex 7.3 — Orbital size of a twice-as-fast planet$
$Ex 7.4 — Mass of Jupiter from Io’s orbit$
$Ex 7.5 — Revolution time of a star in the galaxy$
$Ex 7.6 — Energy of a bound orbiting satellite$
$Ex 7.7 — Does escape speed depend on mass, location, direction, height$
$Ex 7.8 — A comet in an elliptical orbit: which quantities are constant$
$Ex 7.9 — Symptoms afflicting an astronaut in space$
$Ex 7.10 — Gravitational intensity at the centre of a hemisphere$
$Ex 7.11 — Gravitational intensity at an arbitrary point$
$Ex 7.12 — Where the pull of earth and sun balance on a rocket$
$Ex 7.13 — How you would ’weigh the sun’$
$Ex 7.14 — Distance of Saturn from the sun$
$Ex 7.15 — Weight at a height of half the earth’s radius$
$Ex 7.16 — Weight half-way down to the centre of the earth$
$Ex 7.17 — How far a 5 km s⁻¹ rocket goes before returning$
$Ex 7.18 — Speed far away of a body launched at thrice escape speed$
$Ex 7.19 — Energy to rocket a satellite out of earth’s influence$
$Ex 7.20 — Speed of two stars as they collide$
$Ex 7.21 — Force, potential and equilibrium midway between two spheres$

NCERT Exercise 7.1 — Shielding Gravity, Detecting It in Orbit, Lunar Tides

1Exercise question

Step-by-step solution

1(a) No. Gravitational force is independent of the material medium and cannot be screened by a hollow sphere or any other means — unlike electric forces.
2(b) Yes, if the space station is large enough. The astronaut feels weightlessness only because the station falls freely as one body; over a large station the variation of g across its length can make the difference perceptible (tidal effect).
3(c) Tidal effect ∝ 1/r³ while the gravitational force ∝ 1/r². The moon is far nearer to the earth than the sun, so although the sun's pull is stronger, the moon's differential pull (tides) is larger.

Final answer

(a) No (b) yes if the station is large enough (c) tide varies as 1/r³, so the nearer moon dominates.

NCERT Exercise 7.2 — g With Altitude, Depth, Mass; Formula Accuracy

1Exercise question

Step-by-step solution

1(a) Decreases: g_h = g(1 − 2h/R) approximately, so g falls as altitude rises.
2(b) Decreases: inside a uniform sphere g_d = g(1 − d/R), falling linearly to zero at the centre.
3(c) Independent of mass of the body: g = GM/R² contains only the earth's mass M, not the falling body's mass.
4(d) More: the exact difference is ΔV = −GMm(1/r₂ − 1/r₁); mg(r₂ − r₁) assumes constant g, which is only an approximation for heights over which g changes.

Final answer

(a) decreases (b) decreases (c) independent of the mass of the body (d) more accurate.

NCERT Exercise 7.3 — Orbital Size of a Twice-as-Fast Planet

1Exercise question

Step-by-step solution

1T_p = T_e/2 (twice as fast = half the period); let the earth's period be T_e = 1 year and radius 1 AU.
2By Kepler's third law: (R_p/R_e)³ = (T_p/T_e)².
3R_p/R_e = (T_p/T_e)^(2/3) = (1/2)^(2/3) = 0.63.
4So the orbital size is 0.63 times the earth's — smaller by that factor.

Final answer

Planet's orbit is 0.63 times the earth's orbital size.

NCERT Exercise 7.4 — Mass of Jupiter From Io's Orbit

1Exercise question

Step-by-step solution

1For an orbiting satellite: M = 4π²r³/(GT²).
2M_J = 4π²R_Io³/(G T_Io²) with T_Io = 1.769 days and R_Io = 4.22 × 10⁸ m.
3M_Sun = 4π²R_e³/(G T_e²) with T_e = 365.25 days and R_e = 1 AU = 1.496 × 10¹¹ m.
4M_Sun/M_J = (R_e/R_Io)³ × (T_Io/T_e)².
5= [(1.496 × 10¹¹)/(4.22 × 10⁸)]³ × [(1.769)/(365.25)]² ≈ 1045.
6So M_Sun ≈ 1000 M_J — Jupiter is about one-thousandth of the sun's mass.

Final answer

M_Sun/M_J ≈ 1045 ≈ 1000, as required.

NCERT Exercise 7.5 — Revolution Time of a Star in the Galaxy

1Exercise question

Step-by-step solution

1Galactic mass M = 2.5 × 10¹¹ × 2 × 10³⁰ = 5 × 10⁴¹ kg.
2Orbital radius r = 50,000 ly and 1 ly = 9.46 × 10¹⁵ m, so r = 4.73 × 10²⁰ m.
3For a star orbiting the galactic centre: T = √(4π²r³/(GM)).
4T = √(4 × (3.14)² × (4.73)³ × 10⁶⁰/(6.67 × 10⁻¹¹ × 5 × 10⁴¹)) s.
5T ≈ 1.12 × 10¹⁶ s.
6Converting to years: T = 1.12 × 10¹⁶/(365 × 24 × 60 × 60) ≈ 3.55 × 10⁸ years.

Final answer

T ≈ 3.55 × 10⁸ years for one revolution.

NCERT Exercise 7.6 — Energy of a Bound Orbiting Satellite

1Exercise question

Step-by-step solution

1(a) Negative of its kinetic energy: for a circular orbit, V = −2K and E = K + V = −K.
2(b) Less: the orbiting satellite already carries kinetic energy appropriate to its orbit; only a little extra energy (taking it from E = −GMm/2r to 0) is needed, while a stationary object at the same height starts with no kinetic energy and needs more.

Final answer

(a) kinetic energy — total energy E = −K (b) less — the satellite already has orbital kinetic energy.

NCERT Exercise 7.7 — Does Escape Speed Depend on Mass, Location, Direction, Height

1Exercise question

Step-by-step solution

1Escape speed v_esc = √(2GM/R) = √(2gR) — the body's mass cancels out.
2(a) No: v_esc is independent of the mass of the body.
3(b) No: otherwise identical points at the same latitude give the same v_esc.
4(c) No: direction does not enter √(2GM/R).
5(d) Yes: the launch height changes R (the distance from the centre), and v_esc = √(2GM/R) decreases with increasing height.

Final answer

(a) no (b) no (c) no (d) yes — it depends only on the distance from the earth's centre.

NCERT Exercise 7.8 — A Comet in an Elliptical Orbit: Which Quantities Are Constant

1Exercise question

Step-by-step solution

1The gravitational force on the comet is central (always along the sun-line), so the torque about the sun is zero and angular momentum is conserved: (c) yes.
2No dissipative forces act, so the total mechanical energy is constant: (f) yes.
3As the comet swings near the sun, r falls, so its speed rises (equal areas in equal times) — linear speed varies: (a) no.
4Angular speed ω = L/mr² with varying r: (b) no.
5Kinetic energy ½mv² and potential energy −GMm/r both vary along the orbit: (d) no, (e) no.

Final answer

(a) no (b) no (c) yes (d) no (e) no (f) yes — only angular momentum and total energy are constant.

NCERT Exercise 7.9 — Symptoms Afflicting an Astronaut in Space

1Exercise question

Step-by-step solution

1(a) Not swollen feet: legs normally bear the body weight, so in weightlessness the fluid that pooled to counter gravity redistributes — swollen feet do not afflict an astronaut.
2(b) Swollen face: fluid shifts to the head region in weightlessness, causing facial swelling.
3(c) Headache: caused by the mental strain and fluid shift — a real affliction.
4(d) Orientational problem: space has no absolute up-down, so orientation is genuinely difficult.

Final answer

(b) swollen face, (c) headache, (d) orientational problem.

NCERT Exercise 7.10 — Gravitational Intensity at the Centre of a Hemisphere

1Exercise question

Step-by-step solution

1Inside a full spherical shell the intensity is zero because the pulls from opposite elements cancel.
2When the upper half of the shell is removed, that symmetry is broken — the pull of the remaining (lower) hemisphere is no longer cancelled.
3The net force at the centre O now points downward.
4Gravitational intensity = force per unit mass, so it too points downward: direction (iii) c.

Final answer

Option (iii): intensity at O points downward (arrow c).

NCERT Exercise 7.11 — Gravitational Intensity at an Arbitrary Point

1Exercise question

Step-by-step solution

1Inside a complete spherical shell the net intensity is zero at every interior point.
2Removing the upper half leaves the net pull of the lower hemisphere acting downward at any interior point P.
3Hence the intensity at P points downward: arrow (ii) e.

Final answer

Option (ii): intensity at P points downward (arrow e).

NCERT Exercise 7.12 — Where the Pull of Earth and Sun Balance on a Rocket

1Exercise question

Step-by-step solution

1Let x be the distance of the balance point from the earth's centre; the distance from the sun is then r − x with r = 1.5 × 10¹¹ m.
2Equate forces: GMₛm/(r − x)² = GMₑm/x².
3(r − x)/x = √(Mₛ/Mₑ) = √(2 × 10³⁰/6 × 10²⁴) = √(3.33 × 10⁵) ≈ 577.4.
4r − x = 577.4x ⇒ 1.5 × 10¹¹ = 578.4x.
5x = 1.5 × 10¹¹/578.4 ≈ 2.59 × 10⁸ m from the earth's centre.

Final answer

x ≈ 2.59 × 10⁸ m from the centre of the earth.

NCERT Exercise 7.13 — How You Would 'Weigh the Sun'

1Exercise question

Step-by-step solution

1For the earth orbiting the sun, equate the gravitational pull to the required centripetal force: GMₛm/r² = m(4π²r/T²).
2Mₛ = 4π²r³/(GT²).
3r = 1.5 × 10¹¹ m, T = 365.25 × 24 × 60 × 60 s, G = 6.67 × 10⁻¹¹ N m² kg⁻².
4Mₛ = 4 × (3.14)² × (1.5 × 10¹¹)³/[6.67 × 10⁻¹¹ × (3.156 × 10⁷)²].
5Mₛ ≈ 2 × 10³⁰ kg — the known solar mass.

Final answer

Use M = 4π²r³/(GT²) with the earth's period and orbital radius: Mₛ ≈ 2.0 × 10³⁰ kg.

NCERT Exercise 7.14 — Distance of Saturn From the Sun

1Exercise question

Step-by-step solution

1By Kepler's third law: rₛ³/rₑ³ = Tₛ²/Tₑ².
2rₛ = rₑ(Tₛ/Tₑ)^(2/3) = 1.5 × 10¹¹ × (29.5)^(2/3).
3(29.5)^(2/3) ≈ 9.55.
4rₛ = 1.5 × 10¹¹ × 9.55 = 14.32 × 10¹¹ m = 1.43 × 10¹² m.

Final answer

Saturn is about 1.43 × 10¹² m from the sun.

NCERT Exercise 7.15 — Weight at a Height of Half the Earth's Radius

1Exercise question

Step-by-step solution

1At height h above the surface, g′ = g/(1 + h/R)².
2h = R/2 gives g′ = g/(1.5)² = (4/9)g.
3Weight is proportional to g: W′ = (4/9)W = (4/9) × 63.
4W′ = 28 N.

Final answer

Gravitational force on the body = 28 N.

NCERT Exercise 7.16 — Weight Half-Way Down to the Centre of the Earth

1Exercise question

Step-by-step solution

1At depth d inside a uniform sphere: g_d = g(1 − d/R).
2d = R/2 gives g_d = g(1 − ½) = g/2.
3Weight W′ = m g_d = (1/2) m g = W/2 = 250/2.
4W′ = 125 N.

Final answer

The body would weigh 125 N.

NCERT Exercise 7.17 — How Far a 5 km s⁻¹ Rocket Goes Before Returning

1Exercise question

Step-by-step solution

1By conservation of energy between the surface and the highest point (v = 0 there):
2½mv² − GMm/R = −GMm/(R + h).
3Simplify: ½v² = GM[1/R − 1/(R + h)] = GMh/[R(R + h)].
4v² = 2gRh/(R + h) with g = GM/R².
5h = Rv²/(2gR − v²) = 6.4 × 10⁶ × (5 × 10³)²/[2 × 9.8 × 6.4 × 10⁶ − (5 × 10³)²].
6h ≈ 1.6 × 10⁶ m.
7Distance from the earth's centre = R + h = 6.4 × 10⁶ + 1.6 × 10⁶ = 8.0 × 10⁶ m.

Final answer

The rocket reaches 8.0 × 10⁶ m from the earth's centre (about 1.6 × 10⁶ m above the surface).

NCERT Exercise 7.18 — Speed Far Away of a Body Launched at Thrice Escape Speed

1Exercise question

Step-by-step solution

1Escape speed v_esc = 11.2 km s⁻¹; projection speed v_p = 3v_esc.
2Launch energy = ½mv_p² − GMm/R. But ½mv_esc² = GMm/R.
3Conservation: ½mv_p² − ½mv_esc² = ½mv_f² (at infinity, potential energy = 0).
4v_f = √(v_p² − v_esc²) = √(9v_esc² − v_esc²) = √8 × v_esc.
5v_f = 2.828 × 11.2 ≈ 31.68 km s⁻¹.

Final answer

v_f ≈ 31.68 km s⁻¹ far away from the earth.

NCERT Exercise 7.19 — Energy to Rocket a Satellite Out of Earth's Influence

1Exercise question

Step-by-step solution

1At height h the radius of the orbit is r = R + h = 6.8 × 10⁶ m, and the orbital speed is v = √(GM/r).
2Total energy of the satellite in orbit: E = ½mv² − GMm/r = −GMm/2r (the negative binding energy).
3Energy required to escape = 0 − E = GMm/2r.
4= (1/2) × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200/(6.8 × 10⁶).
5≈ 5.9 × 10⁹ J.

Final answer

Energy needed ≈ 5.9 × 10⁹ J.

NCERT Exercise 7.20 — Speed of Two Stars as They Collide

1Exercise question

Step-by-step solution

1Initial separation r = 10⁹ km = 10¹² m; speeds negligible, so total energy = −GM²/r.
2Just before collision, separation between centres = 2R = 2 × 10⁷ m, and each star has speed v.
3Total kinetic energy = 2 × ½Mv² = Mv²; total potential energy = −GM²/2R.
4Energy conservation: Mv² − GM²/2R = −GM²/r.
5v² = GM(1/2R − 1/r) = 6.67 × 10⁻¹¹ × 2 × 10³⁰ × [1/(2 × 10⁷) − 1/10¹²].
6v² ≈ 13.34 × 10¹⁹ × 5 × 10⁻⁸ = 6.67 × 10¹².
7v ≈ 2.58 × 10⁶ m s⁻¹.

Final answer

Collision speed ≈ 2.58 × 10⁶ m s⁻¹.

NCERT Exercise 7.21 — Force, Potential and Equilibrium Midway Between Two Spheres

1Exercise question

Step-by-step solution

1At the midpoint X each sphere is at distance r/2 = 0.5 m, pulling the object in opposite directions — the net force is zero.
2Potential at X = −GM/(r/2) − GM/(r/2) = −4GM/r.
3V = −4 × 6.67 × 10⁻¹¹ × 100/1 = −2.67 × 10⁻⁸ J kg⁻¹.
4The object is in equilibrium (net force zero).
5But shifting it slightly towards one sphere increases the pull from that side — it is flung there. Hence the equilibrium is unstable.

Final answer

Force = 0; potential = −2.67 × 10⁻⁸ J kg⁻¹; the object is in equilibrium, but it is unstable.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Newton's law of gravitation

F = \frac{G m _{1} m _{2}}{r ^{2}}

Escape speed

v_{e} = \frac{2 GM}{R}

Orbital speed

v_{or b} = \frac{GM}{r} = g R

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

v_e = √(2GM/R) = √(2gR) ≈ 11.2 km s⁻¹ for the Earth, and it does not depend on the mass of the escaping body.
Escape speed and orbital speed differ by exactly √2: v_e = √2 · v_orb.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Physics Chapter 7 (Gravitation)?

There are 21 exercise questions in this chapter, numbered 7.1 to 7.21. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Gravitation Class 11 Physics?

The formulas this chapter's questions actually turn on are: Newton's law of gravitation, Escape speed, Orbital speed. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Gravitation important for JEE Main and NEET?

Very high weightage — gravitation is a favourite unit in NEET and JEE Main, and escape-speed and orbital-speed questions are close to guaranteed.

Next Chapters

Continue learning

Previous: Systems of Particles and Rotational Motion Next: Mechanical Properties of Solids Gravitation revision notes All Class 11 Physics NCERT Solutions

Interactive Quiz

Chapter MCQ practice test

Instant scoring with complete solutions — test your mastery in under 15 minutes.

Active Recall Practice

Chapter MCQ Mock Test

Evaluate how well you have retained the concepts, formulas, and reaction mechanisms from this chapter. Questions adhere strictly to latest CBSE, JEE & NEET trends.

15 questions (of 25)~23 minutesInstant Score & Solutions

Same solutions, live doubt-clearing help

Reading a solution is step one — getting a doubt resolved in real time is what clears it. ClassApna runs small-batch CBSE, JEE & NEET coaching with daily doubt sessions and mock tests.

Book a Free Demo Class Explore Plans & Batches

Small batches · 1-on-1 personal mentorship · Live online & offline centre