NCERT Solutions for Class 11 Physics Chapter 7: Gravitation

These Class 11 Physics Chapter 7 solutions cover Gravitation with every NCERT exercise (7.1–7.21) reproduced verbatim and solved step by step, with units shown and final answers cross-checked. The chapter builds from Kepler’s laws to Newton’s universal law of gravitation, the variation of g with height and depth, gravitational potential energy, escape speed, and the energy of Earth satellites — all updated for session 2026–27.

Class: 11 Subject: Physics Chapter: 7 Chapter Name: Gravitation Exercises: 7.1 – 7.21 Session: 2026–27
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Class 11 Physics Chapter 7 Gravitation – Overview

Chapter 7, Gravitation, explains the force that holds the Solar System together and keeps us bound to the Earth. It begins with the historical journey from Ptolemy’s geocentric model to the heliocentric model and Kepler’s three laws (orbits, areas, periods). Newton turned these laws into the universal law of gravitation — every mass attracts every other with a force proportional to the product of the masses and inversely proportional to the square of their separation. The chapter then derives how the acceleration due to gravity g changes with height and depth, defines gravitational potential energy for a point mass, obtains the escape speed (about 11.2 km s−1 for Earth), and analyses the energy of an orbiting satellite, which is always negative for a bound orbit. These ideas let us “weigh” the Earth and the Sun and explain tides, weightlessness in orbit and why the Moon has no atmosphere.

Key Concepts & Definitions

Kepler’s laws: (1) every planet moves in an ellipse with the Sun at one focus; (2) the line from the Sun to a planet sweeps equal areas in equal times (a consequence of conservation of angular momentum); (3) the square of the orbital period is proportional to the cube of the semi-major axis (T2 ∝ a3).

Universal law of gravitation: every body attracts every other body with a force F = Gm1m2/r2, directed along the line joining the centres.

Gravitational constant G: a universal constant, G = 6.67 × 10−11 N m2 kg−2, first measured by Cavendish.

Acceleration due to gravity (g): the acceleration of a freely falling body near Earth’s surface, g = GME/RE2 ≈ 9.8 m s−2; it decreases both above and below the surface and is maximum at the surface.

Gravitational potential energy: for two masses a distance r apart, V = −Gm1m2/r (taken zero at infinity).

Escape speed: the minimum speed needed for a body to escape a planet’s gravity, ve = √(2GM/R) = √(2gR).

Energy of an orbiting satellite: total energy E = −GmME/2(RE+h) is negative; the kinetic energy is positive and half the magnitude of the (negative) potential energy.

Important Formulas

Newton’s law: F = G·m1m2 / r2

Acceleration due to gravity (surface): g = GME / RE2

At height h (h « RE): g(h) = g(1 − 2h/RE);  exactly, g(h) = GME/(RE+h)2

At depth d: g(d) = g(1 − d/RE)

Kepler’s third law: T2 = (4π2/GM)·R3

Gravitational potential energy: V = −Gm1m2 / r

Orbital speed: V = √(GME / (RE+h))

Escape speed: ve = √(2GME / RE) = √(2gRE) ≈ 11.2 km s−1

Total energy of satellite (circular orbit): E = −Gm ME / 2(RE+h)

NCERT Solutions for Class 11 Physics Chapter 7 – Exercises

All questions below are reproduced verbatim from the NCERT textbook. Answers are original and expert-checked; numerical results are verified against NCERT.

7.1 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?

ANSWER (a) No. Gravitational shielding is impossible. Unlike electric forces, there is no negative gravitational mass to cancel the field, so a hollow sphere cannot block the gravitational influence of matter outside it. (The field inside a uniform shell is zero, but the shell does not screen external bodies.) (b) Yes, if the space station is large enough. In a large station the value of g varies measurably from one part to another (tidal variation), so the astronaut could detect gravity by sensing this difference; in a small ship the variation is too tiny to notice. (c) The tidal effect depends on the inverse cube of the distance, whereas gravitational force depends on the inverse square. Although the Sun’s force is larger, the Moon is far closer; because of the steeper inverse-cube dependence the Moon’s tidal effect on Earth is greater than the Sun’s.

7.2 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.

ANSWER (a) decreases — g(h) = g(1 − 2h/RE) falls with height. (b) decreases — g(d) = g(1 − d/RE) falls with depth. (c) mass of the body — g = GME/RE2 depends on the Earth’s mass but not on the falling body’s mass. (d) more — the formula −GMm(1/r2 − 1/r1) is exact, while mg(r2 − r1) is only an approximation valid for small height differences.

7.3 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?

ANSWER By Kepler’s third law, T2 ∝ R3, so R ∝ T2/3. The planet goes around twice as fast, so its period is half that of the Earth: Tp = TE/2. Rp/RE = (Tp/TE)2/3 = (1/2)2/3 = 0.63. Its orbit is smaller than the Earth’s by a factor of 0.63 (about 0.63 times the Earth’s orbital radius). ✓

7.4 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.

ANSWER For Io about Jupiter: T = 1.769 d = 1.769 × 86400 = 1.528 × 105 s, R = 4.22 × 108 m. MJ = 4π2R3 / (GT2) = 4(9.87)(4.22×108)3 / [(6.67×10−11)(1.528×105)2] Numerator: 39.48 × 7.515 × 1025 = 2.967 × 1027. Denominator: 6.67×10−11 × 2.335×1010 = 1.558 × 100. MJ = 2.967 × 1027 / 1.558 = 1.90 × 1027 kg. The Sun’s mass MS = 2 × 1030 kg, so MJ/MS = 1.90×1027 / 2×1030 ≈ 1/1000. Hence Jupiter’s mass is about one-thousandth that of the Sun. ✓

7.5 Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105 ly.

ANSWER Mass enclosed: M = 2.5 × 1011 × 2 × 1030 = 5 × 1041 kg. Radius: r = 50000 ly = 5 × 104 × 9.46 × 1015 m = 4.73 × 1020 m. Kepler’s third law: T2 = 4π2r3/(GM) = 39.48 × (4.73×1020)3 / [(6.67×10−11)(5×1041)] = 39.48 × 1.058×1062 / (3.335×1031) = 4.176×1063 / 3.335×1031 = 1.252 × 1032 s2. T = √(1.252×1032) = 1.119 × 1016 s = 1.119×1016 / (3.156×107) yr = 3.54 × 108 years. ✓

7.6 Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

ANSWER (a) kinetic energy. For a circular orbit, total energy E = −GmME/2(RE+h) and kinetic energy K = +GmME/2(RE+h), so E = −K; the total energy is the negative of the kinetic energy. (b) less. A stationary object at that height already has total energy = potential energy only (negative), while the orbiting satellite already has positive kinetic energy too, so its total energy is higher (less negative). Hence less extra energy is needed to free the orbiting satellite.

7.7 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?

ANSWER (a) No — escape speed ve = √(2gR) is independent of the body’s mass. (b) No (to a good approximation) and (c) No — escape speed does not depend on the direction of projection. (Strictly, escape speed depends only on the gravitational potential at the launch point, which varies very slightly with latitude/height.) (d) Yes — it depends on the height of the launch location, because the gravitational potential (and hence the escape speed) changes with distance from the Earth’s centre.

7.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

ANSWER Constant: (c) angular momentum and (f) total energy — angular momentum is conserved because gravity is a central force, and total mechanical energy is conserved because gravity is conservative. Not constant: (a) linear speed, (b) angular speed, (d) kinetic energy and (e) potential energy — all of these change along the elliptical orbit (the comet moves fastest near perihelion and slowest near aphelion).

7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.

ANSWER (b) swollen face, (c) headache and (d) orientational problem are likely. In the weightlessness of orbit, body fluids redistribute towards the head, causing a swollen (puffy) face and headache; the lack of a clear up/down reference causes orientational problems. (a) swollen feet is not likely, since blood does not pool in the feet without gravity.

7.10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0.

ANSWER (iii) c. Complete the hemisphere into a full spherical shell. At the centre of a full shell the gravitational intensity is zero. So the upper half (the given hemisphere) and the missing lower half must produce equal and opposite intensities at the centre. The arrow that points along the axis of symmetry, towards the missing half — option c — gives the direction of the field due to the hemispherical shell.

7.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

ANSWER (ii) e. By the same “complete-the-sphere” argument, the field at point P due to the full shell is zero, so the field of the hemisphere at P is equal and opposite to that of the missing half. The correct direction is the arrow e.

7.12 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).

ANSWER Let the neutral point be at distance x from the Earth’s centre; it is (r − x) from the Sun, with r = 1.5 × 1011 m. Setting the two pulls equal: GMEm/x2 = GMSm/(r − x)2  ⇒  (r − x)2/x2 = MS/ME. MS/ME = (2×1030)/(6×1024) = 3.333 × 105; so (r − x)/x = √(3.333×105) = 577.4. r/x = 1 + 577.4 = 578.4 ⇒ x = r/578.4 = 1.5×1011 / 578.4 = 2.6 × 108 m from the Earth’s centre. ✓

7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.

ANSWER For the Earth orbiting the Sun, gravity provides the centripetal force, giving Kepler’s third law: MS = 4π2r3/(GT2). Here r = 1.5 × 108 km = 1.5 × 1011 m and T = 1 year = 3.156 × 107 s. MS = 39.48 × (1.5×1011)3 / [(6.67×10−11)(3.156×107)2] = 39.48 × 3.375×1033 / (6.67×10−11 × 9.96×1014) = 1.332×1035 / 6.644×104 = 2.0 × 1030 kg. ✓

7.14 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?

ANSWER By Kepler’s third law, RS/RE = (TS/TE)2/3. TS/TE = 29.5, so (29.5)2/3 = (29.52)1/3 = (870.25)1/3 = 9.54. RS = 9.54 × 1.50 × 108 km = 1.43 × 109 km = 1.43 × 1012 m. ✓

7.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

ANSWER At height h, weight W(h) = W × RE2/(RE+h)2. Here h = RE/2, so RE+h = (3/2)RE. W(h) = 63 × RE2 / [(3/2)RE]2 = 63 × (1)/(9/4) = 63 × 4/9. W(h) = 252/9 = 28 N. ✓

7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

ANSWER At depth d, g(d) = g(1 − d/RE). Half way to the centre means d = RE/2. g(d) = g(1 − 1/2) = g/2, so the weight becomes half. W(d) = 250 × 1/2 = 125 N. ✓

7.17 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.

ANSWER Let the rocket rise to a maximum distance r from the Earth’s centre (height h = r − RE), where its speed is zero. Conserving energy: ½mv2 − GMEm/RE = −GMEm/r. ⇒ 1/r = 1/RE − v2/(2GME). v2/(2GME) = (5000)2 / [2(6.67×10−11)(6.0×1024)] = 2.5×107 / 8.004×1014 = 3.123×10−8 m−1. 1/RE = 1/(6.4×106) = 1.5625×10−7 m−1. So 1/r = 1.5625×10−7 − 0.3123×10−7 = 1.2502×10−7 m−1. r = 1/1.2502×10−7 = 8.0 × 106 m from the Earth’s centre (so it rises about 1.6 × 106 m above the surface). ✓

7.18 The escape speed of a projectile on the earth’s surface is 11.2 km s−1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

ANSWER By energy conservation, far from Earth (potential energy → 0): ½mvi2 − ½mve2 = ½mvf2, since the escape speed corresponds to just reaching infinity with zero speed. vf2 = vi2 − ve2 = (3ve)2 − ve2 = 9ve2 − ve2 = 8ve2. vf = √8 × ve = 2√2 × 11.2 = 2.828 × 11.2 = 31.7 km s−1. ✓

7.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10−11 N m2 kg−2.

ANSWER Orbital radius: r = RE + h = 6.4×106 + 0.4×106 = 6.8 × 106 m. Total energy in orbit: E = −GMEm/(2r) = −(6.67×10−11)(6.0×1024)(200) / (2 × 6.8×106). Numerator: 6.67×10−11 × 6.0×1024 × 200 = 8.004 × 1016. Denominator: 1.36 × 107. E = −8.004×1016 / 1.36×107 = −5.885 × 109 J. To escape, total energy must become zero, so energy required = 0 − E = 5.9 × 109 J. ✓

7.20 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

ANSWER Initial separation ri = 109 km = 1012 m; at collision the centres are 2R apart, rf = 2 × 104 km = 2 × 107 m. Let each star reach speed v. M = 2×1030 kg. Energy conservation (initial KE ≈ 0): −GM2/ri = 2(½Mv2) − GM2/rf. Mv2 = GM2(1/rf − 1/ri) ⇒ v2 = GM(1/rf − 1/ri). 1/rf = 1/(2×107) = 5×10−8; 1/ri = 10−12 (negligible). So 1/rf − 1/ri ≈ 5×10−8 m−1. v2 = (6.67×10−11)(2×1030)(5×10−8) = 6.67 × 1012. v = 2.58 × 106 ≈ 2.6 × 106 m s−1. ✓

7.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

ANSWER Each sphere has m = 100 kg; the mid-point is at distance r = 0.5 m from each centre. Gravitational force: the two spheres pull a test mass in opposite directions with equal magnitude, so the net force at the mid-point is zero. Gravitational potential (scalar, both add): V = −Gm/r − Gm/r = −2Gm/r = −2(6.67×10−11)(100)/(0.5). V = −2 × 6.67×10−9 / 0.5 = −2.668×10−8 ≈ −2.7 × 10−8 J kg−1. Equilibrium: the force is zero, so an object placed there is in equilibrium. But the equilibrium is unstable — a slight displacement towards either sphere increases the pull from the nearer sphere, moving the object further away from the mid-point. ✓

Extra Practice Questions

Short Answer Type Questions

Q1. State Newton’s universal law of gravitation and give its mathematical form.

ANSWEREvery body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, F = Gm1m2/r2, where G = 6.67 × 10−11 N m2 kg−2, directed along the line joining the two masses.

Q2. Why is the value of g maximum at the Earth’s surface?

ANSWERAbove the surface, g decreases as g(h) = g(1 − 2h/RE) because distance from the centre increases. Below the surface, g decreases as g(d) = g(1 − d/RE) because only the inner sphere contributes. Since g falls off in both directions, it is maximum exactly at the surface.

Q3. Define escape speed and write its expression for Earth.

ANSWEREscape speed is the minimum speed a body must be given so that it just escapes the gravitational pull of a planet and never returns. For Earth, ve = √(2GME/RE) = √(2gRE) ≈ 11.2 km s−1. It is independent of the body’s mass and direction of projection.

Q4. Why does the Moon have no atmosphere?

ANSWERThe escape speed on the Moon is only about 2.3 km s−1, roughly five times smaller than on Earth, because the Moon’s mass and radius are small. Gas molecules at the Moon’s surface have thermal speeds exceeding this escape speed, so they escape into space and the Moon cannot retain an atmosphere.

Q5. Why is the total energy of a satellite in a bound orbit negative?

ANSWERFor a circular orbit, the kinetic energy is +GmME/2(RE+h) and the potential energy is −GmME/(RE+h). Their sum, the total energy, is −GmME/2(RE+h), which is negative. A negative total energy means the satellite is bound; if it were zero or positive, the satellite would escape to infinity.

Long Answer Type Questions

Q1. State Kepler’s three laws of planetary motion and explain how the second law follows from conservation of angular momentum.

ANSWERLaw of orbits: every planet moves in an ellipse with the Sun at one focus. Law of areas: the line joining a planet to the Sun sweeps equal areas in equal time intervals. Law of periods: the square of the orbital period is proportional to the cube of the semi-major axis (T2 ∝ a3). The second law follows from angular momentum conservation: gravity is a central force directed along the line joining the Sun and the planet, so it exerts no torque about the Sun and the angular momentum L remains constant. The area swept per unit time is dA/dt = L/(2m); since L and m are constant, dA/dt is constant, meaning equal areas are swept in equal times. This makes a planet move faster near the Sun (perihelion) and slower far from it (aphelion).

Q2. Derive the expression for the variation of acceleration due to gravity with depth below the Earth’s surface.

ANSWERTreat the Earth as a uniform sphere of mass ME, radius RE and density ρ. At the surface, g = GME/RE2 = (4/3)πGρRE. At depth d, only the inner sphere of radius (RE − d) attracts the body (the outer shell exerts zero force). Its mass is Ms = (4/3)π(RE − d)3ρ, so g(d) = GMs/(RE − d)2 = (4/3)πGρ(RE − d). Dividing by the surface value, g(d) = g(RE − d)/RE = g(1 − d/RE). Thus g decreases linearly with depth and becomes zero at the centre (d = RE).

Q3. Explain weightlessness experienced by an astronaut in an orbiting satellite.

ANSWERWeightlessness in a satellite is not because gravity is absent — at the height of a typical low orbit, g is still nearly as large as at the surface. The satellite and everything in it, including the astronaut, are in continuous free fall towards the Earth; the gravitational force provides exactly the centripetal force for the circular orbit. Since the astronaut and the floor of the satellite accelerate together at the same rate, the floor exerts no normal (contact) force on the astronaut. The apparent weight, which is the normal reaction, is therefore zero, and the astronaut feels weightless — the same effect felt momentarily in a freely falling lift.

MCQs & Assertion–Reason

1. The gravitational force between two masses is inversely proportional to:

(a) the distance between them    (b) the square of the distance    (c) the cube of the distance    (d) the product of masses

2. The acceleration due to gravity at the centre of the Earth is:

(a) maximum    (b) 9.8 m s−2    (c) zero    (d) infinite

3. The escape speed from the Earth’s surface is approximately:

(a) 7.9 km s−1    (b) 11.2 km s−1    (c) 2.3 km s−1    (d) 42 km s−1

4. Kepler’s second law (law of areas) is a direct consequence of the conservation of:

(a) linear momentum    (b) energy    (c) angular momentum    (d) mass

5. The value of the universal gravitational constant G is:

(a) 9.8 N kg−1    (b) 6.67 × 10−11 N m2 kg−2    (c) 6.67 × 1011 N m2 kg−2    (d) 3 × 108 m s−1

6. The total energy of a satellite in a bound circular orbit is:

(a) positive    (b) zero    (c) negative    (d) equal to its kinetic energy

7. At a height equal to the radius of the Earth above its surface, the value of g becomes:

(a) g/2    (b) g/4    (c) g/9    (d) 2g

8. The gravitational potential energy of a two-mass system is taken as zero when the separation is:

(a) zero    (b) equal to the Earth’s radius    (c) infinity    (d) one metre

9. The orbital speed of a satellite close to the Earth’s surface is about:

(a) 11.2 km s−1    (b) 7.9 km s−1    (c) 2.3 km s−1    (d) 3 km s−1

10. Inside a uniform spherical shell, the gravitational force on a particle is:

(a) maximum at the centre    (b) zero everywhere    (c) directed outward    (d) equal to mg

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The escape speed of a body does not depend on the mass of the body.

Reason: Escape speed depends only on the mass and radius of the planet, given by ve = √(2GM/R).

A-R 2. Assertion: An astronaut in an orbiting satellite feels weightless.

Reason: There is no gravitational force acting on the astronaut at the height of the orbit.

A-R 3. Assertion: A planet moves faster when it is nearer to the Sun.

Reason: The angular momentum of the planet about the Sun is conserved.

A-R 4. Assertion: The value of g decreases as we go below the Earth’s surface.

Reason: Only the mass of the sphere of radius less than the body’s distance from the centre contributes to gravity.

A-R 5. Assertion: Gravitational shielding is possible using a hollow conducting sphere.

Reason: There is no negative mass to cancel the gravitational field of external matter.

Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(D).

Common Mistakes to Avoid

Watch out for these

  • Confusing the formulas for height and depth: at height use g(1 − 2h/RE), but at depth use g(1 − d/RE) — the factor of 2 appears only for height.
  • Forgetting that gravitational potential energy is negative (V = −Gm1m2/r) and that potential is a scalar (add algebraically), while force is a vector (add as vectors).
  • Mixing up orbital speed (√(gRE) ≈ 7.9 km s−1) with escape speed (√(2gRE) ≈ 11.2 km s−1); they differ by a factor √2.
  • Writing the time period in days instead of seconds (or distance in km instead of m) in Kepler’s-law calculations — always convert to SI first.
  • Assuming “weightlessness” means zero gravity — it is actually free fall, where the normal reaction is zero but g is still large.
  • Thinking a hollow sphere can shield gravity the way a conductor shields electric fields — gravitational shielding is impossible.

How to score full marks in this chapter

Memorise the standard data (g = 9.8 m s−2, RE = 6.4 × 106 m, G = 6.67 × 10−11, escape speed 11.2 km s−1) and always convert to SI before substituting. For numericals, write the formula first, then substitute with units, and box your final answer with its unit. Use energy conservation for escape/projectile problems and Kepler’s third law (T2 ∝ R3) for orbit-size and period comparisons — ratios save time. For theory questions, link Kepler’s second law to angular-momentum conservation and the satellite’s negative energy to its bound orbit, exactly as the NCERT text does.

Frequently Asked Questions

What is Class 11 Physics Chapter 7 Gravitation about?

Chapter 7, Gravitation, covers Kepler’s laws, Newton’s universal law of gravitation, the gravitational constant G, the variation of acceleration due to gravity with height and depth, gravitational potential energy, escape speed, and the energy of Earth satellites — with all NCERT exercises (7.1–7.21) solved.

How many exercises are there in Class 11 Physics Chapter 7?

The NCERT chapter has 21 numbered exercise questions, 7.1 to 7.21, mixing conceptual choices and numericals. All of them are reproduced verbatim and solved step by step on this page.

What is the escape speed from the Earth?

The escape speed from the Earth’s surface is about 11.2 km s−1, given by ve = √(2GME/RE) = √(2gRE). It is independent of the mass and direction of projection of the body.

Are these Class 11 Physics Chapter 7 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Physics are free and follow the official NCERT textbook for session 2026–27, with every numerical answer verified.

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