Gravitation

1

Kepler's first law states that planets move in:Easy

A. Circular orbits around the Sun

B. Elliptical orbits with Sun at one focus

C. Parabolic orbits around the Sun

D. Hyperbolic orbits around the Sun

✅ Solution

Kepler's First Law (Law of Orbits): All planets move in elliptical orbits with the Sun situated at one of the foci. This was a deviation from the Copernican circular orbit model. A circle is a special case of an ellipse.

Answer: B

2

Kepler's second law (Law of Areas) is a consequence of conservation of:Moderate

A. Linear momentum

B. Kinetic energy

C. Angular momentum

D. Total mechanical energy

✅ Solution

Gravitational force is a central force (directed along r). For any central force, angular momentum L = r × p is conserved. Since ΔA/Δt = L/(2m) = constant, equal areas are swept in equal times.

Answer: C — Angular momentum

3

Kepler's third law states T² ∝ a³. If the orbital radius of a planet is doubled, its time period becomes:Moderate

A. 2T

B. 4T

C. 2√2 T

D. √2 T

✅ Solution

T² ∝ a³ → T ∝ a^(3/2). If a → 2a: T' ∝ (2a)^(3/2) = 2^(3/2) × a^(3/2) = 2√2 × T

Answer: C — 2√2 T

4

The value of the universal gravitational constant G is:Easy

A. 6.67 × 10⁻⁸ N m²/kg²

B. 6.67 × 10⁻¹¹ N m²/kg²

C. 9.8 m/s²

D. 6.67 × 10¹¹ N m²/kg²

✅ Solution

First measured by Henry Cavendish in 1798 using a torsion balance experiment. G = 6.67 × 10⁻¹¹ N m²/kg²G is a universal constant — same everywhere in the universe.

Answer: B

5

The acceleration due to gravity on the surface of Earth is given by:Easy

A. g = GME/RE

B. g = GME/RE²

C. g = GME²/RE

D. g = RE²/GME

✅ Solution

From F = mg = GMEm/RE²: g = GME/RE²This is independent of the mass of the object. All objects fall with the same acceleration at a given point.

Answer: B

6

As we go to higher altitude above Earth's surface, acceleration due to gravity:Easy

A. Increases

B. Decreases

C. Remains the same

D. First increases then decreases

✅ Solution

g(h) = GME/(RE+h)² = g(1 − 2h/RE) for h << REAs h increases, g decreases. g is maximum on Earth's surface.

Answer: B — Decreases

7

As we go below the Earth's surface, acceleration due to gravity:Moderate

A. Increases

B. Decreases

C. Becomes zero at the centre

D. Both B and C

✅ Solution

g(d) = g(1 − d/RE)As d increases (going deeper), g decreases linearly. At the centre (d = RE), g = 0. So g decreases AND becomes zero at centre.

Answer: D — Both B and C

8

The gravitational potential energy of two point masses m₁ and m₂ separated by distance r is:Easy

A. +Gm₁m₂/r

B. −Gm₁m₂/r

C. Gm₁m₂/r²

D. −Gm₁m₂/r²

✅ Solution

V = −Gm₁m₂/r (taking V = 0 at r → ∞)Gravitational PE is always negative because gravity is attractive. Work must be done against gravity to separate the masses to infinity.

Answer: B

9

The escape speed from Earth's surface is:Easy

A. √(gRE)

B. √(2gRE)

C. 2gRE

D. √(gRE/2)

✅ Solution

v_e = √(2GME/RE) = √(2gRE) ≈ 11.2 km/sEscape speed does NOT depend on the mass of the projected object or the direction of projection.

Answer: B — √(2gRE) ≈ 11.2 km/s

10

The orbital speed of a satellite at height h above Earth's surface is:Moderate

A. V = GME/(RE+h)

B. V = √[GME/(RE+h)]

C. V = GME(RE+h)

D. V = √(gRE)

✅ Solution

By equating centripetal force = gravitational force: mV²/(RE+h) = GMEm/(RE+h)² → V = √[GME/(RE+h)]V decreases as h increases.

Answer: B

11

The total energy of an orbiting satellite is:Moderate

A. Positive

B. Zero

C. Negative

D. Equal to kinetic energy

✅ Solution

E = KE + PE = GMEm/2(RE+h) − GMEm/(RE+h) = −GMEm/2(RE+h)Total energy is always negative for a bound satellite (closed orbit). KE = |E|, PE = 2E.

Answer: C — Negative

12

For a satellite in circular orbit, which relation between KE and PE is correct?Hard

A. KE = PE

B. KE = −PE/2

C. KE = 2|PE|

D. KE = PE/2

✅ Solution

KE = GMEm/2(RE+h)PE = −GMEm/(RE+h)KE = −PE/2 or |KE| = |PE|/2This is the Virial theorem. Total E = KE + PE = −KE = PE/2.

Answer: B — KE = −PE/2

13

A geostationary satellite has a time period of:Easy

A. 12 hours

B. 24 hours

C. 27.3 days

D. 365 days

✅ Solution

A geostationary satellite orbits in the equatorial plane with the same period as Earth's rotation — 24 hours. It appears stationary over a fixed point on Earth. Used for communications and weather monitoring.

Answer: B — 24 hours

14

The gravitational force on a particle inside a uniform hollow spherical shell is:Moderate

A. Maximum at the centre

B. Zero everywhere inside

C. Directed toward the centre

D. Equal to the surface value

✅ Solution

For a particle inside a uniform hollow spherical shell, gravitational forces from all parts of the shell cancel out completely. The net gravitational force is zero everywhere inside. (Gravitational shielding, unlike electrical, is NOT possible from outside sources.)

Answer: B — Zero everywhere inside

15

Newton's universal law of gravitation states that F ∝ m₁m₂/r². If the separation between two masses is doubled, the gravitational force becomes:Easy

A. Doubled

B. Halved

C. One-fourth

D. Four times

✅ Solution

F ∝ 1/r². If r → 2r: F' ∝ 1/(2r)² = 1/(4r²) = F/4Force decreases to one-fourth.

Answer: C — One-fourth

16

An astronaut in a satellite experiences weightlessness because:Moderate

A. Gravity is zero in space

B. The satellite is far from Earth

C. Both astronaut and satellite are in free fall

D. The satellite moves very fast

✅ Solution

Gravity is NOT zero in orbit — it's what keeps the satellite in orbit. The astronaut experiences weightlessness because both the astronaut and the satellite are in "free fall" toward Earth with the same acceleration (g). There is no normal reaction — hence apparent weightlessness.

Answer: C

17

Perihelion is the point where a planet is:Easy

A. Farthest from the Sun

B. At equal distance on both sides

C. Closest to the Sun

D. Moving with minimum speed

✅ Solution

Perihelion (P) = closest point to Sun. Aphelion (A) = farthest from Sun. At perihelion, the planet moves fastest (from Kepler's 2nd law — equal areas in equal time). At aphelion, it moves slowest.

Answer: C — Closest to the Sun

18

The time period of a satellite orbiting very close to Earth's surface (h ≈ 0) is approximately:Moderate

A. 24 hours

B. 27.3 days

C. 85 minutes

D. 365 days

✅ Solution

T₀ = 2π√(RE/g) = 2π√(6.4×10⁶/9.8) ≈ 5078 s ≈ 85 minThis is the minimum possible orbital period for any Earth satellite.

Answer: C — ≈ 85 minutes

19

Does escape speed depend on the mass of the projected object?Easy

A. Yes, directly proportional

B. Yes, inversely proportional

C. No, it is independent of mass

D. Yes, proportional to √m

✅ Solution

v_e = √(2GME/RE) = √(2gRE)Escape speed depends on Earth's mass and radius, but NOT on the mass of the escaping object. A feather and a rocket need the same escape speed from Earth's surface.

Answer: C — Independent of mass

20

The gravitational potential at a point at distance r from mass M is:Moderate

A. −GM/r²

B. +GM/r

C. −GM/r

D. −GMm/r

✅ Solution

Gravitational potential U(r) = potential energy per unit mass. U(r) = −GM/r (J/kg)Compare: Gravitational PE = −GMm/r (for mass m). Gravitational potential = PE/unit mass = −GM/r.

Answer: C — −GM/r

21

A planet at aphelion moves slower than at perihelion. This is because:Moderate

A. Gravity is weaker at aphelion

B. Angular momentum is conserved: L = mrv = constant

C. Mass increases at aphelion

D. Kinetic energy is constant

✅ Solution

Both A and B are correct, but B is the fundamental reason. L = mrv = constant → at aphelion (r large) → v smallr_A × v_A = r_P × v_P, so v_A/v_P = r_P/r_A < 1.

Answer: B

22

The Moon has no atmosphere because:Moderate

A. No gravity on Moon

B. Escape speed on Moon is too high for gas molecules

C. Escape speed on Moon (≈2.3 km/s) is low; gas molecules escape

D. Moon is too cold for gas molecules

✅ Solution

Moon's escape speed ≈ 2.3 km/s (about 5× smaller than Earth's 11.2 km/s). Gas molecules formed on the Moon's surface have thermal speeds exceeding this low escape speed and escape the Moon's gravitational pull easily.

Answer: C

23

Cavendish's experiment first determined:Easy

A. The value of g

B. The mass of the Sun

C. The universal gravitational constant G

D. The radius of Earth

✅ Solution

Henry Cavendish (1798) used a torsion balance with lead spheres to first measure G = 6.67×10⁻¹¹ N m²/kg². This also allowed estimation of Earth's mass. That's why Cavendish is said to have "weighed the Earth."

Answer: C

24

If a body weighs W on Earth's surface, its weight at height h = RE above the surface is:Hard

A. W/2

B. W/4

C. W/√2

D. 2W

✅ Solution

At h = RE, distance from centre = RE + RE = 2RE: g(h) = GME/(2RE)² = GME/4RE² = g/4Weight = mg(h) = mg/4 = W/4

Answer: B — W/4

25

In Kepler's data (Table 7.1), the quantity T²/a³ for all planets is approximately constant at:Moderate

A. 2.96 × 10⁻³⁴ y² m⁻³

B. 6.67 × 10⁻¹¹ y² m⁻³

C. 9.8 y² m⁻³

D. 1.0 y² m⁻³

✅ Solution

From NCERT Table 7.1, for all planets T²/a³ ≈ 2.96–3.01 × 10⁻³⁴ y² m⁻³. This constant = 4π²/GM_sun. The near-constant value confirms Kepler's Third Law.

Answer: A — ≈ 2.96 × 10⁻³⁴ y² m⁻³

1

Suppose a planet goes around the Sun twice as fast as Earth. What is its orbital radius compared to Earth's? (Use Kepler's third law) Moderate3 Marks

✅ Solution

Twice as fast means orbital period T_P = T_E/2 = 0.5 years. By Kepler's 3rd law: T² ∝ a³ → (T_P/T_E)² = (a_P/a_E)³ (0.5)² = (a_P/a_E)³ → 0.25 = (a_P/a_E)³ a_P/a_E = (0.25)^(1/3) = (1/4)^(1/3) ≈ 0.63

Orbital radius ≈ 0.63 times Earth's orbital radius

2

Mars has a moon Phobos with orbital period 7 hrs 39 min and orbital radius 9.4 × 10³ km. Calculate the mass of Mars. (G = 6.67×10⁻¹¹ N m²/kg²) Moderate3 Marks

✅ Solution

T = 7h 39min = 459 min = 459 × 60 = 27540 s R = 9.4 × 10³ km = 9.4 × 10⁶ m From Kepler's 3rd law: T² = 4π²R³/(GM_Mars) M_Mars = 4π²R³/(GT²) = 4×(3.14)²×(9.4×10⁶)³ / (6.67×10⁻¹¹ × (27540)²) = 4×9.86×8.3×10²⁰ / (6.67×10⁻¹¹ × 7.58×10⁸) ≈ 6.48 × 10²³ kg

Mass of Mars ≈ 6.48 × 10²³ kg

3

Calculate the mass of Earth using: g = 9.81 m/s², RE = 6.37 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg². Easy3 Marks

✅ Solution

From g = GME/RE²: ME = gRE²/G = 9.81 × (6.37×10⁶)² / (6.67×10⁻¹¹) = 9.81 × 4.058×10¹³ / 6.67×10⁻¹¹ = 3.98×10¹⁴ / 6.67×10⁻¹¹ ME ≈ 5.97 × 10²⁴ kg

Mass of Earth ME ≈ 5.97 × 10²⁴ kg

4

Saturn's year is 29.5 Earth years. How far is Saturn from the Sun if Earth is 1.5 × 10⁸ km away? Moderate3 Marks

✅ Solution

T_S/T_E = 29.5; By Kepler's 3rd law: (T_S/T_E)² = (R_S/R_E)³ (29.5)² = (R_S/R_E)³ → 870.25 = (R_S/R_E)³ R_S/R_E = (870.25)^(1/3) ≈ 9.55 R_S = 9.55 × 1.5×10⁸ = 1.43 × 10⁹ km

Saturn is ≈ 1.43 × 10⁹ km from the Sun

5

A body weighs 63 N on Earth's surface. Find the gravitational force on it at a height h = RE/2. (RE = 6.4×10⁶ m) Moderate3 Marks

✅ Solution

At h = RE/2, distance from centre = RE + RE/2 = 3RE/2 F(h) = GMEm/(3RE/2)² = GMEm/(9RE²/4) = (4/9) × GMEm/RE² F(h) = (4/9) × W = (4/9) × 63 = 28 N

Gravitational force at h = RE/2 is 28 N

6

A body weighs 250 N on Earth's surface. How much does it weigh halfway down to Earth's centre? Moderate3 Marks

✅ Solution

At depth d = RE/2 (halfway down): g(d) = g(1 − d/RE) = g(1 − RE/2RE) = g/2 Weight = mg(d) = mg × (1/2) = W/2 = 250/2 = 125 N

Weight halfway down = 125 N

7

Find the potential energy of a system of four particles each of mass m placed at vertices of a square of side l. Also find the gravitational potential at the centre. (NCERT Example 7.3) Hard3 Marks

✅ Solution

4 side pairs, each distance l, and 2 diagonal pairs, each distance √2 l W = −4×Gm²/l − 2×Gm²/√2l = −Gm²/l (4 + √2) = −5.41 Gm²/l Centre is at r = l√2/2 = l/√2 from each corner U(centre) = −4×Gm/(l/√2) = −4√2 Gm/l

W = −5.41 Gm²/l; Gravitational potential at centre = −4√2 Gm/l

8

Calculate the escape speed from Earth's surface. Given g = 9.8 m/s², RE = 6.4 × 10⁶ m. Easy3 Marks

✅ Solution

v_e = √(2gRE) = √(2 × 9.8 × 6.4×10⁶) = √(125.44 × 10⁶) = √(1.2544 × 10⁸) v_e = 1.12 × 10⁴ m/s = 11.2 km/s

Escape speed from Earth's surface = 11.2 km/s

9

A satellite orbits Earth at height 400 km. How much energy must be expended to rocket the satellite out of Earth's gravitational influence? Mass of satellite = 200 kg, ME = 6×10²⁴ kg, RE = 6.4×10⁶ m, G = 6.67×10⁻¹¹. Moderate3 Marks

✅ Solution

r = RE + h = 6.4×10⁶ + 0.4×10⁶ = 6.8×10⁶ m Energy of orbiting satellite: E = −GMEm/2r E = −6.67×10⁻¹¹ × 6×10²⁴ × 200 / (2 × 6.8×10⁶) = −8.004×10¹⁶ / 1.36×10⁷ = −5.88×10⁹ J Energy to escape = 0 − E = +5.88×10⁹ J

Energy required ≈ 5.88 × 10⁹ J ≈ 5.9 GJ

10

A 400 kg satellite is in circular orbit of radius 2RE. It is transferred to orbit of radius 4RE. Find the energy required. What are the changes in KE and PE? (ME = 6×10²⁴ kg, RE = 6.37×10⁶ m, g = 9.81 m/s²) [NCERT Example 7.8] Moderate3 Marks

✅ Solution

E_i = −GMEm/4RE; E_f = −GMEm/8RE ΔE = E_f − E_i = GMEm(1/4RE − 1/8RE) = GMEm/8RE ΔE = gmRE/8 = 9.81 × 400 × 6.37×10⁶ / 8 ≈ 3.13 × 10⁹ J ΔKE = −3.13×10⁹ J (KE decreases at higher orbit) ΔPE = 2×ΔE = −6.25×10⁹ J (PE becomes less negative, i.e., increases)

ΔE = 3.13×10⁹ J; ΔKE = −3.13×10⁹ J; ΔPE = −6.25×10⁹ J

11

A rocket is fired vertically from Earth's surface with speed 5 km/s. How far from Earth does it go before returning? ME = 6×10²⁴ kg, RE = 6.4×10⁶ m, G = 6.67×10⁻¹¹. Moderate3 Marks

✅ Solution

Use energy conservation: ½mv² − GMEm/RE = −GMEm/r_max ½v² = GME(1/RE − 1/r_max) → 1/r_max = 1/RE − v²/2GME GME = 6.67×10⁻¹¹ × 6×10²⁴ = 4.002×10¹⁴ m³/s² 1/r_max = 1/(6.4×10⁶) − (5×10³)²/(2×4.002×10¹⁴) = 1.5625×10⁻⁷ − 3.123×10⁻⁸ = 1.25×10⁻⁷ → r_max = 8×10⁶ m Height = r_max − RE = 8×10⁶ − 6.4×10⁶ = 1.6×10⁶ m

Maximum height above surface ≈ 1.6 × 10⁶ m = 1600 km

12

The escape speed on Earth is 11.2 km/s. A body is projected at 3 × 11.2 = 33.6 km/s. What is its speed far away from Earth? (Ignore Sun and other planets.) Hard3 Marks

✅ Solution

Energy conservation: ½mv_i² − GMEm/RE = ½mv_f² (at infinity, PE = 0) v_f² = v_i² − 2GME/RE = v_i² − v_e² (since v_e² = 2gRE = 2GME/RE) v_f² = (33.6)² − (11.2)² = 1128.96 − 125.44 = 1003.52 (km/s)² v_f = √1003.52 ≈ 31.67 km/s

Speed far away ≈ 31.67 km/s ≈ 31.7 km/s

13

A rocket is fired from Earth toward the Sun. At what distance from Earth's centre is the gravitational force on the rocket zero? M_sun = 2×10³⁰ kg, ME = 6×10²⁴ kg, orbital radius = 1.5×10¹¹ m. Hard3 Marks

✅ Solution

Let the zero-force point be at distance r from Earth. Distance from Sun = (L − r) where L = 1.5×10¹¹ m. GME/r² = GM_sun/(L−r)² → M_sun/ME = (L−r)²/r² (L−r)/r = √(M_sun/ME) = √(2×10³⁰/6×10²⁴) = √(3.33×10⁵) ≈ 577 L−r = 577r → L = 578r → r = 1.5×10¹¹/578 ≈ 2.59×10⁸ m

Distance from Earth ≈ 2.59 × 10⁸ m ≈ 2.6 × 10⁵ km

14

The martian orbit is 1.52 times the orbital radius of Earth. What is the length of a Martian year in days? Easy3 Marks

✅ Solution

By Kepler's 3rd law: T_M²/T_E² = R_MS³/R_ES³ T_M² = T_E² × (R_MS/R_ES)³ = (365)² × (1.52)³ = 133225 × 3.512 = 468,000 days² T_M = √468000 ≈ 684 days

Length of Martian year ≈ 684 days

15

Two stars each of solar mass (2×10³⁰ kg) approach each other. When they are 10⁹ km apart, their speeds are negligible. Find the speed at which they collide (radius of each star = 10⁴ km). G = 6.67×10⁻¹¹. Hard3 Marks

✅ Solution

Initial KE ≈ 0; Initial PE = −GM²/r_i where r_i = 10⁹ km = 10¹² m At collision: r_f = 2R = 2×10⁴ km = 2×10⁷ m (centres just touching) Energy conservation (each star has speed v at collision): 0 − GM²/r_i = 2×(½Mv²) − GM²/r_f v² = GM(1/r_f − 1/r_i) = 6.67×10⁻¹¹×2×10³⁰×(1/2×10⁷ − 1/10¹²) ≈ 6.67×10⁻¹¹×2×10³⁰×5×10⁻⁸ = 6.67×10¹² m²/s² v = √(6.67×10¹²) ≈ 2.58 × 10⁶ m/s ≈ 2580 km/s

Collision speed ≈ 2.58 × 10⁶ m/s ≈ 2580 km/s

📋 General Instructions

Section A — Multiple Choice Questions

Section B — Numerical Problems