For the best learning experience, attempt all questions before checking the answers. Reviewing the detailed solutions afterward will sharpen your concepts and ensure complete clarity on Gravitation.

Q1. An apple of mass m is attracted toward Earth by a gravitational force of magnitude F. According to Newton’s third law of motion, what is the magnitude of the gravitational force that the apple simultaneously exerts on Earth?

A. Zero (no force on Earth)
B. Much smaller than F (negligible)
C. Much larger than F (Earth’s mass makes the force bigger)
D. Equal to F

Correct Answer: D

Explanation: Newton’s third law states that if Earth pulls on the apple with force F, the apple pulls on Earth with the same force in the opposite direction. Using F = ma, this same force causes large acceleration in the apple but an extremely tiny one in Earth due to its enormous mass. As a result, the change in Earth’s velocity is practically zero.

Q2. Imagine Earth’s mass is spread out as a large hollow spherical shell of the same total mass M and radius R. How would the gravitational force on a spacecraft located just outside this hollow Earth (at distance R) compare to the force from a solid Earth of mass M?

A. Zero force for the hollow Earth
B. Weaker than for solid Earth
C. Stronger than for solid Earth
D. The same as for a solid Earth of equal mass

Correct Answer: D

Explanation: The shell theorem states that outside a spherical shell, the gravitational effect is the same as if all the mass were concentrated at the center. A solid Earth has its mass effectively concentrated at the center. A hollow Earth with the same total mass exerts an identical gravitational effect on an external object, as if its entire mass were concentrated at its center, according to the shell theorem.

Q3. Two objects attract each other with gravitational force F. If one object’s mass is tripled and the distance between them is doubled, what is the new force in terms of F?

A. ¾ F
B. ½ F
C. 3/2 F
D. 3 F

Correct Answer: A

Explanation: According to Newton’s law of gravitation, F = Gm₁m₂/r². If one mass becomes 3m₂, the numerator becomes 3 times larger. If the distance doubles (r → 2r), then the denominator becomes (2r)² = 4r², making the force 1/4 as strong. Combining both effects: F' = (3 × Gm₁m₂)/(4r²) = (3/4) × F. So the new force is ¾ of the original force.

Q4. As a space probe moves farther from Earth, what happens to the gravitational potential energy U and the work done by Earth’s gravity? (Take U = 0 at infinite separation.)

A. U increases (becomes less negative); gravity does negative work.
B. U increases; gravity does positive work.
C. U decreases (becomes more negative); gravity does positive work.
D. U decreases; gravity does negative work.

Correct Answer: A

Explanation: Gravitational potential energy is U = –GMm/r. As r increases, U increases toward zero. Gravity pulls inward, so as the probe moves outward, gravity does negative work (Gravitational force and displacement are 180° apart here. The dot product of force and displacement will yield a negative value).

Q5. Why do astronauts aboard the International Space Station (ISS) feel weightless, even though gravity at that altitude is about 90% of that on Earth?

A. There is virtually no gravity in low-Earth orbit.
B. The ISS’s spin cancels gravity.
C. They are in free-fall, so no normal force acts on them.
D. A centrifugal force balances gravity.

Correct Answer: C

Explanation: Both the ISS and astronauts fall toward Earth together, so no normal reaction force acts on them. This free-fall state causes weightlessness. The ISS has a large tangential velocity component due to its orbital motion. When this tangential velocity is combined with a much smaller vertical component of velocity (directed toward Earth), the resulting velocity vector traces a curved, circular path, maintaining a stable orbit.

Q6. Two satellites are in circular orbits around Earth – one low, one high. How do their orbital speeds compare?

A. The lower satellite moves faster
B. Both move at the same speed
C. The higher satellite moves faster
D. It depends on their masses

Correct Answer: A

Explanation: Orbital speed v = √(GM/r). A smaller orbital radius means a higher speed.

Q7. Planet X is Earth-like but with twice Earth’s mass and twice its radius. How would gravity at the surface compare to Earth’s?

A. Nearly the same
B. About half as much
C. About twice as much
D. About four times as much

Correct Answer: B

Explanation: g = GM/R². Let M become 2M and R become 2R. Then g' = G(2M)/(2R)² = 2GM/4R² = (1/2)(GM/R²). So, g becomes half its original value.

Thus, doubling the mass increases g by 2, but doubling the radius reduces it by 4, leading to an overall factor of 1/2.

Q8. If you climbed to an altitude equal to one Earth radius above Earth’s surface (2 Rₑ from Earth’s center), what fraction of your surface weight would you experience?

A. ½
B. 1/3
C. 1/4
D. 0

Correct Answer: C

Explanation: Gravity decreases with the square of distance. Weight is proportional to 1/r². If you double the distance (r → 2r), then weight becomes W ∝ 1/(2r)² = 1/4r². So, the weight becomes 1/4 of its original value.

Q9. An astronaut weighs 600 N on Earth. How much would the astronaut weigh on the Moon? (Moon’s gravity ≈ 1/6 of Earth’s.)

A. ~600 N
B. ~300 N
C. ~100 N
D. ~3600 N

Correct Answer: C

Explanation: Weight = mg. On the Moon, g_moon = g_earth / 6, so W = 600 × (1/6) = 100 N.

Q10. At approximately what altitude would your weight drop to half of its surface value? (Rₑ = 6.37×10⁶ m.)

A. ~1000 km
B. ~2600 km
C. ~6370 km
D. No finite altitude

Correct Answer: B

Explanation: Using g = GM/(R + h)² and g₀ = GM/R², we take the ratio: g/g₀ = (R / (R + h))². Setting g/g₀ = 0.5 and solving:

(R / (R + h))² = 0.5 ⇒ R / (R + h) = √0.5 ⇒ (R + h) = R / √0.5 = 1.414R ⇒

h = 0.414R.

For R = 6370 km, h ≈ 0.414 × 6370 ≈ 2640 km.

Q11. Planet P orbits the Sun at a distance 4 times Earth’s orbital radius (a = 4 AU). What is its orbital period?

A. 4 years
B. 8 years
C. 16 years
D. 64 years

Correct Answer: B

Explanation: Kepler’s third law states T² ∝ a³ or T² / a³ = constant (it is 1 for every planet in our solar system if units of measurement are years and AU).
Let Earth’s period Te = 1 year, ae = 1 AU.
Let Planet P’s orbital distance ap = 4 AU. Then:
1² / 1³ = Tp² / 4³ ⇒ Tp² = 64 ⇒ Tp = √64 = 8 years.

Q12. A planet Q has an orbital period of 27 Earth years. Approximately how far from the Sun is it?

A. 3 AU
B. 9 AU
C. 27 AU
D. 81 AU

Correct Answer: B

Explanation: Using Kepler’s third law: Te² / ae³ = Tp² / ap³.
Let Te = 1 year, ae = 1 AU, and Tp = 27 years. Substituting: (1²)/(1³) = (27²)/(ap³) ⇒ ap³ = 729 ⇒ ap = ∛729 = 9 AU.

Q13. If you triple the distance between two point masses, the gravitational force between them becomes:

A. 9 times greater
B. 3 times greater
C. One third as much
D. One ninth as much

Correct Answer: D

Explanation: Since F ∝ 1/r², tripling r means F ∝ 1/(3r)² = 1/9r². Therefore, the force is reduced by a factor of 9.

Q14. A satellite in a stable circular orbit moves with:

A. Constant speed and constant acceleration
B. Changing speed and zero acceleration
C. Constant speed and zero acceleration
D. Variable speed and constant acceleration

Correct Answer: A

Explanation: Speed remains constant because the satellite maintains uniform circular motion. However, since its direction continuously changes, there is a constant inward (radial) acceleration called centripetal acceleration, which is always directed toward the center of the orbit.

Q15. Which of the following quantities does not affect the gravitational acceleration at the surface of a planet?

A. The planet’s mass
B. The planet’s radius
C. The test mass’s value
D. The gravitational constant

Correct Answer: C

Explanation: Surface gravity depends on planet’s mass and radius but is independent of the object’s mass. From F = ma and F = GMm/R², setting them equal gives a = GM/R². The test mass m cancels.

Q16. If the mass of a planet suddenly doubled but its radius stayed the same, the surface gravity would:

A. Remain the same
B. Double
C. Halve
D. Become four times as large

Correct Answer: B

Explanation: g = GM/R². If M doubles, g doubles.

Q17. The period of a satellite in a circular orbit 20,000 kilometers above Earth's surface is closest to:

A. 24 hours
B. 8.5 hours
C. 12 hours
D. 2 hours

Correct Answer: C

Explanation: Using T = 2π√(r³/GM), the period at 20,000 km altitude is approximately 12 hours.

Q18. A 10 kg object is moved from 1,000 kilometers to 2,000 kilometers above Earth’s surface. The change in gravitational potential energy is:

A. –3.13×10⁷ J
B. 6.44×10⁷ J
C. 3.13×10⁷ J
D. –4.03×10⁷ J

Correct Answer: B

Explanation: ΔU = U_f - U_i = GMm (1/r₁ – 1/r₂).

Q19. Satellite X is twice as far from Earth's center as satellite Y. The ratio of the orbital speeds vₓ / v_y is:

A. 1
B. 2
C. √2
D. 1/√2

Correct Answer: D

Explanation: v ∝ 1/√r. If satellite X is at 2r and Y is at r, then vₓ/v_y = 1/√2.

Q20. An object at rest far from Earth falls toward Earth. When it reaches a distance r from Earth's center (ignoring atmosphere), its speed is:

A. √(2GM/r)
B. √(GM/r)
C. GM/r²
D. 2GM/r

Correct Answer: A

Explanation: Total energy conservation: 0 = ½mv² – GMm/r ⇒ v = √(2GM/r). This matches the escape velocity formula in reverse.

Q21. If Earth's mass was unchanged but its radius shrank to half its original value, the gravitational acceleration at the new surface would be:

A. Same as before
B. Twice as much
C. Four times as much
D. Half as much

Correct Answer: C

Explanation: When R → R/2,
g' = GM/(R/2)² = 4(GM/R²) = 4g.

Q22. A planet’s orbital period doubles; its average distance from the star increases by a factor of:

A. 2
B. 2²ᐟ³
C. 2³ᐟ²
D. 4

Correct Answer: B

Explanation: Kepler’s law states T² ∝ r³.
So r = T²ᐟ³. Given T = 2, we get r = 2²ᐟ³ ≈ 1.59.

Q23. A person stands on a scale inside an elevator. When the elevator accelerates upward, the scale reading increases. This phenomenon is physically equivalent to which of the following scenarios?

A. The elevator moving downward at a constant velocity.
B. The elevator moving upward at a constant velocity.
C. The elevator being placed in a stronger gravitational field.
D. The elevator being placed in a weaker gravitational field.

Correct Answer: C

Explanation: Upward acceleration increases the normal force experienced by the person. Using Newton’s second law, N = m(g + a), the scale reads a higher value, mimicking stronger gravity.

Q24. What is the approximate acceleration due to gravity on Mars if Mars has a mass of 6.42 × 10²³ kg and a radius of 3.39 × 10⁶ m? (G = 6.67 × 10⁻¹¹ N·m²/kg²)

A. 3.72 m/s²
B. 9.81 m/s²
C. 1.62 m/s²
D. 0.64 m/s²

Correct Answer: A

Explanation: g = GM/R² = (6.67×10⁻¹¹)(6.42×10²³)/(3.39×10⁶)² ≈ 3.72 m/s².

Q25. An astronaut with a mass of 75 kg is in an elevator accelerating downward at 2.0 m/s². What is the apparent weight? (g = 9.8 m/s²)

A. 735 N
B. 150 N
C. 585 N
D. 885 N

Correct Answer: C

Explanation: N = m(g - a) = 75 × (9.8 – 2.0) = 585 N.

Q26. Gravitational potential energy is often defined as zero at an infinite distance from a mass. What can we say about U at any finite distance?

A. It is always positive.
B. It is always negative.
C. It can be positive or negative depending on speed.
D. It is always zero.

Correct Answer: B

Explanation: U = –GMm/r < 0 for all finite r. A negative U means the object is bound and requires energy to escape to infinity (U = 0).

Q27. An object is launched with half its escape velocity. What is its final total mechanical energy?

A. Positive
B. Negative
C. Zero
D. Cannot be determined

Correct Answer: B

Explanation: Escape velocity corresponds to zero total energy. With half escape speed, kinetic energy is ¼ of the required amount, so total energy remains negative.

Q28. For a satellite in a stable circular orbit around a planet, what provides the centripetal force?

A. The satellite's inertia.
B. The planet's magnetic field.
C. Gravity between satellite and planet.
D. Atmospheric drag.

Correct Answer: C

Explanation: The gravitational attraction of the planet supplies the inward (centripetal) force for orbital motion.

Q29. Two satellites of different masses are in the same circular orbit. Which quantity differs?

A. Orbital speed
B. Orbital period
C. Gravitational force
D. None of the above

Correct Answer: C

Explanation: Orbital speed v = √(GM/r) and period T = 2π√(r³/GM) are independent of the satellite’s mass. But F = GMm/r² depends on m.

Q30. A geostationary satellite appears fixed because:

A. Its speed is zero.
B. Its period matches Earth's rotation.
C. It is outside Earth's gravity.
D. It has an elliptical orbit.

Correct Answer: B

Explanation: A geostationary satellite orbits in 24 hours, matching Earth’s rotation, and remains fixed relative to a point on the equator.

Q31. An exoplanet orbits at 4 times Earth's orbital radius. If Earth’s period is 1 year, what is the exoplanet’s orbital period?

A. 2 years
B. 4 years
C. 8 years
D. 16 years

Correct Answer: C

Explanation: Kepler’s law T² ∝ r³ gives T = √(4³) = √64 = 8 years.

Q32. Kepler's Second Law (equal areas in equal time) is due to:

A. Conservation of linear momentum
B. Conservation of kinetic energy
C. Conservation of angular momentum
D. Conservation of potential energy

Correct Answer: C

Explanation: No external torque acts on the planet-Sun system. Angular momentum remains constant, causing planets to move faster near the Sun and slower when farther away.

Q33. Kepler's Third Law states T²/R³ = constant. What does this reveal?

A. Total mass of planets
B. Universal constant G
C. Mass of the central body
D. Average density of planets

Correct Answer: C

Explanation: The constant is 4π²/(GM) and depends on the mass M of the central body, not on the orbiting planet.

Q34. A 600 kg satellite is in a circular orbit 800 km above Earth’s surface. What is its total mechanical energy?

A. −1.667×10¹⁰ J
B. –1.98×10⁹ J
C. –3.10×10⁹ J
D. –4.12×10⁹ J

Correct Answer: A

Explanation: Total energy E = –GMm/(2r), where r = 6.37×10⁶ + 8×10⁵ = 7.17×10⁶ m. Substituting values gives approximately −1.667×10¹⁰ J.

Q35. A satellite is shifted from an orbit of radius 8000 km to 12000 km. What is the ratio of its orbital speeds v₂/v₁?

A. √(2/3)
B. √(3/2)
C. 2/3
D. 3/2

Correct Answer: A

Explanation: Orbital speed v ∝ 1/√r. Thus, v₂/v₁ = √(r₁/r₂) = √(8000/12000) = √(2/3).

Q36. A 900 kg satellite in a circular orbit at radius 7.0×10⁶ m is transferred to an elliptical orbit with apogee at 1.2×10⁷ m. What is its speed at perigee?

A. 7.2 km/s
B. 7.8 km/s
C. 8.5 km/s
D. 9.3 km/s

Correct Answer: C

Explanation: Using angular momentum conservation vₚrₚ = vₐrₐ and energy methods, the perigee speed calculates to about 8.5 km/s.

Q37. An astronaut weighs 700 N on Earth. What is her weight at a distance of 2 Earth radii from Earth’s center?

A. 175 N
B. 280 N
C. 350 N
D. 490 N

Correct Answer: A

Explanation: Weight ∝ 1/r². At 2Rₑ, weight = 700/4 = 175 N.

Q38. The Moon orbits Earth at radius 3.84×10⁸ m in 27.3 days. Estimate Earth’s mass.

A. 5.1×10²⁴ kg
B. 6.0×10²⁴ kg
C. 5.97×10²⁴ kg
D. 4.7×10²⁴ kg

Correct Answer: C

Explanation: From T² = 4π²r³/(GM), solving for M with T = 27.3 days gives about 5.97×10²⁴ kg.

Q39. An unknown body moves from r₁ = 1.0×10⁸ km to r₂ = 4.0×10⁸ km from the Sun. What happens to its speed?

A. Reduces by a factor of 2
B. Reduces by a factor of 4
C. Reduces by √2
D. Reduces by 1/2

Correct Answer: B

Explanation: Angular momentum conservation v₁r₁ = v₂r₂ ⇒ v₂/v₁ = 1/4. Speed reduces by a factor of 4.

Q40. At what height above Earth will an object weigh 64% of its surface weight?

A. 2.1×10⁶ m
B. 3.0×10⁶ m
C. 4.0×10⁶ m
D. 5.2×10⁶ m

Correct Answer: A

Explanation: From W/W₀ = (R / (R + h))² = 0.64, solving gives h = 0.25R = 2.1×10⁶ m.

Q41. How much energy is needed to move a 1200 kg spacecraft from a 300 km orbit to infinity?

A. 4.2×10⁹ J
B. 6.4×10⁹ J
C. 8.0×10⁹ J
D. 1.2×10¹⁰ J

Correct Answer: B

Explanation: Escape energy equals the magnitude of total mechanical energy E = –GMm/(2r). For r = Rₑ + 300×10³ m, substituting values yields ≈ 6.4×10⁹ J.

Q42. A spacecraft doubles its kinetic energy in orbit. What happens to the orbit?

A. Escapes Earth
B. Moves to higher circular orbit
C. Enters elliptical orbit with larger apogee
D. Period remains unchanged

Correct Answer: C

Explanation: Doubling kinetic energy disrupts circular balance. The spacecraft transitions to an elliptical orbit with a higher apogee, as total energy becomes less negative but still bound.

Q43. A 1000 kg satellite is moved from 6000 km to 12000 km orbit. What is the change in total mechanical energy?

A. 1.2×10⁹ J
B. 5.27×10⁹ J
C. 5.6×10⁹ J
D. 6.7×10⁹ J

Correct Answer: B

Explanation: ΔE = GMm/2 (1/r₁ – 1/r₂). Substituting values gives approximately 5.27×10⁹ J.

Q44. A planet orbits a star at 0.25 AU. What is its orbital period in Earth years?

A. 0.125
B. 0.25
C. 0.5
D. 0.75

Correct Answer: A

Explanation: Using T² = a³, we get T² = (0.25)³ = 0.015625 ⇒ T = 0.125 years (~46 days).

Q45. A satellite at radius R has velocity v. What is the escape velocity from that point?

A. v
B. √2 × v
C. 2v
D. v / √2

Correct Answer: B

Explanation: v = √(GM/R), while vₑ = √(2GM/R). Thus, vₑ = √2 × v.

Q46. From what altitude must an object be dropped so that it reaches Earth’s surface with speed 11.2 km/s?

A. Infinity
B. 1000 km
C. 10⁷ m
D. 5.5×10⁶ m

Correct Answer: A

Explanation: Escape velocity (11.2 km/s) is reached only by falling from infinity due to energy conservation: v = √(2GM/R).

Q47. A geostationary satellite orbits Earth at radius r. What is the centripetal force acting on it?

A. Zero
B. Equal to weight
C. Less than weight
D. Greater than weight

Correct Answer: B

Explanation: For circular orbits, gravitational force F_g = GMm/r² provides the exact centripetal force F_c = mv²/r required to maintain orbit.

Q48. How much work is required to lift a 500 kg object from Earth’s surface to 1000 km altitude?

A. 2.5×10⁹ J
B. 4.2×10⁹ J
C. 4.0×10⁹ J
D. 5.2×10⁹ J

Correct Answer: B

Explanation: Work done by an external force is W = GMm (1/R – 1/(R + h)). Substituting R = 6.37×10⁶ m and h = 1×10⁶ m gives approximately 4.2×10⁹ J.

Q49. A satellite in elliptical orbit has kinetic energy K₁ at perigee and K₂ at apogee. Which is true?

A. K₁ = K₂
B. K₁ > K₂
C. K₁ < K₂
D. Depends on mass

Correct Answer: B

Explanation: Closer to Earth (perigee), the satellite moves faster, so kinetic energy K₁ is greater than K₂.

Q50. The potential energy of a 2000 kg object in orbit is –5.6×10⁹ J. What is its kinetic energy?

A. –5.6×10⁹ J
B. –2.8×10⁹ J
C. 5.6×10⁹ J
D. 2.8×10⁹ J

Correct Answer: D

Explanation: For a circular orbit, K = –U/2. Thus, K = –(–5.6×10⁹)/2 = 2.8×10⁹ J.

Q51. Two satellites of equal mass orbit Earth. Satellite A is in low Earth orbit, Satellite B is twice as far. Which statement is true?

A. A has more total energy
B. B has more kinetic energy
C. A has higher orbital speed
D. B has shorter period

Correct Answer: C

Explanation: Orbital speed decreases with radius as v ∝ 1/√r. Thus, the closer satellite (A) moves faster.

Q52. Two point masses each of mass m are separated by distance r. If both masses are doubled and their separation is halved, how does the gravitational force change?

A. It stays the same.
B. It doubles.
C. It increases by a factor of 8.
D. It increases by a factor of 16

Correct Answer: D

Explanation: F = Gm₁m₂/r². Doubling each mass (2m) and halving r (r/2) gives
F' = G(2m)(2m)/(r/2)² = 16Gm²/r², i.e., 16 times stronger.

Q53. A small test mass is simultaneously attracted by two massive objects. Which statement best describes the net gravitational force on the test mass?

A. The forces add as scalars; directions are irrelevant.
B. The forces subtract because gravity can be repulsive.
C. The net force is the vector sum of the individual forces.
D. Only the larger force acts because it “screens” the weaker one.

Correct Answer: C

Explanation: Gravity follows the superposition principle; the net force is obtained by vector addition of all gravitational forces.

Q54. According to the shell theorem, what is the gravitational force on a particle located at a distance r from the center of a uniform solid sphere of radius R (r < R)?

A. Zero
B. Constant
C. Decreases with 1/r²
D. Proportional to r

Correct Answer: D

Explanation: Only mass inside radius r contributes to force. Inside a uniform sphere, F = GmMr/R³, which is directly proportional to r.

Q55. Which combination of factors explains why the measured acceleration due to gravity (g) at Earth’s surface differs slightly from GM/R²?

A. Only the rotation of Earth.
B. Only the fact that Earth is not a perfect sphere.
C. The non-uniform density of Earth, its oblate shape and its rotation.
D. Only the latitude at which the measurement is taken.

Correct Answer: C

Explanation: Rotation (centrifugal effects), Earth’s flattening at the poles, and density variations all slightly alter the measured value of g.

Q56. For a satellite of mass m in a circular orbit of radius r around Earth, which relation among kinetic energy (K), potential energy (U) and total energy (E) is correct?

A. U = 0, K = E
B. U = –K, so E = 0
C. U = –2K, so E = K = –U/2
D. U = –4K, so E = 3K

Correct Answer: C

Explanation: In a circular orbit, K = +GMm/(2r) and U = –GMm/r. Thus, U = –2K, and total energy E = –GMm/(2r) = K + U.

Q57. In Newtonian gravity, what does the sign of the total mechanical energy (E) of a two-body system indicate?

A. E > 0 for a bound orbit, E = 0 for unbound motion.
B. E = 0 for a bound orbit, E < 0 for unbound motion.
C. E < 0 for a bound orbit, E = 0 for parabolic escape, and E > 0 for hyperbolic orbits.
D. The sign of E has no physical meaning.

Correct Answer: C

Explanation: Negative energy indicates a bound orbit. Zero corresponds to the escape energy (parabolic trajectory), and positive energy corresponds to unbound hyperbolic motion.

Q58. How much work does Earth’s gravity do on a 1 kg object moved slowly from 3 R to 2 R (measured from Earth’s center)?

A. 0
B. GM/(6R)
C. GM/R
D. GM/(2R)

Correct Answer: B

Explanation: Work done W = –ΔU = –GMm[(1/2R) – (1/3R)] = GM/(6R).

Q59. Which of the following objects has the largest escape velocity?

A. Moon
B. Earth
C. Jupiter
D. Mars

Correct Answer: C

Explanation: Escape velocity vₑ = √(2GM/R). Jupiter’s large mass gives the highest vₑ (~59.5 km/s) among the listed bodies.

Q60. Earth’s orbital period is 1 year and orbital radius is 1 AU = 1.50×10¹¹ m. Using Kepler’s third law, estimate the mass of the Sun. (G = 6.67×10⁻¹¹ N·m²/kg².)

A. 4.0×10²⁹ kg
B. 2.0×10³⁰ kg
C. 2.0×10³² kg
D. 4.0×10³⁰ kg

Correct Answer: B

Explanation: Using T² = (4π² / GM) r³ ⇒ M = (4π² r³) / (G T²). Substituting r = 1.5×10¹¹ m and T = 1 year gives ~1.99×10³⁰ kg.

Q61. A spacecraft of mass 1500 kg is moved from Earth’s surface to an orbit at 2000 km altitude. What is the work needed against gravity? (Use G = 6.67×10⁻¹¹ N·m²/kg², Mₑ = 5.97×10²⁴ kg, Rₑ = 6.37×10⁶ m)

A. 3.2×10¹⁰ J
B. 4.5×10¹⁰ J
C. 6.1×10¹⁰ J
D. 2.24×10¹⁰ J

Correct Answer: D

Explanation: W = ΔU = U_final – U_initial (Note: work by gravity is –ΔU, but work by an external force equals +ΔU). Since U = –GMm/r:
W = GMm (1/R₁ – 1/R₂), where R₁ = Rₑ and R₂ = Rₑ + 2×10⁶ m.

Q62. A satellite lowers in orbit from 8000 km to 6000 km radius (around Earth). What fraction of its total mechanical energy is lost?

A. 25%
B. 33%
C. 50%
D. 66%

Correct Answer: B

Explanation: Total energy for a circular orbit is E = –GMm / (2r). Since E ∝ –1/r,
fractional loss = (1/6000 – 1/8000) ÷ (1/8000) = 1/3, i.e., 33%.

Q63. As a space probe moves farther from Earth, what happens to the gravitational potential energy U and the work done by Earth’s gravity? (Take U = 0 at infinite separation.)

A. U increases (becomes less negative); gravity does negative work.
B. U increases; gravity does positive work.
C. U decreases (becomes more negative); gravity does positive work.
D. U decreases; gravity does negative work.

Correct Answer: A

Explanation: U = –GMm/r increases (becomes less negative) as r increases. Gravity opposes outward motion, thus doing negative work.

Q64. Two stars of equal mass orbit each other at separation 4×10⁹ m. Compute gravitational potential energy of the system. (Each star: 2×10³⁰ kg)

A. –1.3×10⁴¹ J
B. –3.3×10⁴¹ J
C. –6.67×10⁴⁰ J
D. –9.4×10⁴¹ J

Correct Answer: C

Explanation: U = –GM₁M₂ / r. Substituting M₁ = M₂ = 2×10³⁰ kg and r = 4×10⁹ m gives –6.67×10⁴⁰ J.

Q65. How long does it take a comet with a = 25 AU orbit to complete one revolution? (Earth’s orbital period = 1 yr)

A. 125 yr
B. 1250 yr
C. 12 yr
D. 1.25 yr

Correct Answer: A

Explanation: Using Kepler’s third law T² = a³, T² = (25)³ = 15625 ⇒ T = √15625 = 125 years.

Q66. A 500 kg probe orbits at 500 km altitude. It fires retrograde to drop to 300 km. What work must be done to change its orbit?

A. –4.5 × 10⁸ J
B. –1.3×10⁸ J
C. –1.7×10⁸ J
D. –2.1×10⁸ J

Correct Answer: A

Explanation: ΔE = –GMm (1/2r₁ – 1/2r₂). The difference in mechanical energy between the two circular orbits gives the work needed (negative here since orbit lowers).

Q67. A hypothetical Earth‐sized planet has twice Earth's mass but same density. What is escape speed from surface?

A. 14 km/s
B. ~11.2 km/s
C. √3 × 11.2
D. 2 × 11.2

Correct Answer: A

Explanation: vₑ = √(2GM/R). With M' = 2Mₑ and constant density, R' = 2^(1/3)Rₑ ≈ 1.26Rₑ. Thus,
vₑ' ≈ 1.26 × 11.2 = 14 km/s.

Q68. A satellite in circular orbit experiences g_eff = 4 m/s². What is orbital radius from Earth?

A. ~5.0×10⁶ m
B. ~7.0×10⁶ m
C. ~9.9×10⁶ m
D. ~1.1×10⁷ m

Correct Answer: C

Explanation: g_eff = GM / r² ⇒ r = √(GM / g_eff). Substituting values gives ~9.9×10⁶ m.

Q69. A spacecraft must launch from planet X (mass 2 Earth‐masses, radius 1.5×Rₑ). What escape velocity?

A. ~11.2 km/s
B. ~13.8 km/s
C. ~12.93 km/s
D. ~18.5 km/s

Correct Answer: C

Explanation: vₑ = √(2GM/R). For planet X,
vₑ = √[2G(2Mₑ)/(1.5Rₑ)] = √(4/3) × 11.2 ≈ 12.93 km/s.

Q70. A satellite in low‐Earth orbit has specific mechanical energy ε = –30 MJ/kg. What is orbital radius?

A. ~6.6 ×10⁶ m
B. ~1×10⁷ m
C. ~2×10⁷ m
D. ~3×10⁷ m

Correct Answer: A

Explanation: Specific energy ε = –GM / (2r) ⇒ r = –GM / (2ε). Substituting values gives ~6.6×10⁶ m.

Q71. A 1000 kg probe moves from 2 Rₑ to 4 Rₑ radius. How much work done by gravity? (Rₑ = Earth's radius)

A. –1.56 ×10¹⁰ J
B. –5.0×10⁹ J
C. –2.0×10¹⁰ J
D. –4.0×10¹⁰ J

Correct Answer: A

Explanation: W = –ΔU = GMm (1/r₂ – 1/r₁). For r₁ = 2Rₑ and r₂ = 4Rₑ, W = –GMm/(4Rₑ) ≈ –1.56×10¹⁰ J.

Q72. A satellite receives a small kick that increases its total energy to ¼ of its current magnitude. What happens?

A. Stays bound in elliptical orbit with larger semi‐major axis
B. Escapes Earth
C. Drops to lower circular orbit
D. Period remains same

Correct Answer: A

Explanation: Total energy E = –(GMm)/(2a). Reducing the magnitude of E (less negative) increases a, resulting in a larger elliptical orbit but the satellite remains bound (E < 0).