Take session-wise tests or a full 100-question mock on Gravitation for NEET — with 90-second per-question timer, answer review, subtopic breakdown, and detailed solutions.
4 sectional sessions (25 Qs each) + 1 Full Mock (100 Qs). NEET 4/−1 marking. 90 sec per question timer.
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1. Newton's law of gravitation states that gravitational force is:
Explanation: F = Gm₁m₂/r². Force is inversely proportional to r².
2. Universal gravitational constant G has SI units:
Explanation: G = Fr²/m₁m₂. Units = N·m²/kg² = N·m²/kg².
3. Gravitational force between two 1 kg masses separated by 1 m (G = 6.67×10⁻¹¹ N·m²/kg²):
Explanation: F = G×1×1/1² = G = 6.67×10⁻¹¹ N.
4. If separation between two masses doubles, gravitational force becomes:
Explanation: F ∝ 1/r². r→2r: F → F/4 = one-fourth.
5. Three masses m are placed at vertices of equilateral triangle (side a). Gravitational force on one mass:
Explanation: Two forces each Gm²/a² at 60° to each other. Resultant = √(F²+F²+2F²cos60°) = F√3 = √3 Gm²/a².
6. Value of g at surface of Earth is approximately:
Explanation: Standard value of acceleration due to gravity at Earth's surface: g ≈ 9.8 m/s².
7. g at height h above Earth (R = radius, h << R):
Explanation: g_h ≈ g(1 − 2h/R) for h g(1 − 2h/R).
8. g at depth d below Earth's surface:
Explanation: g_d = g(1 − d/R). At depth d, g decreases linearly.
9. g at centre of Earth:
Explanation: At Earth's centre, d = R. g_d = g(1−R/R) = 0.
10. Which planet has highest surface gravity? (Use proportional reasoning: g ∝ M/R²)
Explanation: Jupiter has the highest mass and correspondingly high g ≈ 25 m/s² at its surface.
11. Gravitational field intensity at point r from mass M:
Explanation: g = GM/r² — same formula as acceleration due to gravity. = GM/r².
12. If both masses are doubled and separation halved, force becomes:
Explanation: F = G(2m₁)(2m₂)/(r/2)² = 4Gm₁m₂×4/r² = 16 times original. 16 times.
13. g varies with latitude because Earth:
Explanation: g is maximum at poles (minimum r, no centrifugal effect) and minimum at equator — due to rotation and oblate shape.
14. Gravitational force is:
Explanation: Gravitational force is always attractive only between masses.
15. Gravitational field is a:
Explanation: Gravitational field has both magnitude and direction — it is a vector field.
16. Gravitational force obeys:
Explanation: Gravitational forces between two bodies are equal and opposite — Newton's 3rd law applies.
17. Height at which g = g/9 (radius = R):
Explanation: g/(1+h/R)² = g/9. (1+h/R)² = 9. 1+h/R = 3. h = 2R.
18. Inside a hollow spherical shell, gravitational field is:
Explanation: Inside a hollow sphere, gravitational field = 0 (shell theorem).
19. Value of G is approximately:
Explanation: Universal gravitational constant G = 6.67×10⁻¹¹ N·m²/kg².
20. g on Moon is approximately 1/6 of Earth's g. On Moon, weight of 60 kg person (g=10 m/s²):
Explanation: W_moon = mg_moon = 60×(10/6) = 60×1.67 ≈ 100 N.
21. Gravitational force is a:
Explanation: Gravity acts at a distance — it is a non-contact (field) force.
22. A tunnel through Earth's diameter — period of oscillation of body in tunnel:
Explanation: Body in tunnel undergoes SHM with T = 2π√(R/g) ≈ 84 min = 2π√(R/g).
23. Gravitational field due to a point mass at distance r varies as:
Explanation: g = GM/r² ∝ 1/r².
24. Gravitational force is weakest of four fundamental forces. Relative strength compared to electromagnetic:
Explanation: Gravity is approximately 10³⁶ times weaker than electromagnetic force.
25. g is independent of:
Explanation: g = GM/R². It depends on M and R of Earth, and G. It is independent of the mass of falling body.
26. Gravitational PE of mass m at distance r from M (reference at infinity):
Explanation: Gravitational PE = −GMm/r (negative because gravity is attractive and work done against field).
27. Gravitational PE at Earth's surface (mass m, radius R, mass M):
Explanation: PE_surface = −GMm/R.
28. Escape velocity from Earth's surface is approximately:
Explanation: Escape velocity from Earth ≈ 11.2 km/s.
29. Escape velocity formula:
Explanation: v_e = √(2GM/R) = √(2gR).
30. If mass of planet doubles and radius doubles, escape velocity:
Explanation: v_e = √(2GM/R). M→2M, R→2R: v_e = √(2G×2M/2R) = √(2GM/R) = same.
31. Orbital velocity of satellite at radius r (from centre, mass M):
Explanation: Orbital velocity v₀ = √(GM/r).
32. For a satellite in circular orbit, orbital speed is _______ than escape velocity:
Explanation: v_e = √(2GM/R) = √2 × v₀. So v₀ = v_e/√2 — orbital speed is √2 times less.
33. A satellite at height h above Earth. Orbital speed v₀ compared to surface orbital speed:
Explanation: v₀ = √(GM/(R+h)) = √(gR²/(R+h)) = √(gR²/(R+h)).
34. Kepler's first law states that planets move in:
Explanation: Kepler's 1st law: Planets move in elliptical orbits with the Sun at one focus.
35. Kepler's second law (law of areas) states:
Explanation: Kepler's 2nd law: equal areas are swept in equal times — due to conservation of angular momentum.
36. Kepler's third law: T² ∝ r³. If planet B has orbital radius 4 times planet A, T_B/T_A:
Explanation: T² ∝ r³. (T_B/T_A)² = (r_B/r_A)³ = 4³ = 64. T_B/T_A = √64 = 8.
37. Time period of a satellite at radius r:
Explanation: T = 2π√(r³/GM) = 2π√(r³/GM). (From Kepler's 3rd law.)
38. Period of satellite at height h = R above Earth surface (g, R given):
Explanation: r = R+h = 2R. T = 2π√(r³/GM) = 2π√(8R³/gR²) = 2π√(8R/g) = 2π√(8R/g).
39. Geostationary satellite orbits at approximately:
Explanation: Geostationary orbit ≈ 36,000 km above the equator. Period = 24 hours.
40. At perihelion, a planet moves ______ than at aphelion:
Explanation: By angular momentum conservation: at perihelion (closest), v is maximum — planet moves faster.
41. Total energy of satellite in circular orbit:
Explanation: KE = GMm/2r, PE = −GMm/r. Total E = KE + PE = GMm/2r − GMm/r = −GMm/2r.
42. Escape velocity is independent of:
Explanation: v_e = √(2GM/R). It depends on M, R, G — but NOT on the mass of the escaping body.
43. A satellite in orbit has total energy:
Explanation: Total E = −GMm/2r negative.
44. Energy required to move satellite from orbit r₁ to r₂ (r₂ > r₁):
Explanation: ΔE = E₂−E₁ = −GMm/2r₂ − (−GMm/2r₁) = GMm(1/r₁−1/r₂)/2 = GMm(1/r₁−1/r₂)/2.
45. When satellite orbit radius increases, its speed:
Explanation: v₀ = √(GM/r). As r increases, v₀ decreases.
46. Kepler's third law is derived from:
Explanation: Kepler's 3rd law (T² ∝ r³) follows directly from Newton's law of gravitation.
47. Moon has no atmosphere because:
Explanation: Moon's escape velocity ≈ 2.4 km/s — less than typical gas molecule speeds. Atmosphere escapes. Low escape velocity.
48. Work done to move mass m from surface to height h (h << R):
Explanation: For h mgh.
49. A satellite in circular orbit experiences:
Explanation: In circular orbit, gravity provides centripetal force. Only gravitational force acts.
50. A body is weightless in a satellite because:
Explanation: Weightlessness in satellite is due to free fall — both body and satellite accelerate at g toward Earth.
51. Binding energy of satellite in orbit (radius r, mass m, planet mass M):
Explanation: Binding energy = −(Total E) = −(−GMm/2r) = GMm/2r. Energy needed to free the satellite.
52. KE of satellite in circular orbit (mass m, radius r):
Explanation: mv²/r = GMm/r². v² = GM/r. KE = ½mv² = GMm/2r.
53. Gravitational potential at distance r from mass M:
Explanation: V = −GM/r. Gravitational potential is always negative.
54. Relation between gravitational field (g) and gravitational potential (V):
Explanation: g = −dV/dr. Gravitational field is negative gradient of potential.
55. For a satellite to be boosted to higher orbit, it must be given:
Explanation: Higher orbit has higher total energy (less negative). Satellite must gain positive energy.
56. If Earth-Sun distance doubles, Earth's orbital period becomes:
Explanation: T² ∝ r³. T₂ = T₁×(r₂/r₁)^(3/2) = 1×2^(3/2) = 2√2 years = 2√2 years.
57. Gravitational potential inside uniform solid sphere at distance r < R from centre:
Explanation: Inside solid sphere: V = −GM(3R²−r²)/(2R³) = −GM(3R²−r²)/2R³.
58. Angular momentum of satellite in circular orbit:
Explanation: In circular orbit, no net torque (gravity is central). Angular momentum = constant.
59. Geostationary satellite has period:
Explanation: Geostationary = stationary relative to Earth. Period = 24 hours (same as Earth's rotation).
60. Equipotential surfaces due to a point mass are:
Explanation: V = −GM/r = constant on surfaces of equal r. These are concentric spheres.
61. According to Kepler's 2nd law, planet moves fastest when:
Explanation: At perihelion (closest point), by conservation of angular momentum, planet moves fastest.
62. A satellite is in orbit at r. To make it spiral inward (reduce orbit), we should:
Explanation: Decreasing speed reduces kinetic energy — satellite falls to lower orbit. Decrease its speed.
63. Gravitational potential energy between two masses at very large separation:
Explanation: As r→∞, PE = −GMm/r → 0 (reference level at infinity).
64. Orbital speed of Earth (mass M_sun, distance r):
Explanation: v_orbital = √(GM_sun/r). Both expressions √(GM/r) and 2πr/T are correct. Primary formula: √(GM_sun/r).
65. Polar satellites orbit at height ~800 km and have period approximately:
Explanation: Polar satellites (low Earth orbit ~800 km) have period of approximately ~100 minutes.
66. Work done by gravitational force moving mass m from r₁ to r₂ (r₂ > r₁):
Explanation: W = −ΔPE = −[−GMm/r₂ − (−GMm/r₁)] = GMm(1/r₁−1/r₂). (Negative as r increases.)
67. Mars is ~1.52 AU from Sun. Its orbital period (Earth's = 1 year, 1 AU = Earth-Sun distance):
Explanation: T² ∝ r³. T = (1.52)^(3/2) ≈ 1.52^1.5 ≈ 1.87 years.
68. In elliptical orbit, where is satellite's KE maximum?
Explanation: At perigee (closest point), speed is maximum → KE is maximum.
69. Orbital speed of satellite near Earth's surface (g=10 m/s², R=6.4×10⁶ m):
Explanation: v₀ = √(gR) = √(10×6.4×10⁶) ≈ 8 km/s ≈ 7.9 km/s.
70. Gravitational potential at surface of Earth (M, R, G):
Explanation: V_surface = −GM/R = −GM/R (= −gR since g = GM/R²).
71. For a satellite in circular orbit, PE = −2KE. Total energy:
Explanation: TE = KE + PE = KE − 2KE = −KE. Total energy equals negative of KE.
72. Kepler's third law is also called:
Explanation: Kepler's 3rd law (T² ∝ r³) is called the Law of Periods or Harmonic Law.
73. Escape velocity from Moon (M_moon = M/81, R_moon = R/3.7) compared to Earth:
Explanation: v_e(Moon) = √(2GM_moon/R_moon) ≈ 2.4 km/s.
74. No work is done in moving a mass along an:
Explanation: On an equipotential surface, potential is constant → ΔV = 0 → W = mΔV = 0.
75. A satellite of mass m revolves in circular orbit r₁. To transfer to orbit r₂ > r₁, minimum energy required:
Explanation: ΔE = E₂−E₁ = −GMm/2r₂ − (−GMm/2r₁) = GMm(1/r₁−1/r₂)/2.
76. Weight of body at Earth's centre:
Explanation: At Earth's centre, g = 0. W = mg = 0.
77. If Earth's radius decreases by 1% (mass constant), g changes by:
Explanation: g = GM/R². If R decreases by 1%: Δg/g = −2ΔR/R = +2% (g increases by 2%).
78. Gravitational PE of Earth-Moon system (M_E, M_M, distance d):
Explanation: PE = −GM_E M_M/d (negative, bound system).
79. Astronaut weighs 800 N on Earth. Weight on planet (g_planet = g/4):
Explanation: W ∝ g. W_planet = 800 × (1/4) = 200 N.
80. An artificial satellite orbiting at height R (R = Earth radius). Orbital period (g, R):
Explanation: r = 2R. T = 2π√(r³/GM) = 2π√(8R³/gR²) = 2π√(8R/g).
81. Gravitational field is zero at a point between two masses m and 4m separated by distance d. Point from m:
Explanation: Gm/x² = G(4m)/(d−x)². (d−x)²/x² = 4. (d−x)/x = 2. d = 3x. x = d/3.
82. Gravitational constant G does NOT depend on:
Explanation: G is a universal constant — independent of masses, temperature, position, medium.
83. Minimum energy to launch satellite from surface into orbit at height h (mass m, g, R):
Explanation: ΔE = (PE at h − PE at surface) + KE in orbit = mgh + GMm/2(R+h)... standard result: mgh + GMm/2(R+h) (approx for h
84. A body is projected from Earth with v = v_e/2. It:
Explanation: v return to Earth.
85. Ratio of escape velocity to orbital velocity for any planet:
Explanation: v_e = √(2GM/R), v_o = √(GM/R). v_e/v_o = √2.
86. g at height h = R (R = Earth radius). g_h/g_surface:
Explanation: g_h = GM/(R+h)² = GM/(2R)² = GM/4R² = g/4. g_h/g = 1/4.
87. Tidal forces are due to:
Explanation: Tidal forces arise from differential gravitational force across an extended body (nearer side attracted more).
88. A satellite close to Earth surface (h ≈ 0). Period T₀ = 2π√(R/g). If g = 10 m/s², R = 6.4×10⁶ m, T₀:
Explanation: T = 2π√(6.4×10⁶/10) = 2π×800 ≈ 5026 s ≈ 84 min.
89. Chandrayaan/lunar orbiter period ~2 hours at ~100 km altitude. This orbit is:
Explanation: Close to Moon's surface with ~2 h period — this is a low lunar orbit.
90. Black hole: escape velocity ≥ speed of light (c). Schwarzschild radius r_s for mass M:
Explanation: Setting v_e = c: √(2GM/r) = c. r_s = 2GM/c² (Schwarzschild radius).
91. Why does satellite not need fuel to stay in orbit?
Explanation: Gravity provides centripetal force. No tangential force is required for circular motion.
92. Gravitational force is a central force because:
Explanation: A central force acts along the line joining the two interacting masses.
93. For a double star system (masses M and 2M, separation d), they orbit their common COM. Angular velocity:
Explanation: Total mass = 3M. ω²×r = GM_other/d². For circular orbits about COM: ω = √(3GM/d³)... standard: ω = √(G×3M/d³) = √(3GM/d³).
94. If Earth suddenly stops rotating, weight of a body at equator:
Explanation: Rotation reduces apparent weight at equator (centrifugal effect). If rotation stops, this reduction disappears — weight increases.
95. Gravitational self-energy of uniform sphere (M, R):
Explanation: Gravitational self-energy of uniform sphere = −3GM²/5R.
96. Speed of satellite does NOT depend on:
Explanation: v₀ = √(GM/r). Depends on M, G, r — NOT on mass of the satellite.
97. Period of geostationary satellite is equal to:
Explanation: Geostationary satellite period = Earth's rotation period ≈ 24 hours.
98. Energy needed to lift 1 kg from surface to infinity (g=10 m/s², R=6.4×10⁶ m):
Explanation: E = GMm/R = mgR = 1×10×6.4×10⁶ = 6.4×10⁷ J.
99. At what latitude is g maximum?
Explanation: At poles: Earth radius is minimum (oblate) and no centrifugal reduction. g is maximum at poles.
100. An object projected with exactly escape velocity will:
Explanation: At escape velocity, total E = 0. Object reaches r = ∞ with v = 0 — reaches infinity with zero final velocity.