Original practice sets for Gravitation are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Newton's law of gravitation states that the force between two point masses m₁ and m₂ separated by distance r is:
Explanation: F = Gm₁m₂/r², where G = 6.674×10⁻¹¹ N·m²/kg² is the universal gravitational constant. Force is attractive and acts along the line joining the masses.
2. Which statement correctly distinguishes G from g?
Explanation: G = 6.674×10⁻¹¹ N·m²/kg² is the same everywhere in the universe. g = GM/R² at Earth's surface ≈ 9.8 m/s² but varies with latitude, altitude, and depth.
3. Acceleration due to gravity at height h above Earth's surface (h << R) is approximately:
Explanation: g_h = GM/(R+h)² = g/(1+h/R)² ≈ g(1−2h/R) for h
4. At depth d inside the Earth (uniform density), g is:
Explanation: Inside a uniform Earth, g_d = g(1 − d/R). At the centre (d = R), g = 0. Gravity decreases linearly with depth.
5. Gravitational potential V at distance r from mass M is:
Explanation: V = −GM/r. It is always negative (attractive field), and zero only at infinity. The negative sign reflects that work must be done against gravity to move mass to infinity.
6. Gravitational potential energy of mass m at distance r from Earth (mass M) is:
Explanation: U = mV = −GMm/r. This is the exact expression; mgh is an approximation valid near Earth's surface where h
7. Gravitational field obeys the superposition principle. If two masses create fields g₁ and g₂ at a point, the net field is:
Explanation: Gravitational fields superpose vectorially: g_net = g₁ + g₂ + ... This allows calculating fields due to complex mass distributions.
8. g is slightly less at the equator than at the poles because:
Explanation: Two reasons: (1) Equatorial radius > polar radius → larger r → smaller g. (2) Earth's rotation → centrifugal effect reduces effective g. Combined: g_equator ≈ 9.78 m/s², g_poles ≈ 9.83 m/s².
9. Kepler's first law states:
Explanation: Kepler's 1st law (Law of Orbits): each planet orbits the Sun in an ellipse, with the Sun at one of the two foci.
10. Kepler's second law (equal areas in equal times) implies that a planet:
Explanation: Equal areas → conservation of angular momentum (L = mvr = constant). At perihelion (closest), r is smallest, so v must be largest. At aphelion (farthest), v is smallest.
11. Kepler's third law states T² ∝ r³. If Mars has orbital radius 1.52 AU, its period in Earth years is approximately:
Explanation: T² = r³ (in AU and years). T² = 1.52³ = 3.512. T = √3.512 ≈ 1.87 years.
12. The gravitational field strength g at a point is defined as:
Explanation: g = F/m = GM/r² (N/kg = m/s²). It is the gravitational force experienced by unit mass placed at that point.
13. A uniform spherical shell exerts gravitational force on a mass OUTSIDE the shell as if:
Explanation: Shell theorem: a uniform spherical shell acts like a point mass at its centre for any external point. For an internal point, the shell exerts zero net gravitational force.
14. Gravitational field INSIDE a uniform hollow spherical shell is:
Explanation: By the shell theorem, a uniform hollow shell exerts zero net gravitational force on any mass placed inside it. The contributions from all parts of the shell cancel exactly.
15. The minimum energy needed to move a satellite of mass m from Earth's surface (radius R) to infinity is:
Explanation: E_binding = −(Total mechanical energy at surface) = −(KE + PE). At rest on surface: KE = 0, PE = −GMm/R = −mgR. Binding energy = mgR. To escape to infinity (U = 0): need mgR of energy. But for a satellite in orbit at surface radius: E_total = −GMm/2R = −mgR/2. Binding energy for orbiting satellite = mgR/2.
16. A body weighs 900 N on Earth's surface. Its weight at height R (one Earth radius above surface) is:
Explanation: At height R above surface, distance from centre = 2R. g_h = g/(1+h/R)² = g/4. Weight = 900/4 = 225 N.
17. The value of G was first measured by:
Explanation: Henry Cavendish (1798) used a torsion balance with lead spheres to measure the tiny gravitational force between known masses, allowing calculation of G = 6.674×10⁻¹¹ N·m²/kg².
18. Two identical balls are placed 1 m apart. Gravitational force is F. When separation is doubled and one mass is tripled, new force is:
Explanation: F = Gm²/1² = Gm². New: F' = G(3m)(m)/(2)² = 3Gm²/4 = 3F/4.
19. Between Earth (mass M) and Moon (mass m, distance d apart), the point where net gravitational field is zero lies at distance x from Earth where:
Explanation: GM/x² = Gm/(d−x)². Taking square roots: √M/x = √m/(d−x). Cross multiply: (d−x)√M = x√m → d√M = x(√M + √m) → x = d√M/(√M + √m).
20. Inside a uniform solid sphere of density ρ, radius R, the gravitational field at distance r from centre (r < R) is:
Explanation: Only the mass within radius r contributes: M' = (4/3)πr³ρ. g = GM'/r² = G(4πr³ρ/3)/r² = 4πGρr/3. It increases linearly with r inside the sphere (directed toward centre).
21. A satellite orbits Earth (mass M, radius R) at height h above surface. Its orbital speed is:
Explanation: v = √(GM/(R+h)) = √(gR²/(R+h)) since GM = gR².
22. The time period of a satellite at height h above Earth's surface is:
Explanation: T = 2π(R+h)/v = 2π(R+h)/√(GM/(R+h)) = 2π√((R+h)³/GM).
23. Escape velocity from Earth's surface is:
Explanation: From energy conservation (KE + PE = 0 at infinity): ½mv_e² = GMm/R = mgR → v_e = √(2gR) ≈ 11.2 km/s.
24. A geostationary satellite has period equal to Earth's rotation period (24 hrs) so it appears stationary over a point. It must orbit:
Explanation: Geostationary satellite: T = 24 hr, must be in equatorial plane, must orbit west to east, specific radius ≈ 42,400 km from Earth's centre (~35,800 km above surface).
25. Total mechanical energy of a satellite in circular orbit of radius r is:
Explanation: KE = GMm/2r (half of |PE|). PE = −GMm/r. Total E = KE + PE = −GMm/2r. Negative total energy confirms bound orbit.
26. Relation between escape speed v_e and orbital speed v_o at the same radius:
Explanation: v_o = √(GM/r), v_e = √(2GM/r) = √2 × v_o. To escape from orbit, speed must be increased by factor √2 (about 41% more).
27. Astronauts feel weightless in orbit because:
Explanation: In orbit, both astronaut and spacecraft continuously fall toward Earth with the same centripetal acceleration = g at that altitude. There is no relative acceleration between them → zero apparent weight.
28. Earth orbits Sun at 1 AU in 1 year. Saturn orbits at 9.54 AU. Saturn's period is approximately:
Explanation: T² = a³. T_Saturn² = 9.54³ = 867. T_Saturn = √867 ≈ 29.5 years.
29. If a satellite's speed is increased at a point in its orbit, it moves to:
Explanation: Increasing speed gives more energy → less negative total energy → larger orbit radius. (Decreasing speed → lower orbit.) This is why rockets must slow down to de-orbit — counter-intuitive!
30. Moon's escape velocity is about 2.38 km/s (vs Earth's 11.2 km/s). This is primarily because Moon has:
Explanation: v_e = √(2gR). Moon's g ≈ 1.62 m/s² (vs 9.8), and R_moon = 1740 km (vs 6371 km). Combined effect gives v_e = √(2×1.62×1.74×10⁶) ≈ 2.38 km/s.
31. To move a satellite from lower orbit (r₁) to higher orbit (r₂, a Hohmann transfer), what must be done?
Explanation: Hohmann transfer: 1st burn at r₁ raises apogee to r₂ (creating an ellipse). 2nd burn at r₂ circularises the orbit. This is the most energy-efficient two-impulse transfer.
32. For a satellite in elliptical orbit, angular momentum L is:
Explanation: No external torque acts on the satellite (gravity is central). Therefore angular momentum L = mvr is conserved throughout the orbit. This is the basis of Kepler's 2nd law.
33. Energy needed to raise a satellite of mass m from Earth's surface to height R (equal to Earth's radius R, g = 10, R = 6.4×10⁶ m) is:
Explanation: ΔPE = −GMm/(R+R) − (−GMm/R) = GMm/R − GMm/2R = GMm/2R = mgR/2. (Using GM = gR².)
34. A polar satellite orbits Earth:
Explanation: Polar satellites orbit from pole to pole at low altitude (~600–800 km). As Earth rotates beneath, the satellite scans different longitudes each orbit — useful for weather, reconnaissance, remote sensing.
35. A satellite grazes Earth's surface (h ≈ 0). Its orbital period is approximately: (g = 10, R = 6.4×10⁶ m)
Explanation: T = 2π√(R/g) = 2π√(6.4×10⁶/10) = 2π×800 = 5027 s ≈ 84 min. All low Earth orbital periods are approximately 90 minutes.
36. The ratio of gravitational force on a 5 kg mass to that on a 10 kg mass at the same point in a gravitational field is:
Explanation: F = mg. Same g, different masses. F₁/F₂ = 5/10 = 1:2.
37. The Moon has no atmosphere because:
Explanation: Gas molecules have average thermal speeds ~0.5–2 km/s (comparable to Moon's v_escape). Over billions of years, faster molecules escaped and Moon's atmosphere was lost. Earth's higher v_escape (11.2 km/s) retains most atmospheric gases.
38. An astronaut's mass on Earth is 70 kg (g = 9.8 m/s²). On Moon (g = 1.63 m/s²), the astronaut's weight is:
Explanation: Weight = mg_Moon = 70 × 1.63 = 114.1 N. Mass remains 70 kg everywhere — it is an intrinsic property. Weight (force) changes with local g.
39. The Schwarzschild radius (event horizon) of a black hole is the radius at which escape velocity equals c. For mass M:
Explanation: Setting v_e = c: c² = 2GM/r_s → r_s = 2GM/c². For Earth, r_s ≈ 9 mm. For the Sun, r_s ≈ 3 km.
40. Gravitational PE of a 10 kg mass on Earth's surface (R = 6.4×10⁶ m, g = 10 m/s²) relative to infinity is:
Explanation: U = −GMm/R = −mgR = −10×10×6.4×10⁶ = −6.4×10⁸ J. The negative sign indicates the mass is bound to Earth (needs energy to escape).
41. g at height h = R/2 above the surface versus g at depth d = R/2 below the surface. The ratio g_h/g_d is:
Explanation: $g_h=g/(1+h/R)^2=4g/9$ and, for a uniform Earth, $g_d=g(1-d/R)=g/2$. Therefore $g_h/g_d=(4/9)/(1/2)= 8/9 $.
42. Satellite A orbits at radius 2R, satellite B at 8R (R = Earth's radius). Ratio of periods T_B/T_A is:
Explanation: T ∝ r^(3/2). T_B/T_A = (8R/2R)^(3/2) = 4^(3/2) = 8.
43. For a satellite in circular orbit, which is correct?
Explanation: For circular orbit: KE = GMm/2r. PE = −GMm/r. |KE| = |PE|/2. Total E = KE + PE = −GMm/2r = −|KE|.
44. A satellite orbits at radius r. Energy required to escape to infinity from this orbit is:
Explanation: Total energy in orbit = −GMm/2r. Energy at infinity = 0. Energy needed = 0 − (−GMm/2r) = GMm/2r.
45. At what height above Earth's surface does g equal g/4? (R = 6400 km)
Explanation: g_h = g/(1+h/R)². g/4 = g/(1+h/R)² → (1+h/R)² = 4 → 1+h/R = 2 → h = R = 6400 km.
46. Earth (mass M_E) and Moon (mass M_M) are at distance d apart. The gravitational force on the Moon is F. If Earth's mass is doubled, the new force is:
Explanation: F = GM_E M_M/d². If M_E → 2M_E: F' = G(2M_E)M_M/d² = 2F.
47. A satellite has orbital radius 4 times that of another satellite with period 8 days. Period of the outer satellite is:
Explanation: T ∝ r^(3/2). T₂/T₁ = (4r/r)^(3/2) = 4^(3/2) = 8. T₂ = 8×8 = 64 days.
48. Masses M and 4M are 3 m apart. Where is the gravitational field zero between them?
Explanation: GM/x² = G(4M)/(3−x)². 1/x² = 4/(3−x)². (3−x)² = 4x². 3−x = 2x → x = 1 m from M.
49. A spacecraft moves from circular orbit r₁ to r₂ (r₂ > r₁) via Hohmann transfer. Total energy change is:
Explanation: E₁ = −GMm/2r₁. E₂ = −GMm/2r₂. ΔE = E₂−E₁ = −GMm/(2r₂) + GMm/(2r₁) = GMm(1/r₁−1/r₂)/2 > 0 (energy input needed for higher orbit).
50. Two masses M and m (initially separated by ∞) are brought to separation r. Work done by gravity is:
Explanation: PE at ∞ = 0. PE at r = −GMm/r. Work done by gravity = ΔPE (lost) = 0 − (−GMm/r) = GMm/r (positive — gravity does positive work as masses approach).
51. A satellite is boosted to a higher orbit. Compared to the original orbit, the satellite in the new orbit has:
Explanation: Higher orbit: r increases → v = √(GM/r) decreases (less speed). But total energy = −GMm/2r increases (less negative = more energy). Counter-intuitive: slower satellite has more total mechanical energy.
52. Tidal forces on Earth are caused by:
Explanation: The side of Earth facing the Moon experiences stronger pull than the centre, and the far side weaker. This differential (tidal) force stretches Earth, creating two tidal bulges and the twice-daily tides.
53. Minimum orbital period of any satellite around a uniform density planet (density ρ) is independent of planet's size and equals:
Explanation: For orbital radius = R (surface): T = 2π√(R³/GM) = 2π√(R³/(G×4πR³ρ/3)) = 2π√(3/(4πGρ)). The period depends only on density, not size.
54. At latitude λ on Earth's surface (angular velocity ω, radius R), effective g is approximately:
Explanation: The centripetal acceleration at latitude λ = ω²R cosλ directed toward the rotation axis. Its component along the local vertical (radially outward) = ω²R cos²λ. Effective g = g − ω²R cos²λ.
55. Einstein's equivalence principle states:
Explanation: Einstein's equivalence principle: a uniformly accelerating frame is locally indistinguishable from a gravitational field. Conversely, a freely falling frame is locally equivalent to a gravity-free inertial frame (the basis of general relativity).
56. The gravitational self-energy (energy of assembly) of a uniform solid sphere (mass M, radius R) is:
Explanation: Gravitational self-energy = −3GM²/5R (negative because energy is released when assembling mass from infinity). This represents the binding energy of the sphere.
57. g decreases from surface value as you go up OR down. The rate of decrease is faster:
Explanation: Going up: g ∝ 1/(R+h)² decreases as 2g/R per unit height initially (rate = 2g/R). Going down: g decreases as g/R per unit depth (rate = g/R). Rate of decrease is twice as fast going up compared to going down, near the surface.
58. For a satellite in orbit, total mechanical energy is negative. This means:
Explanation: Negative total energy means the satellite is gravitationally bound to Earth. To escape, it needs additional energy to make total energy ≥ 0.
59. Planet X has orbital period 8 years and orbital radius 4 AU. Planet Y has period 27 years. Y's orbital radius is:
Explanation: T² ∝ r³. For X: 64 = 4³ = 64 ✓ (ratio 1:1). For Y: 27² = 729. r³ = 729 → r = 9 AU.
60. A person inside a freely falling lift feels weightless because:
Explanation: In free fall, both the person and lift accelerate at g downward. The floor does not push up (N = 0), so the person experiences zero apparent weight.
61. A geosynchronous satellite has orbital period:
Explanation: A geosynchronous (geostationary) satellite has period = Earth's rotation period ≈ 23 h 56 min (sidereal day), so it stays above the same longitude.
62. If Earth's radius shrinks to half but mass stays the same, surface g becomes:
Explanation: g = GM/R². If R → R/2: g' = GM/(R/2)² = 4GM/R² = 4g.
63. Planet P has same mass as Earth but twice the radius. Escape velocity from P compared to Earth:
Explanation: v_e = √(2GM/R). Same M, double R: v_e' = √(2GM/2R) = √(GM/R) = v_e/√2.
64. A satellite orbits at 3R from Earth's centre (R = Earth's radius, g₀ = 10 m/s², R = 6.4×10⁶ m). Orbital speed is:
Explanation: v = √(gR²/r) = √(10×(6.4×10⁶)²/(3×6.4×10⁶)) = √(10×6.4×10⁶/3) = √(2.13×10⁷) ≈ 4620 m/s ≈ 4.6 km/s.
65. A planet moves in an ellipse. At aphelion (distance r_a, speed v_a) and perihelion (r_p, v_p). Which relation holds?
Explanation: Conservation of angular momentum: L = mv_a r_a = mv_p r_p → v_a r_a = v_p r_p. Since r_a > r_p, v_a
66. A uniform ring (mass M, radius a) has gravitational field along its axis at distance x from centre:
Explanation: By integration, the field along the axis of a ring: g = GMx/(a²+x²)^(3/2). It is zero at centre (x=0) and maximises at x = a/√2.
67. The ratio g_pole/g_equator is approximately:
Explanation: g_equator ≈ 9.780 m/s², g_pole ≈ 9.832 m/s². Ratio ≈ 9.832/9.780 ≈ 1.0053 ≈ 1.003 (approximate). The difference is about 0.5%.
68. A satellite is in circular orbit at radius 2R. It loses energy due to air drag and spirals to orbit at R. As it spirals inward, speed:
Explanation: v = √(GM/r). As r decreases, v increases. Losing energy paradoxically increases speed — the satellite falls into a faster, lower orbit, and the loss in PE more than compensates for the gain in KE.
69. A small mass m is placed at the centre of a large hollow spherical shell (mass M). Force on m due to the shell is:
Explanation: By the shell theorem, the gravitational field everywhere inside a uniform shell is zero. So the force on m inside the shell is F = mg_inside = 0.
70. A tunnel is drilled straight through Earth's centre. A ball dropped in from one end undergoes SHM with period:
Explanation: T = 2π√(R/g) = 2π√(6.4×10⁶/10) ≈ 84 min — same as a surface-skimming satellite! This is a remarkable coincidence from Earth's uniform density approximation.
71. Two stars (mass M each) orbit their common centre of mass at separation 2r. Their orbital period is:
Explanation: Each star orbits at radius r (half the separation). Gravitational force = centripetal force: GM²/(2r)² = Mω²r → ω² = GM/4r³. T = 2π/ω = 2π×2√(r³/GM)... Let me redo: GM²/4r² = Mω²r → ω² = GM/4r³ → T = 2π/ω = 4π√(r³/GM). The correct answer depends on exact setup.
72. Light bends around massive objects (gravitational lensing) because:
Explanation: In general relativity, mass curves spacetime. Light follows geodesics (straightest paths) in curved spacetime, which appear curved to distant observers. This was first confirmed during the 1919 solar eclipse.
73. Galaxy rotation curves (stars orbiting galaxy centres at nearly constant speed for all radii) suggest:
Explanation: Kepler/Newton predict v ∝ 1/√r for orbital speed beyond a galaxy's visible mass. Observations show v ≈ constant (flat rotation curve), implying much more mass (dark matter) than visible matter exists.
74. The vis-viva equation for a satellite at radius r in an ellipse (semi-major axis a) is:
Explanation: The vis-viva equation: v² = GM(2/r − 1/a). For circular orbit (r = a): v² = GM/r. For escape (a = ∞): v² = 2GM/r. It unifies circular, elliptical, and escape trajectories.
75. The Moon always shows the same face to Earth because:
Explanation: Tidal forces over billions of years synchronised the Moon's spin and orbital periods (both ≈ 27.3 days). This synchronous rotation keeps one face permanently toward Earth.
76. The L1 Lagrange point between Earth and Sun is significant because:
Explanation: At L1 (between Earth and Sun), the Sun's reduced gravity (partially cancelled by Earth's pull) allows objects to orbit the Sun with Earth's 1-year period despite being at a smaller orbital radius. SOHO and DSCOVR spacecraft orbit here.
77. To escape Earth's gravity from a circular orbit at radius r (speed v₀ = √(GM/r)), the speed must be increased to:
Explanation: Escape speed from any point at radius r: v_e = √(2GM/r) = √2 × √(GM/r) = √2 × v_orbital. The spacecraft must increase speed by factor √2 (increase of about 41%).
78. The geoid is:
Explanation: The geoid is the equipotential surface of Earth's gravitational field corresponding to mean sea level. It is bumpy (not a perfect ellipsoid) due to density variations in Earth's interior.
79. A uniform sphere of density ρ has orbital period T for a surface-skimming satellite. If ρ doubles (same mass, smaller radius), T:
Explanation: T = 2π√(R/g) = 2π√(R/(4πGρR/3)) = 2π√(3/(4πGρ)). T depends only on density: T ∝ 1/√ρ. If ρ doubles, T → T/√2. But surface-skimming orbit: T² ∝ R³/M ∝ (M/ρ)/M = 1/ρ, so T ∝ 1/√ρ → halving ρ doubles T.
80. A rocket in circular orbit at radius r fires its engine briefly backward (retrograde). The rocket then:
Explanation: Retrograde burn: speed decreases → KE decreases → total E becomes more negative → apogee stays at r but perigee drops → elliptical orbit with lower perigee. A second retrograde burn at perigee circularises at the lower orbit.
81. Light emitted from the surface of a massive star is redshifted when observed far away because:
Explanation: In general relativity, photons lose energy (Δf/f = −ΔΦ/c² where ΔΦ is the potential difference) as they climb out of a gravitational potential well → frequency decreases → red shift. This is called gravitational redshift.