68 free MCQs on Gravitation with worked answers and explanations. Universal gravitation, orbital mechanics, gravitational potential, and escape velocity.
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g is maximum at Earth's surface; going up, it falls off as 1/r² (inverse-square); going down, it falls off linearly with depth (since only the mass enclosed within radius r contributes), reaching zero at the centre.
Q1.
Value of gravitational acceleration g at Earth's surface is approximately:
Answer: A. 9.8 m/s<sup>2</sup>
Why: g = 9.8 m/s<sup>2</sup> approximately. For calculations, g = 10 m/s<sup>2</sup> is often used.
Q2.
As we go higher from Earth's surface, the value of g:
Answer: B. Decreases
Why: g = GM/r<sup>2.</sup> As r increases (going higher), g decreases.
Q3.
Universal gravitational constant G has the value:
Answer: A. 6.67 x 10<sup>-11</sup> N m<sup>2</sup>/kg<sup>2</sup>
Why: G = 6.674 x 10<sup>-11</sup> N m<sup>2</sup> kg<sup>-2</sup> (Universal Gravitational Constant).
Q4.
Kepler's second law (equal areas in equal time) is a consequence of:
Answer: B. Conservation of angular momentum
Why: Equal areas swept in equal time means angular momentum is constant: a result of no torque acting on the planet.
Q5.
Escape velocity from Earth's surface is approximately:
Answer: B. 11.2 km/s
Why: Escape velocity = sqrt(2gR) = sqrt(2 x 9.8 x 6.4 x 10<sup>6</sup>) = approx 11.2 km/s.
Q6.
The gravitational force between two masses is proportional to:
Answer: B. Product of masses
Why: F = Gm<sub>1</sub>m<sub>2</sub>/r<sup>2.</sup> Force is proportional to product of masses m<sub>1</sub> x m<sub>2</sub>.
Q7.
Orbital velocity of a satellite near Earth's surface is approximately:
Answer: B. 7.9 km/s
Why: Orbital velocity v<sub>o</sub> = sqrt(gR) = sqrt(9.8 x 6.4 x 10<sup>6</sup>) approx 7.9 km/s.
Q8.
Time period of a geostationary satellite is:
Answer: B. 24 hours
Why: Geostationary satellites have T = 24 hours (same as Earth's rotation), so they appear stationary over the equator.
Q9.
An astronaut in an orbiting satellite experiences weightlessness because:
Answer: C. Both satellite and astronaut are in free fall
Why: Weightlessness occurs because the satellite and astronaut are both in free fall (continuously falling toward Earth as they orbit).
Q10.
Gravitational force is a _____ force:
Answer: C. Long-range, always attractive
Why: Gravity acts over infinite range and is always attractive (unlike electromagnetic force which can be attractive or repulsive).
Q11.
If distance between two masses doubles, gravitational force becomes:
Answer: C. One-fourth
Why: F = Gm<sub>1</sub>m<sub>2</sub>/r<sup>2.</sup> If r doubles, F = Gm<sub>1</sub>m<sub>2</sub>/(2r)<sup>2</sup> = F/4. Force becomes one-fourth.
Q12.
Kepler's third law states that T<sup>2</sup> is proportional to:
Answer: C. r<sup>3</sup>
Why: Kepler's third law: T<sup>2</sup> is proportional to r<sup>3</sup> (square of period proportional to cube of orbital radius).
Q13.
Weight of a body on Moon is about _____ of weight on Earth:
Answer: C. 1/6
Why: Moon's gravity is about 1/6 of Earth's gravity. So weight on Moon = W<sub>earth</sub> / 6.
Q14.
Gravitational PE of a mass m at height h above Earth's surface (for h << R) is:
Answer: A. mgh
Why: Near Earth's surface, gravitational PE = mgh (taking surface as reference, PE is positive above it).
Q15.
The orbits of all planets around Sun are:
Answer: B. Elliptical
Why: Kepler's first law: Planets move in elliptical orbits with the Sun at one focus.
Q16.
At the center of Earth, the value of g is:
Answer: D. Zero
Why: At Earth's center, gravitational pull from all surrounding mass cancels out. g = 0 at the center.
Q17.
A planet closer to the Sun moves _____ compared to a distant planet:
Answer: B. Faster
Why: From Kepler's third law, T<sup>2</sup> proportional to r<sup>3</sup> means inner planets (smaller r) have smaller T (faster orbital speed).
Q18.
Which force keeps the Moon in orbit around Earth?
Answer: C. Gravitational force
Why: Earth's gravitational pull on the Moon provides the centripetal force to keep it in orbit.
Q19.
Relation between escape velocity (ve) and orbital velocity (vo) at Earth's surface:
Answer: C. ve = sqrt(2) x vo
Why: vo = sqrt(gR), ve = sqrt(2gR) = sqrt(2) x vo. Escape velocity is sqrt(2) times orbital velocity.
Q20.
Gravitational force obeys Newton's:
Answer: C. Third law
Why: Every gravitational force has an equal and opposite partner (Newton's 3rd law). Earth pulls Moon; Moon pulls Earth equally.
Q21.
A satellite orbits at height equal to Earth's radius R (i.e., at distance 2R from center). Orbital velocity compared to surface orbital velocity:
Answer: B. 1/sqrt(2) times
Why: vo = sqrt(GM/r). At 2R: vo_new = sqrt(GM/2R) = vo_surface/sqrt(2).
Q22.
At what height h above Earth's surface is g reduced to one-fourth its surface value?
Answer: B. R
Why: g at height h: g' = g/(1+h/R)<sup>2.</sup> For g' = g/4: (1+h/R)<sup>2</sup> = 4. 1+h/R = 2. h = R.
Q23.
Time period of satellite at height h from Earth's surface:
Answer: B. 2*pi*sqrt((R+h)<sup>3</sup>/(GM))
Why: T = 2*pi*sqrt(r<sup>3</sup>/(GM)) where r = R+h. Or T = 2*pi*sqrt((R+h)<sup>3</sup>/GM).
Q24.
Two bodies of mass M and m are at distance r. At what distance from M is the gravitational field zero?
Answer: A. r*sqrt(M)/(sqrt(M)+sqrt(m))
Why: Field from M: GM/x<sup>2.</sup> Field from m: Gm/(r-x)<sup>2.</sup> Setting them equal gives x = r*sqrt(M)/(sqrt(M)+sqrt(m)), the distance from M.
Q25.
If Earth suddenly stopped rotating, weight of a person at equator would:
Answer: A. Increase
Why: Earth's rotation causes centrifugal effect that slightly reduces apparent weight. Without rotation, weight = mg (true weight, slightly more).
Q26.
Orbital velocity of satellite at height h from Earth (radius R, mass M):
Answer: B. sqrt(GM/(R+h))
Why: Orbital velocity: v<sub>o</sub> = sqrt(GM/(R+h)) for a satellite at height h.
Q27.
A satellite in low Earth orbit completes one revolution in about:
Answer: B. 90 min
Why: Low Earth orbit satellites (ISS, etc.) have orbital periods of about 90 minutes.
Q28.
Total mechanical energy of a satellite in circular orbit of radius r (mass m, Earth mass M):
Answer: B. -GMm/(2r)
Why: KE = GMm/(2r), PE = -GMm/r. Total E = KE+PE = GMm/(2r) - GMm/r = -GMm/(2r). (Negative means bound system.)
Q29.
Gravitational potential V at distance r from mass M is:
Answer: B. -GM/r
Why: Gravitational potential V = -GM/r (energy per unit mass). It is always negative.
Q30.
Energy required to send satellite from surface to orbit at radius r=2R (R=Earth's radius, M=Earth's mass):
Answer: A. GMm/(4R)
Why: Surface total energy = -GMm/R. Orbital total energy at 2R = -GMm/(4R). Delta E = -GMm/4R - (-GMm/R) = GMm(1/R - 1/4R) = 3GMm/(4R).
Q31.
At poles, g is slightly greater than at equator because:
Answer: C. Polar radius is smaller so closer to center
Why: Earth is oblate: polar radius is smaller. Since g = GM/r<sup>2</sup>, smaller r at poles means larger g.
Q32.
Gravitational field intensity at surface of Earth (radius R, mass M) is:
Answer: B. GM/R<sup>2</sup>
Why: Gravitational field intensity (force per unit mass) = GM/R<sup>2</sup> = g (acceleration due to gravity at surface).
Q33.
Ratio of orbital velocity to escape velocity at Earth's surface:
Answer: B. 1/sqrt(2)
Why: vo = sqrt(gR), ve = sqrt(2gR). Ratio vo/ve = sqrt(gR)/sqrt(2gR) = 1/sqrt(2).
Q34.
If mass of Earth doubled but radius stayed same, escape velocity would:
Answer: C. Increase by sqrt(2)
Why: ve = sqrt(2GM/R). If M doubles, ve = sqrt(2 x 2GM/R) = sqrt(2) x original ve.
Q35.
A planet has mass 2M and radius 2R. Surface gravity compared to Earth:
Answer: B. Half
Why: g' = G(2M)/(2R)<sup>2</sup> = 2GM/4R<sup>2</sup> = GM/(2R<sup>2</sup>) = g/2. Surface gravity is half of Earth.
Q36.
Black holes have escape velocity equal to or greater than:
Answer: B. Speed of light
Why: A black hole is an object where escape velocity >= speed of light (c = 3 x 10<sup>8</sup> m/s). Not even light can escape.
Q37.
Binding energy of a satellite in orbit at radius r is:
Answer: B. GMm/(2r)
Why: Binding energy = -total energy = -(-GMm/(2r)) = GMm/(2r). It represents the energy needed to free the satellite.
Q38.
For an orbit to be geosynchronous, it must be over:
Answer: B. Equator
Why: A geostationary orbit must be directly over the equator with zero inclination to remain fixed over one point.
Q39.
Value of g at depth d below Earth's surface:
Answer: A. g(1-d/R)
Why: g at depth d = g(1 - d/R). At center (d=R), g = 0. At surface (d=0), g = g.
Q40.
The gravitational PE of a body of mass m at distance r from Earth's center (M = Earth mass) is:
Answer: A. -GMm/r
Why: Gravitational PE = -GMm/r. The negative sign indicates the body is in a bound (attractive) potential well.
Q41.
Planet A has twice the mass and four times the radius of planet B. Ratio of g<sub>A</sub> to g<sub>B</sub>:
Answer: A. 1:8
Why: g = GM/R<sup>2.</sup> g<sub>A</sub>/g<sub>B</sub> = (2M<sub>B</sub>)/(4R<sub>B</sub>)<sup>2</sup> x (R<sub>B</sub>)<sup>2</sup>/M<sub>B</sub> = 2/16 = 1/8.
Q42.
Two planets have time periods in ratio 8:1. Ratio of orbital radii:
Answer: B. 4:1
Why: T<sup>2</sup> proportional to r<sup>3</sup> (Kepler). (T<sub>1</sub>/T<sub>2</sub>)<sup>2</sup> = (r<sub>1</sub>/r<sub>2</sub>)<sup>3.</sup> (8)<sup>2</sup> = (r<sub>1</sub>/r<sub>2</sub>)<sup>3.</sup> 64 = (r<sub>1</sub>/r<sub>2</sub>)<sup>3.</sup> r<sub>1</sub>/r<sub>2</sub> = 4.
Q43.
A satellite changes orbit from r<sub>1</sub> to r<sub>2</sub> (r<sub>2</sub> > r<sub>1</sub>). Change in KE is:
Answer: B. Negative (decreases)
Why: KE = GMm/(2r). As r increases, KE decreases. Energy must be added to raise orbit but KE actually decreases.
Q44.
Angular velocity of Moon around Earth: T = 27 days. Angular velocity in rad/s:
Answer: A. 2.7 x 10<sup>-6</sup> rad/s
Why: omega = 2*pi/T = 2*pi/(27 x 24 x 3600) = 6.28/2332800 approx 2.7 x 10<sup>-6</sup> rad/s.
Q45.
A body is projected vertically upward from Earth's surface with speed ve/2 (ve = escape speed). Max height reached:
Answer: A. R/3
Why: Energy: (1/2)m(ve/2)<sup>2</sup> - GMm/R = -GMm/(R+h). (ve<sup>2</sup>/8) - GM/R = -GM/(R+h). ve<sup>2</sup> = 2gR = 2GM/R. GM/(4R) - GM/R = -GM/(R+h). -3GM/(4R) = -GM/(R+h). R+h = 4R/3. h = R/3.
Q46.
Escape velocity on the Moon (g<sub>moon</sub> = g/6, R<sub>moon</sub> = R/4):
Answer: B. ve*sqrt(6)/4
Why: ve_moon = sqrt(2 x g<sub>moon</sub> x R<sub>moon</sub>) = sqrt(2 x g/6 x R/4) = sqrt(gR/12) = ve x sqrt(1/24) = ve/sqrt(24). Hmm. ve = sqrt(2gR). ve_moon = sqrt(2(g/6)(R/4)) = sqrt(gR/12) = sqrt(gR)/sqrt(12). ve_moon/ve = sqrt(1/12)/sqrt(2) = 1/sqrt(24). So ve_moon = ve/sqrt(24).
Q47.
A satellite is in circular orbit. When it loses energy due to air resistance, its orbital speed:
Answer: B. Increases
Why: When energy decreases, satellite moves to lower orbit (smaller r). Since v = sqrt(GM/r), smaller r means higher speed. Paradoxically, losing energy makes it speed up.
Q48.
A satellite at radius r from Earth center has angular momentum L. If radius decreases to r/4 (no external torque), new angular momentum is:
Answer: C. L
Why: No external torque means angular momentum is conserved. L = constant regardless of r change.
Q49.
A planet is at perihelion (closest) and aphelion (farthest). Speed ratio v<sub>perihelion</sub> : v<sub>aphelion</sub> equals:
Answer: A. r<sub>aph</sub> : r<sub>per</sub>
Why: By conservation of angular momentum: m x v<sub>per</sub> x r<sub>per</sub> = m x v<sub>aph</sub> x r<sub>aph</sub>. So v<sub>per</sub>/v<sub>aph</sub> = r<sub>aph</sub>/r<sub>per</sub>.
Q50.
Gravitational potential at a point is -5 x 10<sup>7</sup> J/kg. Work done to bring 1 kg mass from infinity to this point is:
Answer: B. -5 x 10<sup>7</sup> J
Why: Work done = m x (V<sub>final</sub> - V<sub>infinity</sub>) = 1 x (-5 x 10<sup>7</sup> - 0) = -5 x 10<sup>7</sup> J. Negative means energy released.
Q51.
Tidal forces on Earth due to Moon cause:
Answer: B. Ocean bulges on both sides facing and opposite Moon
Why: Tidal forces stretch Earth along the Moon-Earth axis, creating bulges on both the near AND far sides. Hence two high tides per day.
Q52.
If Earth's mass is M, radius R, orbital velocity at height h = R (i.e., r = 2R) in terms of g and R:
Answer: A. sqrt(gR/2)
Why: vo = sqrt(GM/(2R)). GM = gR<sup>2.</sup> vo = sqrt(gR<sup>2</sup>/(2R)) = sqrt(gR/2).
Q53.
Minimum energy needed to launch a satellite from Earth's surface to circular orbit at radius 2R:
Answer: B. 3GMm/(4R)
Why: E<sub>surface</sub> = -GMm/R. E<sub>orbit</sub>(2R) = -GMm/4R. Delta E = -GMm/4R - (-GMm/R) = GMm(1/R - 1/4R) = 3GMm/(4R).
Q54.
Gravitational field is zero at all interior points of a uniform hollow spherical shell. This is because:
Answer: B. Forces from all shell mass cancel inside due to symmetry
Why: Inside a uniform spherical shell, gravitational contributions from all parts of the shell cancel by symmetry. This is the shell theorem.
Q55.
Two satellites orbit at 2R and 3R from Earth center. Period ratio T<sub>1</sub>:T<sub>2</sub>:
Answer: C. 2*sqrt(2):3*sqrt(3)
Why: T proportional to r<sup>3/2</sup>. T<sub>1</sub>:T<sub>2</sub> = (2R)<sup>3/2</sup> : (3R)<sup>3/2</sup> = 2*sqrt(2) : 3*sqrt(3).
Q56.
A satellite in orbit has KE = K. Its total energy is:
Answer: B. -K
Why: For circular orbit: KE = GMm/(2r) = K. PE = -GMm/r = -2K. Total = KE + PE = K - 2K = -K.
Q57.
Gravitational force between two 1 kg masses 1 m apart is:
Answer: A. 6.67 x 10<sup>-11</sup> N
Why: F = Gm<sub>1</sub>m<sub>2</sub>/r<sup>2</sup> = 6.674 x 10<sup>-11</sup> x 1 x 1 / 1<sup>2</sup> = 6.67 x 10<sup>-11</sup> N.
Q58.
A tunnel is drilled through Earth's center. A ball dropped into it undergoes:
Answer: C. Simple harmonic motion
Why: Inside Earth, g decreases linearly with depth (g' = g(1-d/R) = g x r/R). This gives restoring force F = -kr, which is SHM.
Q59.
Binary stars of mass M each orbit their center of mass at distance d apart. Orbital period T in terms of G, M, d:
Answer: A. 2*pi*sqrt(d<sup>3</sup>/(2GM))
Why: Each star orbits at distance d/2 from center. Centripetal force = GM<sup>2</sup>/d<sup>2.</sup> (d/2)x omega<sup>2</sup> x M = GM<sup>2</sup>/d<sup>2.</sup> omega<sup>2</sup> = 2GM/d<sup>3.</sup> T = 2*pi/omega = 2*pi*sqrt(d<sup>3</sup>/(2GM)).
Q60.
Chandrasekhar limit refers to the maximum mass of:
Answer: C. A white dwarf star
Why: Chandrasekhar limit (~1.4 solar masses) is the maximum mass of a stable white dwarf. Above this, it collapses further.
Q61.
A planet has the same radius as Earth but twice its mass. If Earth escape velocity is 11.2 km/s, the escape velocity from this planet is:
Answer: B. 15.8 km/s
Why: v<sub>e</sub> ∝ √M for fixed R, so v = 11.2√2 ≈ 15.8 km/s.
Q62.
A satellite orbits at radius r with period T. If the orbital radius is increased to 4r, the new period is:
Answer: C. 8T
Why: By Kepler's third law T ∝ r<sup>3/2</sup>, so 4<sup>3/2</sup> = 8; new period = 8T.
Q63.
The acceleration due to gravity at a depth equal to half the Earth radius (compared with the surface value g) is:
Answer: A. g/2
Why: g' = g(1 − d/R) = g(1 − 1/2) = g/2.
Q64.
At a height equal to the Earth radius above the surface, the acceleration due to gravity becomes:
Answer: B. g/4
Why: g' = g R²/(R + h)² = g R²/(2R)² = g/4.
Q65.
The total mechanical energy of a satellite of mass m in a circular orbit of radius r around a planet of mass M is:
Answer: B. −GMm/2r
Why: E = KE + PE = GMm/2r − GMm/r = −GMm/2r.
Q66.
A planet moves in an elliptical orbit. If its aphelion distance is 4 times its perihelion distance, the ratio of its perihelion speed to aphelion speed is:
Answer: B. 4
Why: Conservation of angular momentum: v<sub>p</sub>·r<sub>p</sub> = v<sub>a</sub>·r<sub>a</sub>, so v<sub>p</sub>/v<sub>a</sub> = r<sub>a</sub>/r<sub>p</sub> = 4.
Q67.
A planet has twice the radius of Earth but the same mean density. The surface gravity on the planet, compared with Earth g, is:
Answer: B. 2g
Why: g = (4/3)πGρR, so g ∝ R at constant density; doubling R gives 2g.
Q68.
Two satellites move in the same circular orbit around Earth, one having twice the mass of the other. Their orbital speeds are:
Answer: A. equal
Why: Orbital speed v = √(GM/r) is independent of the satellite mass, so they are equal.