Gravitation NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Gravitation question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Gravitation previous year questions
29 questions. Tap an option to check your answer.
Q1NEET 2021
The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :
Not quite — the answer is D.
v_e = √(2GM/R) with M = (4/3)πR³ρ gives v_e = R√(8πGρ/3), so v_e ∝ R at fixed density. Four times the radius gives 4v. Answering 2v uses v_e ∝ √R, which holds only at fixed mass, not fixed density.
Q2NEET 2021
A particle of mass 'm' is projected with a velocity v = kVe (k < 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is :
Not quite — the answer is D.
Energy conservation: (1/2)mk²(2GM/R) − GMm/R = −GMm/(R + h). This gives 1/(R + h) = (1 − k²)/R, so R + h = R/(1 − k²) and h = Rk²/(1 − k²). Using the flat-earth formula v²/2g would wrongly give h = k²R.
Q3NEET 2020 · Q143
A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ?
Not quite — the answer is B.
Gravitational force varies as 1/r², so at r = R + R/2 = 1.5R the force is 72 × (1/1.5)² = 72 × 4/9 = 32 N. The approximation g(1 − 2h/R) is only valid for h ≪ R and would wrongly give zero here.
Q4NEET 2019 · Q16
A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth?
Not quite — the answer is D.
Inside the earth g(d) = g(1 − d/R); at d = R/2, g is halved, so the weight becomes 100 N. 150 N wrongly uses a quarter reduction.
Q5NEET 2019 · Q29
The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:
Not quite — the answer is C.
W = GMm(1/R − 1/2R) = GMm/2R = ½ mgR. mgR assumes g stays constant up to height R, which is invalid.
Q6NEET 2018 · Q40
If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
Not quite — the answer is D.
g = GM_E/R² depends on G and the Earth's mass, not the Sun's. With G ten times larger, g becomes ten times larger, so the claim that g stays the same is wrong; the larger g also shortens the pendulum period and makes raindrops fall faster.
Q7NEET 2017 · Q157
The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:
Not quite — the answer is D.
For small heights g_h = g(1 − 2h/R) and at depth g_d = g(1 − d/R). Equating gives d = 2h = 2 km. Taking d = h ignores the factor 2 in the height formula.
Q8NEET 2017 (Gujarati paper) · Q141
A satellite of mass m is in circular orbit of radius 3R_E about earth (mass of earth M_E, radius of earth R_E). How much additional energy is required to transfer the satellite to an orbit of radius 9R_E?
Not quite — the answer is C.
Total energy in a circular orbit is −GMm/2r. ΔE = GM_E m/2 × (1/3R_E − 1/9R_E) = GM_E m/2 × 2/(9R_E) = GM_E m/(9R_E). Using only potential energies would double this to 2GM_E m/(9R_E).
Q9NEET 2017 (Gujarati paper) · Q144
Imagine earth to be a solid sphere of mass M and radius R. If the value of acceleration due to gravity at a depth 'd' below earth's surface is same as its value at a height 'h' above its surface and equal to g/4 (where g is the value of acceleration due to gravity on the surface of earth), the ratio of h/d will be:
Not quite — the answer is A.
Below the surface g(1 − d/R) = g/4 gives d = 3R/4. Above the surface g/(1 + h/R)² = g/4 gives 1 + h/R = 2, so h = R. Hence h/d = R/(3R/4) = 4/3. Using the small-height formula g(1 − 2h/R) would be wrong here because h is not small compared with R.
Q10NEET 2016 Phase 2 · Q12
A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g₀, the value of acceleration due to gravity at the earth's surface, is
Not quite — the answer is B.
Total energy of a circular orbit is −GMm/(2r) with r = R + h. Using GM = g₀R² gives E = −mg₀R²/(2(R + h)). The positive expression is the kinetic energy, not the total energy, which must be negative for a bound orbit.
Q11NEET 2016 Phase 1 · Q36
The ratio of escape velocity at earth (vₑ) to the escape velocity at a planet (vₚ) whose radius and mean density are twice as that of earth is:
Not quite — the answer is B.
vₑ = √(2GM/R) and M = (4/3)πR³ρ, so v ∝ R√ρ. For the planet v ∝ 2R × √(2ρ) = 2√2 times earth's value. Hence vₑ:vₚ = 1:2√2; ignoring the density change gives 1:2.
Q12AIPMT 2015
Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet, i.e. T² = Kr³, here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is F = GMm/r², here G is gravitational constant. The relation between G and K is described as:
Not quite — the answer is C.
Gravity provides the centripetal force: GMm/r² = m(2π/T)²r, so T² = (4π²/GM)r³. Hence K = 4π²/GM, i.e. GMK = 4π². GK = 4π² drops the Sun's mass.
Q13AIPMT 2015
Two spherical bodies of mass M and 5 M and radii R and 2 R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is:
Not quite — the answer is D.
Only internal forces act, so the centre of mass stays fixed. It is 5M × 12R/6M = 10R from the small body initially. At contact the centres are R + 2R = 3R apart and the small body is 2.5R from the centre of mass, so it moves 10R − 2.5R = 7.5R.
Q14AIPMT 2014
A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 × 10²⁴ kg) have to be compressed to be a black hole?
Not quite — the answer is B.
Setting escape speed √(2GM/R) equal to c gives R = 2GM/c² = 2 × 6.67 × 10⁻¹¹ × 5.98 × 10²⁴/(9 × 10¹⁶) ≈ 8.9 × 10⁻³ m, which is of the order 10⁻² m.
Q15NEET 2013
A body of mass m is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be
Not quite — the answer is B.
ΔU = GMm/R − GMm/3R = (2/3)GMm/R = (2/3)mgR, using GM = gR². The answer 2mgR comes from wrongly using mgh with h = 2R, which is valid only for heights much smaller than R.
Q16NEET 2013
Infinite number of bodies, each of mass 2 kg, are situated on x-axis at distances 1 m, 2 m, 4 m, 8 m, … respectively, from the origin. The resulting gravitational potential due to this system at the origin will be
Not quite — the answer is D.
V = −Gm(1/1 + 1/2 + 1/4 + 1/8 + …) = −G × 2 × [1/(1 − 1/2)] = −G × 2 × 2 = −4G. Forgetting the 2 kg mass gives −2G-type answers; the sum of the geometric series is 2.
Q17AIPMT 2012 Mains
If ve is escape velocity and v₀ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by
Not quite — the answer is D.
Near the surface, orbital speed v₀ = √(GM/R) and escape speed ve = √(2GM/R), so ve = √2·v₀. The escape speed is the larger one, so v₀ = √2·ve is reversed.
Q18AIPMT 2012 Prelims
The height at which the weight of a body becomes 1/16th of its weight on the surface of earth (radius R) is
Not quite — the answer is D.
g_h = g/(1 + h/R)². For g_h = g/16, 1 + h/R = 4, so h = 3R. 4R is the distance from the centre of the earth, not the height above the surface.
Q19AIPMT 2012 Prelims
A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is
Not quite — the answer is D.
Orbit radii are 6R (geostationary, T = 24 h) and 3R (height 2R). By Kepler's law T ∝ r^(3/2): T = 24 × (3R/6R)^(3/2) = 24/(2√2) = 6√2 h. Using heights (5R and 2R) instead of orbital radii gives a wrong value.
Q20AIPMT 2012 Prelims
A spherical planet has a mass Mp and diameter Dp. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to
Not quite — the answer is B.
g = GM/R² and R = Dₚ/2, so g = GMₚ/(Dₚ/2)² = 4GMₚ/Dₚ². Acceleration does not depend on the particle's mass m, so expressions containing m are forces, not accelerations.
Q21AIPMT 2011 Mains
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The magnitude of the gravitational potential at a point situated at a/2 distance from the centre, will be
Not quite — the answer is D.
Potential due to the central particle at a/2 is −GM/(a/2) = −2GM/a; inside a shell the potential is constant at −GM/a. Total = −3GM/a, magnitude 3GM/a. Taking only the particle's contribution gives 2GM/a, forgetting that the shell's potential inside is not zero.
Q22AIPMT 2011 Mains
A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is
Not quite — the answer is C.
To not return, the particle must reach infinity: ½mu² = GMm/R, so u = √(2GM/R) (= √(2gR)). √(2GM/R²) and √(2gR²) are dimensionally wrong as speeds.
Q23AIPMT 2011 Prelims
A planet moving along an elliptical orbit is closest to the sun at a distance r₁ and farthest away at a distance of r₂. If v₁ and v₂ are the linear velocities at these points respectively, then the ratio v₁/v₂ is
Not quite — the answer is C.
Gravity exerts no torque about the sun, so angular momentum is conserved; at perihelion and aphelion the velocity is perpendicular to the radius, so mv₁r₁ = mv₂r₂ and v₁/v₂ = r₂/r₁. The planet moves faster when closer, so the ratio must exceed 1, which rules out r₁/r₂.
Q24AIPMT 2010 Prelims · Q12
The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be
Not quite — the answer is C.
Orbital speed v = √(GM/r), so v ∝ 1/√r. v_B/v_A = √(4R/R) = 2, so v_B = 2 × 3V = 6V. Using v ∝ 1/r would give 12V.
Q25AIPMT 2010 Prelims · Q13
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre, will be
Not quite — the answer is B.
Inside the shell its potential is constant at −GM/a. The point mass at the centre contributes −GM/(a/2) = −2GM/a. Total V = −2GM/a − GM/a = −3GM/a. Forgetting the shell gives −2GM/a.
Q26AIPMT 2009 Prelims · Q19
The Figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t₁ is the time for the planet to move from C to D and t₂ is the time to move from A to B then:
Not quite — the answer is B.
Kepler's second law: the radius vector sweeps equal areas in equal times, so time is proportional to area swept. Area SCD = 2 × area SAB gives t₁ = 2t₂. The fact that the planet moves faster near the sun is already accounted for by the equal-area law.
Q27AIPMT 2007 Prelims · Q31
Two satellites of earth, S₁ and S₂ are moving in the same orbit. The mass of S₁ is four times the mass of S₂. Which one of the following statements is true?
Not quite — the answer is B.
Orbital speed v = √(GM/r) depends only on the orbit radius, so both move with the same speed (and period). Potential and kinetic energies are proportional to the satellite mass, so they differ by a factor of 4.
Q28AIPMT 2006 Prelims · Q34
The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fvₑ, where vₑ is its escape velocity from the surface of the earth. The value of f is :
Not quite — the answer is B.
Escape velocity from a point at distance r from the centre is √(2GM/r). The platform is at r = R + R = 2R, so the escape velocity is vₑ/√2 and f = 1/√2. Escape becomes easier, not harder, so f cannot exceed 1.
Q29AIPMT 2005 Prelims · Q6
Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then :
Not quite — the answer is A.
g = GM/R² = (4/3)πGρR, so at equal density g ∝ R. A planet with 3 times the radius has g' = 3g. g/9 would follow only if the mass stayed the same.
How often Gravitation came up
Questions from this chapter in each paper on this site.
- NEET 2021: 2
- NEET 2020: 1
- NEET 2019: 2
- NEET 2018: 1
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 2
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 1
- AIPMT 2015: 2
- AIPMT 2014: 1
- NEET 2013: 2
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 3
- AIPMT 2011 Mains: 2
- AIPMT 2011 Prelims: 1
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 1
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 1
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