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  3. Acceleration due to Gravity and Its Variation

Gravitation · JEE & NEET Physics

Acceleration due to Gravity and Its Variation: notes and previous year questions

Where g = GM/R² comes from, and how g changes going up, going down, and with latitude on a spinning Earth.

Acceleration due to Gravity and Its Variation in short

  • g = GM/R² is the same for every falling body: the body's mass cancels.
  • G is a universal constant; g depends on the planet and on where you are.
  • Above the surface g falls as an inverse square of the distance from the centre.
  • For small heights, g falls by 2h/R: don't forget the 2.

1What is g?

The acceleration due to gravity, gg, is the acceleration of a freely falling body. The pull on a body is its weight mgmg, and it equals Newton's pull:

mg=GMmR2 ⇒ g=GMR2mg = \frac{GMm}{R^2}\ \Rightarrow\ g = \frac{GM}{R^2}The body's own mass m cancels, so a heavy and a light ball fall together (with no air).

With the Earth's numbers: g=6.67×10−11×6×1024÷(6.4×106)2≈9.8g = 6.67\times10^{-11} \times 6\times10^{24} \div (6.4\times10^6)^2 \approx 9.8 m/s² (often rounded to 10). A handy swap in problems is GM=gR2GM = gR^2.

Since g∝M/R2g \propto M/R^2, a planet with twice the Earth's mass and twice its radius has g×2÷4=g/2g \times 2 \div 4 = g/2. One with half the mass and half the radius has 2g2g.

Big GSmall g
Value6.67×10−116.67\times10^{-11} N m²/kg²9.8 m/s² at the surface
What it isa universal constantthe acceleration of a falling body
Changes?neverwith height, depth and latitude
Depends onnothingthe planet's M and R

2Going up

At a height hh the distance from the centre is R+hR + h:

gh=GM(R+h)2=g(RR+h)2g_h = \frac{GM}{(R+h)^2} = g\Big(\frac{R}{R+h}\Big)^2

At h=Rh = R: g/4≈2.45g/4 \approx 2.45 m/s². At h=2Rh = 2R: g/9g/9. At h=3Rh = 3R: g/16g/16. For g/2g/2: R+h=2RR + h = \sqrt2 R, so h≈0.414R≈2650h \approx 0.414R \approx 2650 km.

Weight changes the same way, because the mass does not change: 63 N on the ground becomes 63×4/9=2863 \times 4/9 = 28 N at h=R/2h = R/2. And 72 N becomes 18 N at h=Rh = R.

3Small heights

Write gh=g(1+h/R)−2g_h = g(1 + h/R)^{-2}. For small xx, (1+x)n≈1+nx(1 + x)^n \approx 1 + nx, so:

gh≈g(1−2hR)g_h \approx g\Big(1 - \frac{2h}{R}\Big)Good while h is less than about 0.1R ≈ 640 km.

On Mount Everest, h=8.85h = 8.85 km: 2h/R=17.7÷6400≈0.00282h/R = 17.7 \div 6400 \approx 0.0028, so g≈9.77g \approx 9.77 m/s², 0.28% less. A 60 kg climber reads about 170 g less on a scale. At 32 km, gg is 1% less.

4Going down

Inside a hollow shell, the pulls cancel. From any point inside, a small patch on the near side is close but small; the matching patch on the far side is far away but bigger. The area grows as the distance squared and the pull falls as the distance squared, so each pair cancels exactly.

So at a depth dd in a uniform Earth, the shell of rock above pulls nowhere. Only the ball of radius R−dR - d below pulls, with mass M′=M(R−d)3/R3M' = M(R-d)^3/R^3:

gd=GM′(R−d)2=g(1−dR)g_d = \frac{GM'}{(R-d)^2} = g\Big(1 - \frac{d}{R}\Big)A straight-line fall, down to g = 0 at the centre (d = R).

gg is 60% of its surface value at d=0.4R=2560d = 0.4R = 2560 km, and g/2g/2 at d=R/2d = R/2. The real Earth has a denser core, so in deep mines gg at first rises slightly; problems assume uniform density unless told otherwise.

5Up versus down

From the centre, gg rises in a straight line to its largest value at the surface, then falls as 1/r21/r^2 outside.

For the same small distance, going up lowers gg about twice as much as going down (2h/R2h/R against d/Rd/R). At 64 km: 0.98g0.98g up, 0.99g0.99g down. At h=d=R/2h = d = R/2: 4g/9≈0.44g4g/9 \approx 0.44g up, 0.5g0.5g down.

Height giving the same gg as a depth of R/2R/2: g/2=g(R/(R+h))2g/2 = g(R/(R+h))^2, so h=(2−1)R≈0.41Rh = (\sqrt2 - 1)R \approx 0.41R.

6The spinning Earth

At a latitude λ\lambda you go round a circle of radius Rcos⁡λR\cos\lambda once a day. Part of the pull is used up keeping you on that circle, so a scale reads less:

g′=g−ω2Rcos⁡2λg' = g - \omega^2 R\cos^2\lambda

ω=2π/(24 h)≈7.3×10−5\omega = 2\pi/(24\text{ h}) \approx 7.3\times10^{-5} rad/s, so ω2R≈0.034\omega^2 R \approx 0.034 m/s², about 0.35% of gg. At the poles (λ=90°\lambda = 90°) g′=gg' = g, the largest; at the equator g′=g−ω2Rg' = g - \omega^2R, the smallest. The drop is a quarter of the equator's at λ=60°\lambda = 60°. The Earth's flattened shape adds a little more to the difference.

At a height too: g′≈g(1−2h/R)−ω2Rcos⁡2λg' \approx g(1 - 2h/R) - \omega^2R\cos^2\lambda.

Summary

Key ideas

  • g = GM/R² is the same for every falling body: the body's mass cancels.
  • G is a universal constant; g depends on the planet and on where you are.
  • Above the surface g falls as an inverse square of the distance from the centre.
  • For small heights, g falls by 2h/R: don't forget the 2.
  • Inside a hollow shell the pulls cancel, so below the surface only the inner ball pulls.
  • Below the surface g falls in a straight line, to zero at the centre.
  • g is greatest at the surface; for small equal distances, going up lowers it more than going down.
  • A tunnel through the Earth gives SHM with a period of about 84 minutes.
  • The spin lowers g by ω²R cos²λ: most at the equator, not at all at the poles.

Every equation

At the surface
g=GM/R2g = GM/R^2
Handy swap
GM=gR2GM = gR^2
Height h
gh=g (R/(R+h))2g_h = g\,(R/(R+h))^2
Small height
gh≈g(1−2h/R)g_h \approx g(1 - 2h/R)
Depth d
gd=g(1−d/R)g_d = g(1 - d/R)
Mass inside
M′=M(R−d)3/R3M' = M(R-d)^3/R^3
Half g above
h=(2−1)Rh = (\sqrt2 - 1)R
Tunnel
T=2πR/gT = 2\pi\sqrt{R/g}
Latitude
g′=g−ω2Rcos⁡2λg' = g - \omega^2R\cos^2\lambda
Equator
g′=g−ω2Rg' = g - \omega^2 R
Spin term
ω2R≈0.034 m/s2\omega^2 R \approx 0.034\ \text{m/s}^2
Height and latitude
g′≈g(1−2h/R)−ω2Rcos⁡2λg' \approx g(1 - 2h/R) - \omega^2R\cos^2\lambda

Previous year questions with solutions

Real JEE and NEET questions on acceleration due to gravity and its variation. Try each one before you open the solution.

Q1JEE Main 2026One correct option

The height in terms of radius of the earth (R)(R), at which the acceleration due to gravity becomes g9\frac{g}{9}, where gg is acceleration due to gravity on earth's surface, is

____\_\_\_\_ .

  1. A3R\sqrt{3}R
  2. B22R2\sqrt{2}R
  3. C2R2R
  4. D49R\frac{4}{9}R
Show answer and solution

Answer: Option C

Set gh=g(RR+h)2=g9g_{h} = g\left(\dfrac{R}{R + h}\right)^{2} = \dfrac{g}{9}. Then RR+h=13\dfrac{R}{R + h} = \dfrac{1}{3}, so R+h=3RR + h = 3R and h=2Rh = 2R.

The body is 3R3R from the centre but only 2R2R above the surface. B, 22R2\sqrt{2}R, comes from taking the distance as R2+h2\sqrt{R^{2} + h^{2}}, as if the height stood at right angles to the radius; it lies along the radius, so the two simply add. A and D come from no correct working. Check: at h=2Rh = 2R, (R3R)2=19\left(\dfrac{R}{3R}\right)^{2} = \dfrac{1}{9}.

Q2NEET 2025One correct option

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

  1. A32 N
  2. B36 N
  3. C16 N
  4. D27 N
Show answer and solution

Answer: Option D

At h=R3h = \dfrac{R}{3} the body is 4R3\dfrac{4R}{3} from the centre, so the pull is

48×(R4R/3)2=48×916=27 N48 \times \left(\dfrac{R}{4R/3}\right)^{2} = 48 \times \dfrac{9}{16} = 27\ \mathrm{N}

The trap is B, 36 N36\ \mathrm{N}, which multiplies by 34\dfrac{3}{4} without squaring. A, 32 N32\ \mathrm{N}, takes a third off the weight, as if gg fell in proportion to the height. C, 16 N16\ \mathrm{N}, keeps only a third of it.

Q3JEE Main 2025Numerical answer

Acceleration due to gravity on the surface of earth is ' gg '. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ________ g.

Show answer and solution

Answer: 9

The mass is unchanged and g=GMR2g = \dfrac{GM}{R^{2}}. A third of the diameter is a third of the radius, so

g′=GM(R/3)2=9GMR2=9gg' = \dfrac{GM}{(R/3)^{2}} = 9\dfrac{GM}{R^{2}} = 9g

The answer is 99.

The trap is 33, forgetting to square: gg goes as 1R2\dfrac{1}{R^{2}}, not 1R\dfrac{1}{R}. Another is 19\dfrac{1}{9}, the ratio upside down. A smaller planet of the same mass pulls harder at its surface, because its surface is closer to its centre.

Practice questions, easy to hard

Three questions from the acceleration due to gravity and its variation practice ladder: one easy, one medium, one hard.

Q4One correct option

Since g=GMR2g = \dfrac{GM}{R^{2}}, a planet's surface gg is in proportion to its mass and falls as the square of its radius. To compare two planets, write the ratio and let GG cancel:

g2g1=M2M1×(R1R2)2\dfrac{g_{2}}{g_{1}} = \dfrac{M_{2}}{M_{1}} \times \left(\dfrac{R_{1}}{R_{2}}\right)^{2}

If a question gives diameters, the radii are in the same ratio: halving the diameter halves the radius.

A planet has twice the Earth's mass and twice the Earth's diameter. What is gg at its surface, in terms of the Earth's gg?

  1. A2g2g
  2. Bg2\dfrac{g}{2}
  3. Cgg
  4. Dg4\dfrac{g}{4}
Show answer and solution

Answer: Option B

Twice the diameter is twice the radius, so gpg=2×(12)2=12\dfrac{g_{p}}{g} = 2 \times \left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{2}, and gp=g2g_{p} = \dfrac{g}{2}.

The trap is C, gg, expecting the two doublings to cancel. They would only if the radius entered to the first power; it enters squared, so it wins. A counts only the extra mass, and D only the larger radius.

Q5One correct option

Every value of gg below the surface value turns up twice on a line out from the centre: once inside the Earth and once above it. To find the depth that matches a given height, work out gg at the height exactly, then set g(1−dR)g\left(1 - \dfrac{d}{R}\right) equal to it. The rule d=2hd = 2h is for small heights only.

At a height of R2\dfrac{R}{2}, gg is 4g9\dfrac{4g}{9}. At what depth below the surface is gg the same?

  1. A4R9\dfrac{4R}{9}
  2. B5R9\dfrac{5R}{9}
  3. CRR
  4. DR4\dfrac{R}{4}
Show answer and solution

Answer: Option B

g(1−dR)=4g9g\left(1 - \dfrac{d}{R}\right) = \dfrac{4g}{9} gives dR=1−49=59\dfrac{d}{R} = 1 - \dfrac{4}{9} = \dfrac{5}{9}, so d=5R9d = \dfrac{5R}{9}.

The trap is A: 49\dfrac{4}{9} is what is left of gg, not how far down you go. C uses d=2hd = 2h, which fails at a height this large; at the centre gg would be zero. D, R4\dfrac{R}{4}, halves the height instead of doubling it.

Q6One correct option

Kinematics carries over unchanged, with the planet's gg in place of the Earth's. A body thrown straight up at speed uu rises H=u22gH = \dfrac{u^{2}}{2g} and stays up for 2ug\dfrac{2u}{g}; a projectile's range is u2sin⁡2θg\dfrac{u^{2}\sin 2\theta}{g}. All three go as 1g\dfrac{1}{g}: weaker gravity gives higher, longer flights.

A planet has the Earth's density but half its radius. A stone thrown straight up there, with the same speed as on the Earth, rises how high compared with its height HH on the Earth?

  1. AH2\dfrac{H}{2}
  2. B2H2H
  3. C4H4H
  4. D2H\sqrt{2}H
Show answer and solution

Answer: Option B

At the same density, g=43πGρRg = \dfrac{4}{3}\pi G\rho R is in proportion to RR, so half the radius gives g2\dfrac{g}{2}. The height u22g\dfrac{u^{2}}{2g} goes as 1g\dfrac{1}{g}, so it doubles: 2H2H.

The trap is C, 4H4H, from 1R2\dfrac{1}{R^{2}} with the mass held fixed; here the mass is not fixed, since a planet of half the radius and the same density has an eighth of the mass. A turns the ratio upside down: weaker gravity cannot give a lower flight. D takes a square root the formula does not contain.