Variation of g with height and depth

How does g change with height and depth?

In plain words: for a uniform Earth, the acceleration due to gravity is greatest at the surface and becomes smaller whether you go up or go down, as NCERT points out (page 134). Going up, you move away from all the Earth’s mass, and the pull falls off as 1/r^². Going down, the shell of rock above you exerts no net pull on you (its pulls cancel), and less mass is left beneath you, so the pull falls in a straight line, to zero at the centre.

The starting point is the result of the previous topic, acceleration due to gravity: at the surface, g = GM_E/R_E², where M_E is the Earth’s mass and G the gravitational constant.

g above the surface of the Earth

A point mass m at height h is outside the Earth, at distance R_E + h from its centre, so the Earth pulls it as if all its mass were at the centre:

Dividing by m (the force is mg(h) by Newton’s second law), and using GM_E = gR_E²:

For h > 0 this is less than g; at h = 0 it equals the surface value. When h is small compared with R_E, write it as g(1 + h/R_E)^⁻² and use the binomial approximation (1 + x)^⁻² ≈ 1 − 2x for small x:

The approximation is good for heights of a few kilometres or tens of kilometres, but not for heights comparable with the Earth’s radius. At h = R_E the exact value is g/4, while the approximate formula would give −g, which is meaningless.

g below the surface of the Earth

A point mass m at depth d is at distance R_E − d from the centre. Think of the Earth as a sphere of radius R_E − d plus a shell of thickness d around it. The point is inside the shell, so the shell’s net pull on it is zero. It is on the surface of the inner sphere, which pulls as if its mass M_s were at the centre. For uniform density, mass is proportional to the cube of the radius:

Substituting M_s = M_E(R_E − d)^³/R_E³, the force is

Dividing by m and using GM_E = gR_E² again:

This is exact for a uniform Earth, at any depth. At the centre, d = R_E, and g = 0: the pulls of the surrounding mass cancel in every direction.

Height versus depth. For small h and d, g falls by 2gh/R_E going up and by gd/R_E going down. So near the surface g falls twice as fast with height as with depth: the fall at a small height h approximately equals the fall at depth 2h.

The real Earth. The linear depth result assumes uniform density. The exterior inverse-square result needs spherical symmetry, but not uniform density; both neglect rotation. The real Earth is much denser towards its centre (its mean density is about twice that of surface rocks), so in fact g changes little, and even rises slightly, for a long way below the surface before it falls to zero at the centre. The uniform-Earth formula is the NCERT model; in a problem with a different density profile, use the enclosed mass instead.

Solved examples

Take R_E = 6400 km and g = 9.8 m s^⁻² unless stated otherwise.

Common mistakes to avoid

Quick revision summary

• For a uniform Earth, g is greatest at the surface and decreases both upwards and downwards.
• At height h: g(h) = gR_E²/(R_E + h)^², about g(1 − 2h/R_E) for small h.
• At depth d (uniform Earth): g(d) = g(1 − d/R_E), exactly, falling to zero at the centre.
• Near the surface, g falls twice as fast with height as with depth: the fall at a small height h approximately equals that at depth 2h.
• In short, with R the Earth’s radius: g(h) = gR^²/(R + h)^² and g(d) = g(1 − d/R).

Questions students ask

How does g vary with latitude?

The acceleration due to gravity increases with latitude: on the real Earth it is about 9.78 m s^⁻² at the equator and about 9.83 m s^⁻² at the poles. Two effects cause this: the Earth’s bulge puts the equator farther from the centre, and the Earth’s spin uses part of the pull to keep a body moving in its circle round the axis; for a spherical Earth the inward radial component of effective gravity is g − ω^²R_E cos^²λ at latitude λ. This is also the effective magnitude to first order in ω^²R_E/g, not an exact magnitude away from the equator and poles: the centrifugal acceleration has a sideways component too. Here g is the gravitational magnitude without rotation and ω is the Earth’s angular speed. This goes beyond NCERT, which treats the Earth as a non-rotating sphere, and the JEE Main 2027 and NEET (UG) 2026 syllabuses name only the variation of g with altitude and depth.

What would happen to g if the Earth stopped rotating or spun faster?

Holding the spherical Earth’s mass and radius fixed, if it stopped rotating, effective gravity at the equator would rise by ω^²R_E, about 0.034 m s^⁻², while g at the poles, where the spin has no effect, would not change. If it spun faster, g at the equator would fall, reaching zero in this ideal model when ω^²R_E = g, that is, ω = √(g/R_E) ≈ 1.24 × 10^⁻³ rad s^⁻¹, about 17 times today’s value. A day would then last only about 84.6 minutes, the period of a satellite skimming the surface, and bodies on the equator would be weightless.

At what height does g fall by 1%?

The acceleration due to gravity falls by 1% at a height of about 32 km, taking R_E = 6400 km: for small heights the fractional fall is 2h/R_E, and 2h/R_E = 0.01 gives h = R_E/200. The exact formula gives almost the same height, about 32.2 km, because 32 km is small compared with R_E. Going down instead, g falls by 1% at a depth d = R_E/100, about 64 km, for a uniform Earth.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. What is the value of g at a height equal to the Earth’s radius?

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   g(R_E) = gR_E²/(2R_E)^² = g/4, about 2.45 m s^⁻².

2. Q2. At what depth below the surface is g one quarter of its surface value?

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   1 − d/R_E = ¼, so d = 3R_E/4, about 4800 km.

3. Q3. By what percentage is a climber’s weight less at the top of a mountain 8.85 km high than at sea level? Take R_E = 6371 km.

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   The fractional fall is 2h/R_E = 2 × 8.85/6371, about 0.0028, that is about 0.28%. (A 60 kg climber’s weight drops from about 588 N by about 1.6 N.)