Variation of g with Altitude & Depth

The surface value as baseline

The acceleration due to gravity at the Earth's surface follows from Newton's law of gravitation applied to a body of mass $m$ resting on a sphere of mass $M_E$ and radius $R$:

$$g = \frac{G M_E}{R^2} \approx 9.8~\text{m s}^{-2}.$$

This single number is the reference against which every variation is measured. Two facts drive everything below. First, $g$ depends on the distance from the centre of the Earth — so moving away weakens gravity. Second, the surface value uses the entire mass $M_E$, true only while all the Earth's mass stays below you; burrow inside and part of the mass ends up above you and stops pulling you down. These two ideas — distance and enclosed mass — split the problem cleanly into the height case and the depth case. For the surface derivation in full, see acceleration due to gravity.

Variation with height above the surface

Take a body at a height (altitude) $h$ above the surface. Its distance from the centre is now $R + h$, and the whole mass $M_E$ still lies below it. Newton's law gives the exact value:

$$g_h = \frac{G M_E}{(R+h)^2} = g\,\frac{R^2}{(R+h)^2} = \frac{g}{\left(1 + \dfrac{h}{R}\right)^2}.$$

This is the form to use whenever $h$ is comparable to $R$ — satellites, distant orbits, "height equal to half the radius" type problems. NIOS §5.3.1 notes the immediate consequence: at $h = R$, the distance from the centre doubles to $2R$, so $g$ falls to one-quarter of its surface value.

When the altitude is small compared with the radius ($h \ll R$) — aircraft, mountains, towers — the binomial expansion of $(1 + h/R)^{-2}$ keeping only the first-order term gives the working approximation NCERT §7.6 quotes:

$$g_h \approx g\left(1 - \frac{2h}{R}\right), \qquad h \ll R.$$

The factor of 2 is the signature of the inverse-square law: a fractional rise $h/R$ in distance produces twice that fractional drop in $g$. NIOS Example 5.2 illustrates the scale — at a typical cruising altitude of 10 km, with $R = 6400$ km, the value of $g$ falls only to about $9.77~\text{m s}^{-2}$, a change of roughly 0.3%.

Variation with depth below the surface

Now drop a body to a depth $d$ below the surface, so its distance from the centre is $r = R - d$. Two things change at once. The distance shrinks, which on its own would increase $g$. But the spherical shell of matter above the body — the outer shell of thickness $d$ — now contributes nothing: NIOS §5.3.2 states the standard shell result that the net gravitational force from a uniform spherical shell on a point inside it is exactly zero. Only the inner sphere of radius $r = R - d$ pulls the body.

Assuming the Earth has uniform density $\rho$, the mass of that inner sphere is $M' = \tfrac{4}{3}\pi\rho\,(R-d)^3$, and the acceleration at depth $d$ becomes

$$g_d = \frac{G M'}{(R-d)^2} = \frac{4}{3}\pi G \rho\,(R-d).$$

Since the surface value is $g = \tfrac{4}{3}\pi G \rho R$ (set $d = 0$), dividing one by the other removes the constants and leaves the clean linear law NCERT §7.6 gives:

$$g_d = g\left(1 - \frac{d}{R}\right), \qquad 0 \le d \le R.$$

The fall-off is linear in $d$ and the coefficient is just $d/R$ — no factor of 2. The single most important consequence: set $d = R$ (the body has reached the centre) and $g_d = 0$. At the centre of the Earth, matter surrounds you symmetrically in every direction and the pulls cancel exactly.

The g-versus-distance graph

The two laws meet at the surface and behave oppositely on either side of it. Plotting $g$ against distance $r$ from the centre makes the whole story visible in one curve: a straight line rising from zero at the centre to the peak at $r = R$, then an inverse-square tail decaying beyond the surface.

Read the graph from the centre outwards. At $r = 0$, $g = 0$. The line climbs steadily to its maximum at the surface. Beyond the surface, $g$ never increases again — the inverse-square tail only decays, passing through $g/4$ at $r = 2R$. The surface is the unique peak of the entire curve.

Height vs depth — the contrast

Place the three variations side by side. They share a starting point — the surface value $g$ — but differ in mechanism, in the fall-off coefficient, and in how fast they decay.

The decisive comparison: for the same small displacement $x$ from the surface, going up changes $g$ by $-2gx/R$ while going down changes it by only $-gx/R$. Height kills $g$ twice as fast as depth. That is the engine behind the NEET-favourite question "at what depth is $g$ the same as at height $h$" — the answer is $d = 2h$, because depth must travel twice as far to lose the same amount.

Variation with latitude and shape

Even staying exactly on the surface, $g$ is not the same everywhere — it depends on latitude $\lambda$ (the angular distance north or south of the equator). NIOS §5.3.3 traces this to the Earth's daily rotation. Every point on the surface moves in a circle about the rotation axis, and circular motion needs centripetal force. Part of the gravitational pull is "spent" supplying that force, so the effective gravity an object actually experiences is reduced:

$$g' = g - \omega^2 R\cos^2\!\lambda,$$

where $\omega$ is the Earth's angular speed of rotation. The radius of a point's circular path is $R\cos\lambda$, and only the component of the centripetal term along the local vertical reduces $g$, which together produce the $\cos^2\lambda$ dependence.

• At the equator ($\lambda = 0$, $\cos^2\lambda = 1$): the reduction is largest, $g' = g - \omega^2 R$. This is the smallest effective $g$ on Earth.
• At the poles ($\lambda = 90^\circ$, $\cos^2\lambda = 0$): the rotation term vanishes, $g' = g$. The poles lie on the rotation axis, trace no circle, and need no centripetal force — so the effective $g$ is largest there.

NIOS confirms the magnitude is small: with $\omega = 7.27\times10^{-5}~\text{rad s}^{-1}$ and $R \approx 6.37\times10^6$ m, the maximum correction $\omega^2 R$ is only about $0.034~\text{m s}^{-2}$ — a fraction of a percent of $g$.

A second, independent effect reinforces the same trend. The Earth is not a perfect sphere; rotation has flattened it so the equatorial radius is larger than the polar radius. Because $g \propto 1/R^2$, the smaller polar radius gives a larger $g$ at the poles. Both the rotation effect and the shape effect therefore point the same way: $g$ is maximum at the poles and minimum at the equator.

Worked examples