Variation of g with depth and height — derive expressions

Question

Derive expressions for the variation of acceleration due to gravity ( $g$ ) with: (a) Height above the Earth’s surface (b) Depth below the Earth’s surface

Also state what happens to $g$ at the centre of the Earth.

Solution — Step by Step

At the Earth’s surface (radius $R_E$ , mass $M_E$ ), using Newton’s law of gravitation and second law:

g = \frac{GM_E}{R_E^2}

This is our starting formula. $g \approx 9.8\text{ m/s}^2$ at the surface.

At height $h$ above the surface, the distance from Earth’s centre is $(R_E + h)$ :

g_h = \frac{GM_E}{(R_E + h)^2} = \frac{GM_E}{R_E^2\left(1 + \frac{h}{R_E}\right)^2} = \frac{g}{\left(1 + \frac{h}{R_E}\right)^2}

Exact formula:

g_h = g\left(1 + \frac{h}{R_E}\right)^{-2}

Approximation for $h \ll R_E$ (using binomial expansion, keeping first term):

g_h \approx g\left(1 - \frac{2h}{R_E}\right)

As $h$ increases, $g_h$ decreases. At infinite height, $g_h \to 0$ .

At depth $d$ below the surface, the object is at distance $(R_E - d)$ from the centre. By the shell theorem, only the mass of Earth within the sphere of radius $(R_E - d)$ contributes to gravity (the shell above exerts zero net force).

Assuming uniform density $\rho$ :

M_{\text{inner}} = \frac{4}{3}\pi(R_E - d)^3 \cdot \rho

g_d = \frac{G M_{\text{inner}}}{(R_E - d)^2} = \frac{G \cdot \frac{4}{3}\pi(R_E-d)^3\rho}{(R_E-d)^2} = \frac{4}{3}\pi G\rho(R_E - d)

Since $g = \frac{4}{3}\pi G\rho R_E$ at the surface:

\boxed{g_d = g\left(1 - \frac{d}{R_E}\right)}

$g_d$ decreases linearly with depth. At $d = R_E$ (centre of Earth):

g_d = g\left(1 - 1\right) = \mathbf{0}

Gravity is zero at the centre of the Earth.

Why This Works

The height formula uses Newton’s law of gravitation — as distance from Earth’s centre increases, gravitational force (and $g$ ) decreases as inverse square. The mass of Earth remains the same, just the distance changes.

The depth formula uses the shell theorem and the assumption of uniform density. The key insight: at depth $d$ , the shell of Earth above you (from $(R_E - d)$ to $R_E$ ) exerts zero net gravitational force by symmetry (the shell theorem). Only the inner sphere matters. Since the inner sphere has less mass, $g$ is reduced.

The dependence at height is quadratic (inverse square) while at depth it’s linear — this distinction is frequently tested.

Alternative Method

Using dimensional reasoning for the depth formula: $g_d \propto M_{\text{inner}} / r^2 \propto r^3/r^2 = r$ (since $M \propto r^3$ for uniform density). So $g \propto r$ at depth — confirming the linear relationship with distance from centre, and $g \to 0$ as $r \to 0$ .

JEE Main and CBSE Class 11 both test the derivations of both expressions. Key exam facts: (1) $g$ decreases on going up or going down, (2) $g$ decreases faster going up (quadratic) than going down (linear), (3) $g = 0$ at Earth’s centre, (4) for $h \ll R_E$ , use the approximation $g_h \approx g(1 - 2h/R_E)$ .

Common Mistake

Students often apply the height formula for depth problems or vice versa. Remember: height formula is $g_h = g(1 + h/R)^{-2}$ — involves inverse square (distance from centre increases). Depth formula is $g_d = g(1 - d/R)$ — linear (only inner mass contributes). Also, at depth $d$ , the distance from centre is $(R - d)$ , which DECREASES — but $g$ still decreases because mass decreases faster than distance squared.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next