JEE Mains Physics · Gravitation

Acceleration due to Gravity: Depth and Rotation

Below the surface g falls in a straight line, g(1 − d/R), to zero at the centre; the earth's spin lowers g by ω²R cos²λ, most at the equator and not at all at the poles.

All Gravitation notes JEE Mains Physics strategy

Why this matters

Seventeen PYQs, fifteen of them multiple choice, and one from 2026. Six use g inside the earth, seven compare a point below the surface with a point above it, and four are about the earth's spin. Five of the seventeen are statement or assertion questions, so the qualitative facts here are worth as much as the formulas.

Concept 1 of 3: g at a depth below the surface

Below the surface, the shell of earth above you pulls equally in all directions and cancels. Only the sphere beneath you, of radius

R - d

, still pulls. For a uniform earth its mass shrinks as the cube of its radius, so g falls in direct proportion to the distance from the centre.

Definition

$g_d = g\left(1 - \dfrac{d}{R}\right) = \dfrac{GMr}{R^{3}}$ , where $r = R - d$ .
At the centre $g = 0$ : a body there has mass but no weight.
Depth $R/2 \to g/2$ ; depth $R/4 \to 3g/4$ .
The g–r graph rises in a straight line from zero at the centre to its maximum at the surface, then falls as $1/r^{2}$ outside.
The formula assumes uniform density; JEE questions say so or take it for granted.

g at depth d

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

Worked example

At what depth below the surface is g reduced by 20%? Take R = 6400 km.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q95Moderate

Example 1 · Gravitation · Acceleration due to Gravity: Depth and Rotation

Assuming the earth to be a sphere of uniform mass density, a body weighed

300\text{ }N

on the surface of earth. How much it would weigh at R/4 depth under surface of earth?

Depth is linear, height is inverse square

Going down, g falls in proportion to d/R. Going up, it falls as 1/(1 + h/R)². Using the height formula below the surface, or the depth formula above it, gives a wrong option every time.

g is largest at the surface

g rises from zero at the centre to its peak at the surface and falls on both sides of it. A graph that keeps rising inside and outside, or that is flat inside, is wrong.

Drill 5 more on g at a depth below the surface

Concept 2 of 3: Comparing g at a depth with g at a height

Both going down and going up reduce g, but at different rates. For small distances, g falls twice as fast going up, so a height h matches a depth 2h. For large distances the approximations break down: set the two exact formulas equal and solve.

Definition

Small distances: $g_h \approx g(1 - 2h/R)$ and $g_d = g(1 - d/R)$ , so equal g needs $d = 2h$ .
Large distances: solve $1 - \dfrac{d}{R} = \dfrac{1}{(1 + h/R)^{2}}$ exactly.
The same distance x below and above: $\dfrac{g_d}{g_h} = (1 - x/R)(1 + x/R)^{2}$ .
Equal weight at the same distance above and below: $(1 - x)(1 + x)^{2} = 1$ with $x = h/R$ , giving $x^{2} + x - 1 = 0$ .
Moving from a small depth d to the same small height d changes g by about $\dfrac{d}{R}$ of its value.

Equal g below and above

1 - \frac{d}{R} = \frac{1}{(1 + h/R)^{2}} \quad\Rightarrow\quad d \approx 2h\ \ (h \ll R)

Worked example

At what height above the surface is g the same as at a depth of half the earth's radius? Take R = 6400 km.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q17Moderate

Example 2 · Gravitation · Acceleration due to Gravity: Depth and Rotation

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)

d = 2h is only for small distances

The rule comes from the approximation 1 − 2h/R. For a depth of R/2 the matching height is (√2 − 1)R, not R/4. If the distance is a sizeable fraction of R, solve the exact equation.

Both directions reduce g

A statement that g increases as you go down is false: g is largest at the surface. Going up and going down both lower it.

Drill 6 more on comparing g at a depth with g at a height

Concept 3 of 3: Effect of the earth's spin on g

A body on the spinning earth moves in a circle about the axis. Part of gravity is used up to supply that circular motion, so the scale reads less. The circle is largest at the equator and shrinks to a point at the poles, so the effect is greatest at the equator and absent at the poles.

Definition

At latitude $\lambda$ : $g' = g - \omega^{2}R\cos^{2}\lambda$ .
Poles ( $\lambda = 90^{\circ}$ ): $g' = g$ . Equator ( $\lambda = 0$ ): $g' = g - \omega^{2}R$ , the least value.
For the earth, $\omega = 7.27 \times 10^{-5}$ rad/s and $\omega^{2}R \approx 0.034\ \text{m/s}^{2}$ .
Bodies at the equator float when $\omega^{2}R = g$ : $\omega = \sqrt{g/R}$ and the day lasts $T = 2\pi\sqrt{R/g}$ .
The effective g points exactly at the centre only at the poles and the equator.
g also differs from place to place because the earth is not a perfect sphere: it is flattened at the poles, which also makes g larger there.

Effective g at latitude λ

g' = g - \omega^{2}R\cos^{2}\lambda

Worked example

A body reads 80 N on a spring balance at the North Pole. What does the same balance read at the equator? Take

g = 9.8\ \text{m/s}^{2}

\omega = 7.27 \times 10^{-5}

rad/s and

R = 6.4 \times 10^{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 10 · Q10Moderate

Example 3 · Gravitation · Acceleration due to Gravity: Depth and Rotation

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately: (Take :

g = 10{\text{ }ms}^{- 2}

, the radius of earth,

R = 6400 \times10^{3}\text{ }m

, Take

\pi= 3.14

)

No effect at the poles, most at the equator

Statements that reverse this are a regular trap. At the poles the body sits on the axis and does not go round a circle, so rotation changes nothing there.

The floating day equals a grazing orbit's period

T = 2π√(R/g) is about 84 minutes, the same as a satellite skimming the surface. Both come from setting ω²R equal to g.

Drill 3 more on effect of the earth's spin on g

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

g at a depth below the surface
g at depth d
$g_d = g\left(1 - \frac{d}{R}\right)$
Comparing g at a depth with g at a height
Equal g below and above
$1 - \frac{d}{R} = \frac{1}{(1 + h/R)^{2}} \quad\Rightarrow\quad d \approx 2h\ \ (h \ll R)$
Effect of the earth's spin on g
Effective g at latitude λ
$g' = g - \omega^{2}R\cos^{2}\lambda$

Watch out for (6)

Depth is linear, height is inverse square
→ g at a depth below the surface
g is largest at the surface
→ g at a depth below the surface
d = 2h is only for small distances
→ Comparing g at a depth with g at a height
Both directions reduce g
→ Comparing g at a depth with g at a height
No effect at the poles, most at the equator
→ Effect of the earth's spin on g
The floating day equals a grazing orbit's period
→ Effect of the earth's spin on g

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