Why does 'g' decrease with altitude?

The acceleration due to gravity, 'g', is inversely proportional to the square of the distance from the center of the Earth ($g = GM/r^2$). As you move to a higher altitude, your distance 'r' from the Earth's center increases. Since 'r' is in the denominator and squared, even a small increase in 'r' leads to a noticeable decrease in 'g'. This is a direct consequence of Newton's Law of Universal Gravitation, where gravitational force weakens with increasing separation.

Why does 'g' decrease with depth, and become zero at the Earth's center?

When an object is at a certain depth below the Earth's surface, the gravitational force it experiences is only due to the mass of the Earth contained within a sphere whose radius extends from the center to the object's position. The mass of the Earth *outside* this sphere (the 'shell' above the object) exerts no net gravitational force on the object. As you go deeper, the effective mass pulling you downwards decreases, leading to a reduction in 'g'. At the Earth's center, you are surrounded by Earth's mass in all directions, and the net gravitational pull is zero, hence 'g' is zero.

How does Earth's rotation affect 'g'?

The Earth's rotation creates a centrifugal effect that acts outwards, opposing the inward pull of gravity. This effect is maximum at the equator, where the rotational speed is highest, and zero at the poles. Consequently, the effective acceleration due to gravity ('g'') is reduced at the equator ($g' = g - Romega^2$) and is maximum at the poles ($g' = g$). This means objects weigh slightly less at the equator than at the poles due to this rotational effect.

When should I use the approximation $g_h = g(1 - 2h/R)$ for altitude?

The approximation $g_h = g(1 - 2h/R)$ is derived using the binomial expansion and is valid when the height 'h' is much smaller than the Earth's radius 'R' ($h ll R$). Typically, this means when 'h' is less than about 5% of 'R'. For larger heights, the exact formula $g_h = g(1 + h/R)^{-2}$ must be used to maintain accuracy. NEET problems often specify conditions or provide options that guide which formula to use.

Is 'g' truly constant at a given location?

No, 'g' is not truly constant even at a given location over time, though its variations are usually very small and often ignored in basic physics. Factors like tidal forces from the Moon and Sun, changes in local atmospheric pressure, and even very slight seismic activity can cause minuscule fluctuations in 'g'. However, for NEET purposes, 'g' at a specific latitude and altitude is generally considered constant unless otherwise specified.

Variation of g

Physics

NEET UG

Version 1Updated 22 Mar 2026

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The acceleration due to gravity, denoted by 'g', is the acceleration experienced by an object solely under the influence of Earth's gravitational force. Its standard value at the Earth's surface and at sea level is approximately $9.8,\text{m/s}^2$ . However, this value is not constant across the Earth. It varies significantly with altitude (height above the surface), depth (distance below the surfac…

Quick Summary

The acceleration due to gravity, 'g', is the acceleration experienced by objects due to Earth's gravitational pull, with an average value of $9.8,\text{m/s}^2$ at the surface. However, 'g' is not constant and varies significantly with location.

As we move to higher altitudes (height 'h' above the surface), 'g' decreases because the distance from the Earth's center increases. The formula is $g_h = g(1 + h/R)^{-2}$ , which approximates to $g_h approx g(1 - 2h/R)$ for small 'h'.

Similarly, 'g' also decreases as we go deeper (depth 'd' below the surface) because only the mass of the Earth within the inner sphere contributes to the gravitational pull. The formula for depth is $g_d = g(1 - d/R)$ , becoming zero at the Earth's center.

Finally, the Earth's rotation causes 'g' to vary with latitude ( $lambda$ ). The centrifugal effect of rotation reduces the effective 'g' most at the equator ( $lambda=0^circ$ ) and least (zero effect) at the poles ( $lambda=90^circ$ ), given by $g' = g - Romega^2 cos^2lambda$ .

Understanding these three primary variations is essential for NEET.

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Key Concepts

Altitude Variation: Exact vs. Approximate Formula

The exact formula for acceleration due to gravity at height $h$ above the Earth's surface is $g_h = g left(…

Depth Variation and Zero Gravity at Center

When an object is at a depth $d$ below the Earth's surface, the acceleration due to gravity $g_d$ is given by…

Effect of Earth's Rotation on 'g'

The Earth's rotation causes a reduction in the effective acceleration due to gravity, $g'$ , which varies with…

Surface 'g' —

g = G M / R^2

Altitude 'h' —

g_h = g(1 + h/R)^{-2}

Altitude (approx. for $h ll R$) —

g_h \approx g(1 - 2h/R)

Depth 'd' —

g_d = g(1 - d/R)

Center of Earth ($d=R$) —

g_d = 0

Latitude $\lambda$ —

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

Equator ($\lambda=0^circ$) —

g'_{equator} = g - R\omega^2

(minimum 'g' due to rotation)

Poles ($\lambda=90^circ$) —

g'_{poles} = g

(maximum 'g' due to rotation)

All Deep Layers Rotate: Altitude, Depth, Latitude, Rotation.

Altitude:

g

goes Away (decreases).

Depth:

g

goes Down (decreases to zero).

Latitude:

g

is Less at the equator, Larger at poles.

Rotation: Reduces

g

(except at poles).