What is the difference between 'g' and 'G'?

'g' represents the acceleration due to gravity, which is the acceleration experienced by an object due to the gravitational pull of a planet (like Earth). Its value is approximately $9.8, ext{m/s}^2$ on Earth's surface and varies with location. 'G' is the universal gravitational constant, a fundamental constant of nature that quantifies the strength of the gravitational force. Its value is constant everywhere in the universe, approximately $6.67 imes 10^{-11}, ext{N m}^2/ ext{kg}^2$. 'g' depends on the mass and radius of the planet, while 'G' is independent of any specific body.

Why does acceleration due to gravity decrease with altitude?

Acceleration due to gravity, $g$, is inversely proportional to the square of the distance from the center of the attracting body ($g = GM/r^2$). As an object moves to a higher altitude, its distance ($r$) from the Earth's center increases. Since $r$ is in the denominator and squared, even a small increase in altitude leads to a noticeable decrease in $g$. Essentially, the gravitational pull weakens as you move further away from the source of gravity.

Why does acceleration due to gravity decrease with depth?

When an object is taken to a depth $d$ below the Earth's surface, the gravitational force acting on it is only due to the mass of the Earth contained within a sphere of radius $(R_E - d)$. The mass of the Earth *outside* this sphere (the hollow shell) does not contribute to the gravitational force on the object inside. As you go deeper, the effective mass pulling you towards the center decreases, and thus $g$ decreases. At the Earth's center, the effective mass is zero, and $g$ becomes zero.

Is acceleration due to gravity zero at the center of the Earth?

Yes, the acceleration due to gravity is zero at the exact center of the Earth. If an object were placed at the Earth's center, it would be surrounded by Earth's mass uniformly in all directions. The gravitational pull from every part of the Earth would cancel out, resulting in a net gravitational force of zero. Consequently, the acceleration due to gravity at the center would also be zero.

Why is 'g' different at the poles and the equator?

The Earth's rotation and its oblate spheroid shape (bulging at the equator, flattened at the poles) cause 'g' to vary. Due to the bulge, the radius at the equator is slightly larger than at the poles, leading to a smaller $g$ ($g propto 1/R^2$). More significantly, the Earth's rotation creates a centrifugal effect that reduces the effective gravity at the equator more than at the poles. At the poles, there's no centrifugal effect. Both factors combine to make $g$ maximum at the poles and minimum at the equator.

Does the acceleration due to gravity depend on the mass of the falling object?

No, the acceleration due to gravity is independent of the mass of the falling object. This is a crucial concept. The derivation $mg = GMm/R^2$ clearly shows that the mass of the object, $m$, cancels out, leaving $g = GM/R^2$. This means that in a vacuum, a feather and a bowling ball would fall with the exact same acceleration, reaching the ground simultaneously if dropped from the same height. Air resistance is what makes lighter objects fall slower in everyday experience.

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Acceleration due to Gravity

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Version 1Updated 22 Mar 2026

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Acceleration due to gravity, denoted by $g$ , is the acceleration experienced by an object solely due to the gravitational force exerted by a celestial body, typically Earth. It is a vector quantity, always directed towards the center of mass of the attracting body. According to Newton's Universal Law of Gravitation, the gravitational force between two objects is directly proportional to the produc…

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Acceleration due to gravity, denoted by $g$ , is the acceleration experienced by an object solely under the influence of a planet's gravitational force. On Earth's surface, its average value is approximately $9.

8, ext{m/s}^2 $(or often approximated as$ 10, ext{m/s}^2 $for simpler calculations). It is a vector quantity, always directed towards the center of the Earth. The fundamental formula for$ g $at the surface of a planet of mass$ M $and radius$ R $is$ g = GM/R^2 $, where$ G$ is the universal gravitational constant.

A key takeaway is that $g$ is independent of the mass of the falling object. However, $g$ is not constant across the Earth. It decreases with increasing altitude (height above the surface) and with increasing depth (below the surface).

It is maximum at the poles and minimum at the equator, primarily due to the Earth's rotation and its slightly oblate shape. At the Earth's center, $g$ becomes zero. Understanding these variations and the underlying principles is essential for NEET.

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Key Concepts

Derivation of

g = GM/R^2

This fundamental formula connects Newton's Law of Gravitation with Newton's Second Law. The gravitational…

Variation of

g

with Altitude

As an object moves to a height $h$ above the Earth's surface, its distance from the center becomes $(R_E +…

Variation of

g

with Depth

When an object is at a depth $d$ below the Earth's surface, only the mass of the Earth within a sphere of…

Definition: — Acceleration due to gravity,

g

, is the acceleration of an object due to gravitational force.

Standard Value: —

g approx 9.8,\text{m/s}^2

on Earth's surface.

Formula: —

g = \frac{GM}{R^2}

(where

M

is planet mass,

R

is planet radius).

Independence: —

g

is independent of the mass of the falling object.

Variation with Altitude ($h$): —

g_h = \frac{g}{(1 + h/R_E)^2}

. For

h \ll R_E

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

Variation with Depth ($d$): —

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

. At center (

d=R_E

g_d = 0

Variation with Latitude ($\lambda$): —

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

. Max at poles (

\lambda=90^\circ

), Min at equator (

\lambda=0^\circ

Relationship: —

g_{\text{pole}} > g_{\text{equator}}

GRAVITY: G-M-R-Squared, Altitude 2H, Depth D-R, Rotate Cos-Squared.

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