What Is Acceleration Due to Gravity?

Acceleration due to gravity is a fundamental concept in physics describing the rate at which objects accelerate towards the Earth due to the gravitational attraction. It serves as a crucial parameter for solving various motion problems in JEE Main Physics and is directly applied in gravitation-related calculations.

Gravitation and Gravity: Fundamental Concepts

Gravitation is the universal force of attraction between any two masses in the universe, acting along the line joining their centers. When one of these masses is the Earth, the force is termed as gravity. Gravity is responsible for objects falling towards Earth's surface and is always directed towards the Earth's center.

The gravitational force between two point masses is given by Newton’s law of universal gravitation:

$F = \dfrac{G m_1 m_2}{r^2}$, where $G$ is the universal gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the separation between their centers. For more detailed insight on gravitation, see Gravitation Overview.

Acceleration Due to Gravity ($g$): Definition and Basic Formula

Acceleration due to gravity, represented by $g$, is the acceleration experienced by an object solely under the influence of Earth’s gravitational pull. It is a vector quantity, directed towards the center of the Earth, and is independent of the mass or nature of the object.

At Earth’s surface, $g$ is calculated as:

$g = \dfrac{GM}{R^2}$

where $G = 6.674 \times 10^{-11}\, \dfrac{N\,m^2}{kg^2}$, $M$ is the mass of the Earth $(6 \times 10^{24}\, kg)$, and $R$ is the mean radius of Earth $(6.4 \times 10^6\, m)$.

The standard value of acceleration due to gravity at Earth's surface is approximately $9.8\ m/s^2$. For practical purposes, $g$ may also be expressed as $32\ ft/s^2$ or $980\ cm/s^2$.

Relation Between $g$ and Universal Gravitational Constant $(G)$

By equating the gravitational force and the weight of an object of mass $m$ at Earth's surface:

$mg = \dfrac{GMm}{R^2}$

On simplifying, the expression for $g$ becomes:

$g = \dfrac{GM}{R^2}$

This formula establishes that $g$ depends on the mass of the Earth and inversely on the square of Earth's radius.

Factors Affecting Acceleration Due to Gravity

Acceleration due to gravity is not uniform everywhere on Earth. It varies with height, depth, latitude, rotation, and the Earth’s shape. These variations are significant for analyzing precise motion under gravitation.

Variation of $g$ with Altitude (Height Above Earth's Surface)

The value of $g$ decreases as one moves above Earth's surface. If an object is at height $h$ above the surface, the acceleration due to gravity is given by:

$g_h = \dfrac{GM}{(R + h)^2} = g \left(1 + \dfrac{h}{R}\right)^{-2}$

For $h \ll R$, the binomial expansion gives:

$g_h \approx g \left(1 - \dfrac{2h}{R}\right)$

This formula is applied only for heights much smaller than the Earth's radius.

Variation of $g$ with Depth (Below Earth's Surface)

As depth inside the Earth increases, the value of $g$ decreases linearly. At depth $d$ below the surface, the acceleration due to gravity is:

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

At the center of the Earth $(d = R)$, $g$ becomes zero. This linear decrease contrasts with the quadratic decrease observed for heights above the surface.

Comparative Table: $g$ with Height and Depth

Comparison of Height and Depth for Equal Change in $g$

For small changes, the decrease in $g$ at height $h$ above Earth's surface equals the decrease at depth $d = 2h$ below the surface. This relationship helps in solving comparative problems in JEE examinations.

Variation of $g$ due to Earth's Rotation

Earth’s rotation causes a reduction in apparent gravity, especially near the equator. The effective acceleration due to gravity is given by:

$g_{eff} = g - \omega^2 R \cos^2\lambda$

Here, $\omega$ is Earth’s angular velocity and $\lambda$ is the latitude. The effect is maximum at the equator $(\lambda = 0^\circ)$ and zero at the poles $(\lambda = 90^\circ)$.

Variation of $g$ due to Earth's Shape

Earth is an oblate spheroid, flattened at the poles and bulged at the equator. The polar radius is less than the equatorial radius, so gravity is slightly greater at the poles than at the equator. The change is about $0.5\%$.

Therefore, the value of $g$ is maximum at the poles and minimum at the equator.

Acceleration Due to Gravity on Celestial Bodies

The acceleration due to gravity differs on various celestial objects, as it depends on their mass and radius. For example, $g$ on the Moon is about $1.62\ m/s^2$, and on Mars it is around $3.71\ m/s^2$, both significantly less than Earth's value.

For questions involving gravitation on other planets, see Gravitation Mock Test 1.

Units and Dimensional Formula of $g$

The SI unit of acceleration due to gravity is $m/s^2$. In some cases, it may also be expressed in $ft/s^2$ or $in/s^2$ depending on the chosen unit system. The dimensional formula for $g$ is $[M^0LT^{-2}]$.

Solved Example: Acceleration Due to Gravity with Height

If $g$ at Earth's surface is $9.8\ m/s^2$, calculate $g$ at a height $h$ where $h = 0.05R$.

• $g = 9.8\ m/s^2$
• $h = 0.05R$, $R = $ Earth's radius
• $g_h = g (1 - \dfrac{2h}{R}) = 9.8 \times (1 - 0.1) = 8.82\ m/s^2$

Solved Example: Acceleration Due to Gravity with Depth

An object is placed at a depth $d = 1600$ km inside Earth. Find the acceleration due to gravity if $R = 6400$ km and $g = 9.8\ m/s^2$ at the surface.

• $g_d = g (1 - \dfrac{d}{R})$
• $g_d = 9.8 \times \left(1 - \dfrac{1600}{6400}\right) = 9.8 \times 0.75 = 7.35\ m/s^2$

Importance of $g$ in Physics Problems

The value and variations of $g$ influence the motion of falling bodies, projectile motion, weight measurements, and energy calculations. Mastery of these concepts is essential for strong performance in JEE examinations.

For understanding gravitational potential and its link with $g$, refer to Gravitational Potential Energy.

Key Points on Acceleration Due to Gravity

• Standard value of $g$ on Earth is $9.8\ m/s^2$
• $g$ decreases with altitude and depth inside Earth
• Earth’s rotation and shape cause variations in $g$
• $g$ is maximum at the poles and minimum at the equator
• $g$ on Moon and Mars is less than on Earth
• $g$ is a vector, always directed to Earth's center

Practice Further on Acceleration Due To Gravity

To enhance your understanding, solve additional problems from Acceleration Due To Gravity, Gravitation Mock Test 2, and Gravitation Mock Test 3.