In this post, we will list down and **derive the formula of Acceleration due to gravity on the earth’s surface**. In other words, we will derive the formula or equation of g on the earth’s surface.**The 2 formulas we will derve for g ( Acceleration due to gravity on the earth’s surface**)

**are: g = GM / R**

^{2}and g = (4/3)*π*R ρ G**So let’s start with the step by step derivation process.**

**Derive the formula of Acceleration due to gravity on the earth’s surface | derive the formula for g on the earth**

In this section let’s find out the formula or equation of g on the earth’s surface.

Force of gravity (gravitational force value due to earth) acting on a body of mass m on the earth surface is expressed as:**F = GMm/R ^{2}** ____________ (1)

Here R is the radius of the earth (considering it a homogenous sphere)

and M here is the total mass of earth concentrated at its center. G is the gravitational constant.

Now, from Newton’s 2^{nd} Law of Motion,

F = mg.___________________(2)

Just to recapitulate, as a body falls downwards it is continuously acted upon by a force of gravity. The body thus possesses an acceleration, called Acceleration Due to gravity(g).

From equation 1 and 2 we can write

**mg = (GMm) / ** R^{2}

Now, we get the equation or formula of g on earth’s surface as follows*:*

Acceleration due to gravity on the earth’s surface is represented as

**g = GM / ** R^{2} ** ______(3)**

[expression of g on earth’s surface]

Let’s find out ** another expression of g on the **earth’s surface using the

**density of the**

*earth*If the mean density of the earth is *ρ* then the mass of the earth is expressed as:

M = volume X density = (4/3) *π* R^{3} . ρ

(π or Pi = 22/7)**g = G.( (4/3) π R ^{3} ρ) / R^{2} **

*So we get the second expression or formula for g on earth’s surface:*Acceleration due to gravity of earth is represented as**g = (4/3) π R ρ G _____(4)**

[2nd expression of g on earth’s surface]