# Acceleration Due to Gravity

## Acceleration due to Gravity Definition

We define the acceleration due to gravity as the constant acceleration produced on the body when it freely falls under the effect of gravity. It is denoted by $\overrightarrow{g}$ because it has both direction and magnitude.

Let us say a body of mass m is placed at distance R from the Earth whose mass is M.

We know that the earth tries to pull everything towards it by force. So, it pulls the body by force, mg.

Now, using the law of gravitation, we get,

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

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

This is the formula for the acceleration due to gravity.

### What is the Acceleration Due to Gravity?

Let us say, a person of mass m is standing at point B at a distance ‘R’ from the center of the Earth whose mass is M.

According to the Gauss law of gravitation, the Earth is a symmetric solid sphere whose mass is concentrated at the center. If we wish to calculate the gravitational field strength $\overrightarrow{E}$ at point B is $\frac{GM}{R^{2}}$.

This $\overrightarrow{E}$ is directed towards its mass, i.e, at the center. This value remains the same for a person standing outside and at the surface of the Earth.

Here, we can see that $\overrightarrow{E} \propto \frac{1}{R^{2}}$. We can show this by a graph.

This means the value of $\overrightarrow{E}$ is maximum at the surface and decreases with the increase in the distance from the earth.

We know that the earth’s gravitational field is $\overrightarrow{g}$. So, $\overrightarrow{E} = \overrightarrow{g}$.

From the formula, $g = \frac{GM}{R^{2}}$, we can see that $\overrightarrow{E} \propto \frac{1}{R^{2}}$.

This means the value of g varies with the variation in R.

It is maximum at the surface of the earth because of the distance of a person from the center of the earth, which is equal to the radius of the earth Re.

Now, R becomes equal to Re

So, the formula for g becomes: $g = \frac{GM}{R_{e}^{2}}$.

∵ g is maximum at the surface and decreases with the increase in the distance from the earth.

∴ The graph will be the same for g, as we have drawn for $\overrightarrow{E}$.

### Value of g on Earth

We know that formula for the acceleration due to gravity on the surface of the earth is:

$g = \frac{GM_{e}}{R_{e}^{2}}$.

Where,

$G = \text{Universal gravitational constant; whose value is } 6.673 \times 10^{-11} Nm^{2}kg^{-1}$

$M_{e} = 6 \times 10^{24} kg$

$R_{e} = 6.4 \times 10^{6}m$

On putting these values in the above formula, we get the value of g at the surface of the earth as 9.8 ms-2.  However, this value varies with height.

The dimensional formula for g is $[M^{0}LT^{-2}]$.

### Variation of g with Height

Consider earth of mass M, radius Re with center at O.

Let g = acceleration due to gravity at its surface, then $g = \frac{GM}{R_{e}^{2}}$ …(1)

If gh is the acceleration due to gravity at a point h above the surface of the earth, then

$g_{n} = \frac{GM}{(R + h)^{2}}$…(2)

Now, (2) ÷ (1)

$\frac{gh}{g} = \frac{R_{e}^{2}}{(R_{e} + h)^{2}} = (1 + \frac{h}{R_{e}})^{-2}$

∵ h < < Re, therefore, h/R is a very small value. Now expanding the term (1 + h/Re)2 by Binomial theorem, we get,

 $g_{n} = g(1 - \frac{2h}{R_{e}})$

We can see that $g_{e} \propto \frac{1}{R_{e}}$, the value of g decreases with height.

### Variation in g with Depth

From the formula $g = \frac{GM_{e}}{R_{e}^{2}}$, we can see that as we go below the earth’s surface, the distance between us and the center of the earth decreases. Therefore, the value of g decreases. Now, let us understand how it decreases.

Let us say, you are standing at point R and the earth is attracting you by force mg’. Here, we are using g’ in place of g because we have come inside the earth.

Using the law of gravitation

$mg' = \frac{GM}{R^{2}}$

If you observe, you are attracted by the inner surface, while M is the mass of the earth. So, we will use another method here:

If ρ is the uniform density of the inner surface, then

$M = \frac{4}{3} \pi R^{3} \rho$

$\text{So, } g = \frac{G \frac{4}{3} \pi R^{3} \rho}{R^{2}}$ ..(1)

Let the mass of the inner surface be M’, and radius, R - d, then

$g’ = \frac{GM’}{(R - d)^{2}}, \text{and M’ } = \frac{4}{3} \pi (R - d)^{3} \rho$

$g' = \frac{G\frac{4}{3} \pi {(R-d)}\rho ^{3}}{(R-d)^{2}}$…(2)

Now, (3) ÷ (2)

$\frac{g’}{g} = \frac{\frac{G\frac{4}{3} \pi {(R-d)}\rho ^{3}}{(R-d)^{2}}}{\frac{G \frac{4}{3} \pi R^{3} \rho}{R^{2}}}$

On solving, we get,

 $g’ = g (1 - \frac{d}{R})$

∴ The value of g decreases with depth.

Q1: Find the Height above the Earth’s Surface Where g Drops to 64% of g at the Surface.

Ans: Here, gh = 64/100 g

Now, using the formula, gh = g (1 + h/Re)2

64/100 g = g/(1 + h/Re)2

(1 + h/Re)= 10/8

h = 1/4 Re

Q2: Why is the Value of g Maximum at Pole and Minimum at the Equator?

Ans: The expression for the value of g due to the rotation of the earth is:

g' = g - Rω2Cos2ф

Here, ф is the latitude.

At the equator, ф = 0°, Cos ф = Cos 0° = 1, then

g' = g - Rω2

∴ g’ is minimum at the equator.

At the pole, ф = 90°, Cos ф = Cos 90° = 0, then

g' = g - Rω2(0)2

g’ = g, which is maximum

Hence, the value of g is minimum at the equator and maximum at the pole.

Q3: Where is the Gravity Strongest?

Ans: The gravity is strongest at the surface of the earth.

Q4: What does Microgravity Mean?

Ans: Microgravity is a condition in which astronauts feel weightless because of the absence of the force of gravity outside the earth.