In this article, you will learn about the acceleration of an object placed in a field of gravity, at a height h above the surface of the Earth.

Let us consider the figure shown above. Here, we can see a point mass m is positioned at a height h above the surface of the earth. The radius of the earth is denoted by R_{e}. Here, the distance of the object from the centre of the earth is (R_{e} + h). Let us say, the magnitude of the force on the point mass m is denoted by F.

So,

\(F = \frac{GM_{e}m}{(R_{e}+h)^{2}}\)

Where, G is the universal gas constant which is equal to 6.67×10^-11.

M_{e} is the mass of the earth, m is the mass of the object, R_{e} is the radius of the earth and h is the height at which the object is positioned.

Now, the acceleration due to gravitational force experienced by the object can be given as F/m.

So,

\(g = \frac{F}{m}=\frac{GM_{e}}{(R_{e}+h)^{2}}\)

As we know that the expression for acceleration due to gravity for an object at the surface of the earth is given by \( \frac{GM_e}{R_{e}^2} \), we can say that, for an object placed at a height h, the acceleration due to gravity is less as compared to that placed on the surface.

## Below the Surface of the Earth

Let us consider the figure shown above. Here, we can see a point mass m is positioned at a depth d below the surface of the Earth. The distance of the object from the centre of the Earth is given by (R_{e}-d) and the magnitude of the gravitational force acting on the object is F(d).

So,

\(F(d)=\frac{GM_{s}m}{(R_{e}-d)^{2}}\)

Here, the object can be assumed to be present at the concentric sphere of radius (R_{e}-d).Hence, the mass of the earth contributing to the gravitational force is the mass of the inner concentric sphere.

The equivalent relation between the masses can be given as

\(\frac{M_{s}}{M_{e}}=\frac{(R_{e}-d)^{2}}{R_{e}^{3}}\)

Putting the value of M_{s} in the equation for gravitational force, we get

\(F(d)=\frac{GM_{e}m(R_{e}-d)^{2}}{R_{e}^{3}}\)

The acceleration due to gravity of point mass m at the given depth of d from the earth surface is given by

\(g(d)=\frac{GM_{e}(R_{e}-d)}{R_{e}^{3}} = \frac{GM_{3}}{R_{e}^{2}} \times \frac{R_{e}-d}{R_{e}} = \frac{GM_{e}}{R_{e}^{2}} \times 1 – \frac{d}{R_{e}}\)

As we know that the expression for acceleration due to gravity for an object at the surface of the earth is given by

\(\small \frac{GM_e}{R_{e}^2} \), we can say that, for an object placed at a height h, the acceleration due to gravity is less as compared to that placed on the surface.