Acceleration Due To Gravity Above And Below The Earth's Surface

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.

gravity 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 Re. Here, the distance of the object from the centre of the earth is (Re + 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.

Me is the mass of the earth, m is the mass of the object, Re 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

Gravity for an Object 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 (Re-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 (Re-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 Ms 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.


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