Answer to Question #230214 in Physics for frank

Question

Prove that the value of g is maximum at the surface of the earth and decrease with the height or depth

Answer 1

The acceleration due to gravity is the acceleration attained by a body in vacuum while it is free falling or falling under gravity alone. This acceleration is caused by the gravitational attraction of the planet and the body. The acceleration due to gravity depends on the mass of the planet where the object is dropped and the distance between the plane and the object. It is denoted by ‘g’. The acceleration due to gravity for a planet of mass M and radius R is given by,

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

Where, G is the gravitational constant.

The above equation gives the acceleration due to gravity value at the surface of the earth. So what happens if you take the object up or down the surface of the earth?

The change in acceleration due to gravity at a height h above the surface of the earth is given by,

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

From the equation it is clear that as h increases the value of g decreases.

The value of g at a depth ‘d’ from the surface of the earth is given by the formula

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

So the value of g decreases as we go to a depth from the surface of the earth. So the maximum value of g is at the surface of the earth.