Question

A body weighs \(63~N\) on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?

Answer 1

let, \(g_h\) be the acceleration due to gravity at a height \(h\) from the ground.

We know that \(\dfrac{g_h}{g}=\left ( \dfrac{R}{R+h} \right )^2\)

Here, \(h=\dfrac{R}{2}\)

\(\therefore \dfrac{g_h}{g}=\left ( \dfrac{R}{R+R/2} \right )^2=\dfrac{4}{9}\)

If \(W\) be the weight of the body on the surface of Earth and \(W_h\) the weight of the body at height \(h\)

Then, \(\dfrac{W_h}{W}=\dfrac{mg_h}{mg}=\dfrac{g_h}{g}=\dfrac{4}{9}\)

or \(W_h=\dfrac{4}{9}W=\dfrac{4}{9}\times 63=28 N\)