Question

A body weighs $72N$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

Answer 1

The correct option is D $32N$

At the surface of Earth, weight $W=72N$

$\therefore F=mg=72N$

$m=\frac{72}{10}=7.2kg$

Now, at some height, h acceleration due to gravity is given as

$g^{\prime }=\frac{g}{(1+\frac{h}{R}{)}^2}$

Here, $h=\frac{R}{2}$

$g^{\prime }=\frac{g}{(1+\frac{1}{2}{)}^2}=\frac{4}{9}g=4.44m/s^2$

Now, the force experienced by the body at half of the radius of the earth is given by:

$F^{\prime }=m{g}^{\prime }=7.2\times 4.44=31.97N\approx 32N$