Question

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

Answer 1

Formula used: $g=\frac{GM}{R^2}$
Given, $mg=63N$, $h=\frac{R}{2}$
Acceleration due to gravity at height $h$ is given by
$g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2$

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

$={\left(\frac{R}{R+\frac{R}{2}}\right)}^2$

$={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$

$g_h=\frac{4}{9}g$

$\therefore m{g}_h=\frac{4}{9}mg$

$=\frac{4}{9}\times 63$

$=28N$
Final Answer: $28$