Question

A body weighs $72N$ on the surface of the earth. What will be the force of gravity acting on the body at a height equal to half the radius of the earth above the earth's surface?

• A. $32N$
• B. $20N$
• C. $10N$
• D. 0

Answer 1

The correct option is A

$32N$

The value of $g$ at earth's surface is

$g_e=\frac{GM}{R^2}$ ... (1)

The value of $g$ at a height $h$ above the earth's surface is given by

$g^{\prime }=\frac{GM}{{(R+h)}^2}$
$=\frac{GM}{{(R+\frac{R}{2})}^2}$ ... (2) $[\because h=\frac{R}{2}]$

Dividing equation (2) by equation(1), we get

$\frac{g^{\prime }}{g_e}=\frac{4}{9}$

$\Rightarrow g^{\prime }=\frac{4}{9}$ $g_e$ ... (3)

Now, weight of the man on the surface of the earth is given by

$w=m{g}_e$

Weight of the man at a height, $h=\frac{R}{2}$

$w^{\prime }=m{g}^{\prime }$

$\Rightarrow \frac{w^{\prime }}{w}=\frac{g^{\prime }}{g_e}=\frac{4}{9}$

$\Rightarrow w^{\prime }=\frac{4}{9}w$

$\Rightarrow w^{\prime }=\frac{4}{9}\times 72=32N$