A body weighs $72 \mathrm{~N}$ on the surface of the earth. The gravitational force experienced by t

Question

A body weighs $72 \mathrm{~N}$ on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to half the radius of the earth ($R/2$) from its surface is

Professor PC Thomas Classes · Answer

(a) : Let the weight on the surface be $W = mg = 72 \mathrm{~N}$. The acceleration due to gravity at a height $h$ is given by $g_h = g \left(\frac{R}{R+h}ight)^2$. Here, $h = R/2$, so $g_h = g \left(\frac{R}{R + R/2}ight)^2 = g \left(\frac{R}{1.5R}ight)^2 = g \left(\frac{2}{3}ight)^2 = \frac{4}{9}g$. The gravitational force at this height is $W_h = m g_h = m \left(\frac{4}{9}gight) = \frac{4}{9} (mg)$. Substituting the surface weight, $W_h = \frac{4}{9} 	imes 72 \mathrm{~N} = 32 \mathrm{~N}$.