Question

The ratio of acceleration due to gravity at a height 3R above earth's surface to the acceleration due to gravity on the surface o f the earth is (R = radius of earth)

• A. $\frac{1}{9}$
• B. $\frac{1}{4}$
• C. $\frac{1}{16}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

The value of acceleration due to gravity changes with height (ie, altitude). Ifg' is the acceleration due to gravity at a point, at height h above the surface of earth, then
$g'=\frac{GM}{(R+h)^2}$
but, $g=\frac{GM}{R^2}$
$\therefore $ $\frac{g'}{g}=\frac{GM}{(R+h)}=\frac{GM}{(R+h)^2}$
Here $g'=\frac{GM}{(R+h)}=\frac{GM}{(R+3R)^2}$
= $\frac{GM}{(4R)^2}=\frac{GM}{16R^2}=\frac{g_e}{16}$