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 of 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)

Acceleration due to gravity decreases with height.
The value of acceleration due to gravity changes with height (ie, altitude). If $g^{\prime }$ is the acceleration due to gravity at a point, at height $h$ above the surface of earth, then
$g^{\prime }=\frac{GM}{{(R+h)}^2}$
but, $g=\frac{GM}{R^2}$
$\therefore \frac{g^{\prime }}{g}=\frac{GM}{{(R+h)}^2}\times \frac{R^2}{GM}=\frac{R^2}{{(R+h)}^2}$
Here, $g^{\prime }=\frac{GM}{{(R+h)}^2}=\frac{GM}{{(R+3R)}^2}$
$=\frac{GM}{{\left(4R\right)}^2}=\frac{GM}{16R^2}$
$=\frac{ge}{16}$
Note: At a depth, value of $g$ also decreases. Its value at centre of earth is zero, value of acceleration due to gravity is maximum at the earth's surface. Also value of acceleration due to gravity increases as we go from equator to the pole.