Question

The ratio of acceleration due to gravity at a height 3R above earth's surface to the acceleration due to the gravity on the surface of the earth is:

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

Answer 1

The correct option is C $\frac{1}{16}$

Acceleration due to gravity
$g=\frac{GM}{R_e^2}$ $\cdots \left(1\right)$
$g^{\prime \prime }=\frac{GM}{(R_e+H{)}^2}$
$g^{\prime \prime }=\frac{GM}{(R_e+3R_e{)}^2}[\therefore H=3R_e]$
$g^{\prime \prime }=\frac{1}{16}\frac{GM}{R_e}$ $\cdots \left(2\right)$
By (1) and (2)
$\frac{g}{g^{\prime \prime }}=\frac{1}{16}$