Question

Acceleration due to gravity at earth's surface is $10m{s}^{-2}$ . The value of acceleration due to gravity at the surface of a planet of mass $\frac{1}{5}$ th and radius $\frac{1}{2}$ of the earth is -

• A. $4m{s}^{-2}$
• B. $6m{s}^{-2}$
• C. $8m{s}^{-2}$
• D. $12m{s}^{-2}$

Answer 1

Answer: (C)

$g_p=\frac{G{M}_p}{R_P^2}$
$M_p=\text{mass of planet}$
$R_p=\text{radius of planet }$
$g_p=\text{Acceleration due to gravity on planet}$
$g_p=\frac{G\left(\frac{M_e}{5}\right)}{{\left(\frac{R_e}{2}\right)}^2}$
$=G\times \frac{1}{5}\times M_e\times \frac{4}{R_e^2}$
$=\frac{4}{5}g=8m{s}^{-2}$