Question

A planet having a same density as that of earth but its radius is $4$ times bigger than the radius of the earth. If the acceleration due to gravity on the surface of earth is $g$ and that on the surface of the planet is $g^{\prime }$, then

• A. $g^{\prime }=\frac{g}{4}$
• B. $g^{\prime }=16g$
• C. $g^{\prime }=9g$
• D. $g^{\prime }=4g$

Answer 1

The correct option is D $g^{\prime }=4g$

If the mass and radius of the new planet be $m^{\prime }$ and $R^{\prime }$ for same density as earth, then,
$R^{\prime }=4R\&m^{\prime }=64m$ $\text{as}[m\propto V],\text{where}V\propto R^3]$
For earth:
$g=\frac{Gm}{R^2}----\left(1\right)$
for new planet:
$g^{\prime }=\frac{G{m}^{\prime }}{R^2}=\frac{G\left(64m\right)}{(4R{)}^2}\Rightarrow 4\left(\frac{GM}{R^2}\right)=4g$
$\therefore g^{\prime }=4g$