Question

If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that the earth, the gravitational acceleration on the surface of that planet is:

Answer 1

Acceleration due to gravity $g=\frac{GM}{R^2}$

Mass of the planet, $M=Vd=d\times \frac{4}{3}\pi R^3$ where $d$ is the density of the planet

$⟹$ $g=\frac{4}{3}\pi dGR$ (for earth)

Given : $d^{\prime }=d$ and $R^{\prime }=0.2R$

$\therefore$ $g^{\prime }=\frac{4}{3}\pi d^{\prime }G{R}^{\prime }=\frac{4}{3}\pi dG\left(0.2R\right)$

$⟹$ $g^{\prime }=0.2g$