Question

A planet's density is $3$ times that of the Earth. But the acceleration due to gravity on its surface is exactly the same as on the Earth's surface. The radius of the planet in terms of the Earth's radius $R$ is

• A. $2R$
• B. $3R$
• C. $\frac{R}{3}$
• D. none of the above

Answer 1

The correct option is C $\frac{R}{3}$

In general we know that, force of gravity between two objects (M and m) at a distance r is given by:

$F=G\frac{Mm}{r^2}$

Also $F=mg$

$\therefore g=\frac{GM}{r^2}$

Now, acceleration due to gravity on the planet's surface is exactly the same as on the Earth's surface:

$g_p=g_e$

$\frac{G{M}_p}{r_p^2}=\frac{G{M}_e}{r_e^2}$

$\frac{M_p}{r_p^2}=\frac{M_e}{r_e^2}$ - (1)

Given: ${\rho }_p=3\times {\rho }_e$

$\frac{4/3\pi r_p^3}{r_p^2}\times 3=\frac{4/3\pi r_e^3}{r_e^2}$

$\therefore 3r_p=r_e$

$r_p=\frac{R}{3}$