Question

The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is it the radius of planet would be
A.$2R$
B.$4R$
C.$\dfrac{1}{4}R$
D.$\dfrac{1}{2}R$

Answer 1

Use the relation between the acceleration due to gravity on the surface of the planet, mass of the planet and radius of the planet. Derive the equation for the mass of the planet and earth in terms of density of the planet and the earth. Use the given condition for density of the planet and calculate the radius of the planet in terms of radius of the earth.

Formula used:
The expression for the acceleration due to gravity $g$ is
$g = \dfrac{{GM}}{{{R^2}}}$ …… (1)
Here, $G$ is the universal gravitational constant, $M$ is the mass of the planet and $R$ is the radius of the planet.
The equation for the density $\rho$ of an object is
$\rho = \dfrac{M}{V}$ …… (2)
Here, $M$ is the mass of the object and $V$ is the volume of the object.
The volume $V$ of a spherical object is given by
$V = \dfrac{4}{3}\pi {R^3}$ …… (3)
Here, $R$ is the radius of the spherical object.

Complete step by step answer:
We have given that the density ${\rho _p}$ of the planet is twice the density $\rho$ of the earth.
${\rho _p} = 2\rho$
Derive the equation for mass of the earth equations (2) and (3).
Substitute for $V$ in equation (2) and rearrange it for the mass $M$ of the earth.
$\rho = \dfrac{M}{{\dfrac{4}{3}\pi {R^3}}}$
$\Rightarrow \rho = \dfrac{{3M}}{{4\pi {R^3}}}$
$\Rightarrow M = \dfrac{{4\pi \rho {R^3}}}{3}$
Rewrite the above equation for mass ${M_p}$ of the planet.
$\Rightarrow {M_p} = \dfrac{{4\pi {\rho _p}R_p^3}}{3}$
Rewrite equation (1) for acceleration due to gravity $g$ on the surface of the earth.
$g = \dfrac{{GM}}{{{R^2}}}$
Substitute $\dfrac{{4\pi \rho {R^3}}}{3}$ for $M$ in the above equation.
$g = \dfrac{{G\left( {\dfrac{{4\pi \rho {R^3}}}{3}} \right)}}{{{R^2}}}$
$\Rightarrow g = \dfrac{{4\pi G\rho R}}{3}$
Rewrite equation (1) for acceleration due to gravity ${g_p}$ on the surface of the planet.
${g_p} = \dfrac{{G{M_p}}}{{R_p^2}}$
Substitute $\dfrac{{4\pi {\rho _p}R_p^3}}{3}$ for ${M_p}$ in the above equation.
${g_p} = \dfrac{{G\left( {\dfrac{{4\pi {\rho _p}R_p^3}}{3}} \right)}}{{R_p^2}}$
$\Rightarrow {g_p} = \dfrac{{4\pi G{\rho _p}{R_p}}}{3}$
The acceleration due to gravity on the surface of the planet and the earth are the same.
${g_p} = g$
Substitute $\dfrac{{4\pi G{\rho _p}{R_p}}}{3}$ for ${g_p}$ and $\dfrac{{4\pi G\rho R}}{3}$ for $g$ in the above equation.
$\dfrac{{4\pi G{\rho _p}{R_p}}}{3} = \dfrac{{4\pi G\rho R}}{3}$
$\Rightarrow {\rho _p}{R_p} = \rho R$
Substitute $2\rho$ for ${\rho _p}$ in the above equation.
$\Rightarrow 2\rho {R_p} = \rho R$
$\Rightarrow {R_p} = \dfrac{1}{2}R$
Therefore, the radius of the planet would be $\dfrac{1}{2}R$.

So, the correct answer is “Option D”.

Note:
The students may think that the shape of the planet is not given then how to determine the volume of the planet. The shape of the planet if considered as spherical for the sake of convenience as the shape of the earth is also considered spherical. Hence, the formula for the volume of the spherical object is taken.