Question

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g,$ where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11{\text{kms}}^{-1}$, the escape speed on the surface of the planet in ${\text{kms}}^{-1}$ will be

• A. $2{\text{kms}}^{-1}$
• B. $3{\text{kms}}^{-1}$
• C. $4{\text{kms}}^{-1}$
• D. $5{\text{kms}}^{-1}$

Answer 1

The correct option is B $3{\text{kms}}^{-1}$

Let the gravitational acceleration on the surface of the planet be $g_p$ and mass density be ${\rho }_p$.

According to the problem,

$\frac{g_p}{g}=\frac{\sqrt{6}}{11}$ & $\frac{{\rho }_p}{\rho }=\frac{2}{3}$

The acceleration due to gravity on the surface of a planet of mass $M_p$ and radius $R_p$ is given by,

$g_p=\frac{G{M}_p}{R_p^2}...\left(1\right)$

The mass is related to density ${\rho }_p$ by
$M_p=\frac{4}{3}\pi R_p^3{\rho }_p$

Substitute in above equation and simplify to get,

$R_p=\frac{3g_p}{4\pi G{\rho }_p}...\left(3\right)$

The escape velocity from the surface of the planet is given by,

$v_p=\sqrt{\frac{2G{M}_p}{R_p}}...\left(4\right)$

From $\left(1\right)$ and $\left(4\right)$,

$v_p=\sqrt{2g_p{R}_p}$

Substituting the value from $\left(3\right)$,

$\therefore v_p=\sqrt{\frac{3g_p^2}{2\pi G{\rho }_p}}$

Substitute values in above equation to get,
$\frac{v_p}{v_e}=\sqrt{\frac{g_p^2/g^2}{{\rho }_p/\rho }}$

$\Rightarrow \frac{v_p}{v_e}=\sqrt{\frac{(\sqrt{6}/11{)}^2}{2/3}}$

$\Rightarrow \frac{v_p}{v_e}=\frac{3}{11}$
$\therefore v_p=\frac{3}{11}\times v_e=\frac{3}{11}\times 11=3{\text{kms}}^{-1}.$

Hence, option (c) is correct answer.