Question

The escape velocity from the earth is $11km{s}^{-1}$. The escape velocity from a planet having twice the radius and same mean density as that of earth is

• A. $5.5km{s}^{-1}$
• B. $11km{s}^{-1}$
• C. $22km{s}^{-1}$
• D. None of these

Answer 1

The correct option is C

$22km{s}^{-1}$

Step 1: Given data

Escape Velocity $v_e=11km{s}^{-1}$

Step 2: Formula Used

The earth's escape velocity is given by $v_e=\sqrt{\frac{GM}{R}}\dots \dots \dots \left(1\right)$

Where, Escape Velocity =$v_e$, G = universal gravitational, M = mass of the earth, and R = radius of the earth

The radius of the planet $\left(R_p\right)$is twice the radius of the earth$\left(R_e\right)$

$i.e.R_p=2R_e$

${\rho }_p={\rho }_e$, where ${\rho }_p$ is the density of the planet and ${\rho }_{e}isdensityofearth$

Step 3: Calculation of the escape velocity,

$\begin{array}{l}\begin{array}{c}v=\sqrt{\frac{2GM}{R}}\\ =\sqrt{\frac{2G\times \frac{4}{3}\pi R^3\rho }{R}}\\ =\sqrt{\frac{8}{3}\pi G{R}^2\rho }\end{array}\\ \begin{array}{l}\therefore v\propto \sqrt{\rho R^2}\end{array}\end{array}$

Therefore,

$\begin{array}{l}\begin{array}{l}\frac{{\mathrm{v}}_{\mathrm{p}}}{{\mathrm{v}}_{\mathrm{e}}}=\sqrt{\frac{{\mathrm{\rho }}_{\mathrm{e}}(2{\mathrm{R}}_{\mathrm{e}}{)}^2}{{\mathrm{\rho }}_{\mathrm{e}}{\mathrm{R}}_{\mathrm{e}}^2}}\end{array}\\ \begin{array}{l}\frac{{\mathrm{v}}_{\mathrm{p}}}{11}=\sqrt{\frac{{\mathrm{\rho }}_{\mathrm{e}}(2{\mathrm{R}}_{\mathrm{e}}{)}^2}{{\mathrm{\rho }}_{\mathrm{e}}{\mathrm{R}}_{\mathrm{e}}^2}}=2\end{array}\\ \begin{array}{l}{\mathrm{v}}_{\mathrm{p}}=22\mathrm{km}/\mathrm{s}\end{array}\end{array}\begin{array}{}\end{array}$

Hence, option C is the correct answer.