Question

The ratio of escape velocity at earth $v_e$ to the escape velocity at a planet $v_p$ whose radius and mean density are twice as the one of the earth is:

Answer 1

Step 1: Given

Given, the escape velocity of Earth is $v_e$ and the escape velocity of the planet is $v_p$.

Let ${\rho }_e$ be the density of Earth and ${\rho }_p$ be the density of the planet.
We are given,
${\rho }_p=2{\rho }_e$

Let $r_e$ be the radius of Earth and $r_p$ be the radius of the planet.
We are given,
$r_p=2r_e$

Step 2: Formulas used

We know that the density of an object is given as,
$\rho =\frac{m}{V}$
where $m$ is the mass and $V$ is the volume.

We know that the volume of a sphere is given as,
$V=\frac{4}{3}\mathrm{\pi }{r}^3$
where $r$ is its radius.

We know that the escape velocity is given as,
$v_s=\sqrt{\frac{2GM}{R}}$
where $M$ is the mass of the planet, $R$ is its radius and $G$ is the gravitational constant.

Step 3: Express mass in terms of density

The mass of the earth in terms of its density can be written as,
$m_e={\rho }_e{V}_e$

Similarly, the mass of the planet is,
$\begin{array}{rcl}{m}_p& =& {\rho }_p{V}_p\\ & =& 2{\rho }_e{V}_p\end{array}$

Step 4: Find the ratio of escape velocities

The ratio of the escape velocity of the earth to the planet is,
$\begin{array}{rcl}\frac{v_e}{v_p}& =& \frac{\sqrt{{\displaystyle \frac{2G{m}_e}{r_e}}}}{\sqrt{{\displaystyle \frac{2G{m}_p}{r_p}}}}\\ & =& \frac{\sqrt{{\displaystyle \frac{{\rho }_e{V}_e}{r_e}}}}{\sqrt{{\displaystyle \frac{2{\rho }_e{V}_p}{2r_e}}}}\\ & =& \frac{\sqrt{{\displaystyle \frac{4}{3}\mathrm{\pi }{r_{\mathrm{e}}}^3}}}{\sqrt{{\displaystyle \frac{{\displaystyle 4}}{{\displaystyle 3}}\mathrm{\pi }{\left(r_p\right)}^3}}}\\ & =& \frac{\sqrt{{\displaystyle {r_e}^3}}}{\sqrt{{\displaystyle {\left(2r_e\right)}^3}}}\\ & =& \frac{\sqrt{r_e}}{\sqrt{8{r_e}^3}}\\ & =& \frac{1}{2\sqrt{2}}\end{array}$

Therefore, the ratio of the escape velocity of the earth to the escape speed of the planet is $\frac{1}{2\sqrt{2}}$.