Question

The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is four times and density is nine times that of the earth, is:

• A. $36$$v_e$
• B. $12$$v_e$
• C. $6$ $v_e$
• D. $20$ $v_e$

Answer 1

The correct option is B

$12$$v_e$

Step 1: Given Data

$4\times$ Radius of earth $\left(R\right)=$ Radius of planet $\left(r\right)$

$9\times$ Density of earth $\left(\rho \right)=$ Density of planet $(\rho \text{'})$

Step 2: Formula used

The formula for escape velocity $v_e$is

$v_e=\sqrt{\frac{2GM}{R}}$

$G=$ universal gravitational constant

The density of earth from Newton's law of gravitation,
$\rho$ is the density of earth

$\rho \text{'}$ is the density of planet

$$\begin{gathered} \rho =\frac{M}{{\displaystyle \frac{4}{3}}\pi R³} \\ \rho \text{'}=\frac{m}{{\displaystyle \frac{4}{3}}\pi r³} \end{gathered}$$

$m$ is the mas of planet

$M$ is the mass of earth

$R$ is the radius of earth

$r$ is the radius of planet

Step 3: Calculate the Mass of the Planet

As we know,

$$\begin{gathered} \rho =\frac{3M}{4\pi R³} \\ \rho \text{'}=\frac{3m}{4\pi r³} \end{gathered}$$

$9\times$ Density of earth $\left(\rho \right)=$ Density of planet $(\rho \text{'})$
$$\begin{gathered} 9\times \frac{3M}{4\pi R³}=\frac{3m}{4\pi r³} \\ 9\times \frac{M}{R³}=\frac{m}{\left(4R\right)³}\left[\because r=4R\right] \\ m=64\times 9M \end{gathered}$$

Step 4: Calculate the Escape Velocity of the Planet

We know that escape velocity $v_e=\sqrt{\frac{2GM}{R}}$

For earth $v_e=\sqrt{\frac{2GM}{R}}$

For planet $v{\text{'}}_e=\sqrt{\frac{2Gm}{r}}=\sqrt{\frac{2G\left(64\times 9M\right)}{4R}}=12\sqrt{\frac{2GM}{R}}=12v_e$

$v{\text{'}}_e$is the escape velocity for a planet whose radius is four times and density is nine times that of the earth.

Hence option B is the correct answer.