Question

If earth has a mass nine times and radius twice to the planet P. Then \(\frac{v_e}{3}\sqrt{x}\) ms\(^{-1}\) will be the minimum velocity required by a rocket to pull out of gravitational force of P, where \(v_e\) is escape velocity on earth. The value of \(x\) is

• A. \(2\)
• B. \(3\)
• C. \(18\)
• D. \(1\)

Hint:

\(v_{esc}=\sqrt{2GM/R}\), so it scales as \(\sqrt{M/R}\).

Answer 1

The Correct Option is A

Step 1: Plan:
Write both escape speeds with $g$ and $R$.

Step 2: Steps:
Using $GM = gR^2$, $v = \sqrt{2gR}$. The planet P has mass $\frac19$ and radius $\frac12$ of Earth's, so $g_P = \frac{GM_P}{R_P^2} = g_e\cdot\frac{1/9}{1/4} = \frac{4g_e}{9}$.
$v_P = \sqrt{2g_PR_P} = \sqrt{2\cdot\frac{4g_e}{9}\cdot\frac{R_e}{2}} = \frac{2}{3}\sqrt{g_eR_e}$, while $v_e = \sqrt{2g_eR_e}$. So $v_P = \frac{2}{3}\cdot\frac{v_e}{\sqrt2} = \frac{\sqrt2}{3}v_e$. Hence $x = 2$.

Final Answer:
The value of $x$ is $2$, option (A). \[ \boxed{2} \]