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: Understanding the Concept:
Escape speed is \(v = \sqrt{\frac{2GM}{R}}\), so \(v\propto\sqrt{\frac{M}{R}}\).

Step 2: Compare:
Earth has \(M_e = 9M_P\) and \(R_e = 2R_P\), so \(M_P = \frac{M_e}{9}\) and \(R_P = \frac{R_e}{2}\).
\[ \frac{v_P}{v_e} = \sqrt{\frac{M_P/R_P}{M_e/R_e}} = \sqrt{\frac{1/9}{1/2}} = \sqrt{\frac29} = \frac{\sqrt2}{3} \]

Step 3: Find x:
So \(v_P = \frac{v_e}{3}\sqrt2\), and comparing with \(\frac{v_e}{3}\sqrt{x}\) gives \(x = 2\).

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