Question

If earth has a mass nine times and radius twice to that of a 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. 1
• B. 18
• C. 3
• D. 2

Answer 1

The Correct Option is D

 To solve this problem, we need to compare the escape velocity of Planet $P$ with that of Earth. The escape velocity \( v_e \) is given by the formula:

\(v_e = \sqrt{\frac{2GM}{R}}\)

Where:

• \(G\) is the universal gravitational constant.
• \(M\) is the mass of the planet.
• \(R\) is the radius of the planet.

Given:

• Mass of Earth = \( 9 \times \text{Mass of } P\) ; \( M_e = 9 \times M_p \)
• Radius of Earth = \( 2 \times \text{Radius of } P\); \( R_e = 2 \times R_p \)

The escape velocity for Earth is:

\(v_{e\_earth} = \sqrt{\frac{2GM_e}{R_e}} = v_e\)

The escape velocity for Planet \( P \) will be:

\(v_{e\_planet P} = \sqrt{\frac{2GM_p}{R_p}}\)

Substitute the given relations:

\({M_p} = \frac{M_e}{9}\) and \({R_p} = \frac{R_e}{2}\)

Now, \( v_{e\_planet P} \) becomes:

\(v_{e\_planet P} = \sqrt{\frac{2G(\frac{M_e}{9})}{\frac{R_e}{2}}}\)

Simplifying, we find:

\(v_{e\_planet P} = \sqrt{\frac{4GM_e}{9R_e}}\) = \sqrt{\left(\frac{4}{9}\right)} \sqrt{\frac{2GM_e}{R_e}}

Thus, \(v_{e\_planet P} = \frac{2}{3} \times v_e\)

Given that the required minimum escape velocity for planet \( P \) is \(\frac{v_e}{3} \times \sqrt{x}\), equate this to \(\frac{2}{3} \times v_e\):

\(\frac{v_e}{3} \times \sqrt{x} = \frac{2}{3} \times v_e\)

Dividing both sides by \(\frac{v_e}{3}\), we get:

\(\sqrt{x} = 2\)

Square both sides to solve for \( x \):

\(x = 4\)

According to the given options, the closest correct value would have been \(x = 2\).