If Earth has a mass 9 times and radius 2 times that of a planet P, then Ve3 x m s^-1 will be the minimum velocity required by a rocket to pull out of the gravitational force of P, where Ve is escape…

Given: Earth has mass 9 times and radius 2 times that of planet P. So, MP = ME9, RP = RE2 Find: The value of x in V escape, P = Ve3 x Escape velocity for a planet is given by V escape = 2GMR Therefore, for planet P, V escape, P = 2G MPRP Substitute MP = ME9, RP = RE2 Then, V escape, P = 2G ( ME9) RE2 = 4GME9RE = 23 GMERE On Earth, Ve = 2GMERE Hence, GMERE = Ve 2 Substituting this above, V escape, P = 23 Ve 2 = Ve 23 Comparing with Ve3 x we get x = 2 Therefore, the correct option is A. Note: The solution says D, but the actual working concludes x = 2, so the answer from the solution is A.

If Earth has a mass 9 times and radius 2 times that of a planet P… — Solution | JEE Physics

MCQEasyJEE 2023Escape Velocity

If Earth has a mass $9$ times and radius $2$ times that of a planet P, then $\frac{V_e}{3}\sqrt{x} \, \text{m s}^{-1}$ will be the minimum velocity required by a rocket to pull out of the gravitational force of P, where $V_e$ is escape velocity on Earth. The value of $x$ is:

A
$2$
B
$3$
C
$18$
D
$1$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Earth has mass $9$ times and radius $2$ times that of planet P.

So,

M_P = \frac{M_E}{9}, \qquad R_P = \frac{R_E}{2}

Find: The value of $x$ in

V_{\text{escape, P}} = \frac{V_e}{3}\sqrt{x}

Escape velocity for a planet is given by

V_{\text{escape}} = \sqrt{\frac{2GM}{R}}

Therefore, for planet P,

V_{\text{escape, P}} = \sqrt{\frac{2G M_P}{R_P}}

Substitute

M_P = \frac{M_E}{9}, \qquad R_P = \frac{R_E}{2}

Then,

V_{\text{escape, P}} = \sqrt{\frac{2G \left(\frac{M_E}{9}\right)}{\frac{R_E}{2}}}

= \sqrt{\frac{4GM_E}{9R_E}}

= \frac{2}{3}\sqrt{\frac{GM_E}{R_E}}

Comparison with Earth's Escape Velocity

On Earth,

V_e = \sqrt{\frac{2GM_E}{R_E}}

Hence,

\sqrt{\frac{GM_E}{R_E}} = \frac{V_e}{\sqrt{2}}

Substituting this above,

V_{\text{escape, P}} = \frac{2}{3} \cdot \frac{V_e}{\sqrt{2}}

= \frac{V_e\sqrt{2}}{3}

Comparing with

\frac{V_e}{3}\sqrt{x}

we get

x = 2

Therefore, the correct option is A.

Note: The solution says D, but the actual working concludes $x = 2$ , so the answer from the solution is A.

Common mistakes

Using the ratio in the wrong direction. Earth is $9$ times heavier and $2$ times larger in radius than planet P, so for planet P you must use $M_P = \frac{M_E}{9}$ and $R_P = \frac{R_E}{2}$ , not the reverse.
Forgetting that escape velocity depends on $\sqrt{\frac{M}{R}}$ , not directly on $\frac{M}{R}$ . First simplify the quantity inside the square root, then take the square root carefully.
Comparing the final expression incorrectly with $\frac{V_e}{3}\sqrt{x}$ . Once you obtain $\frac{V_e\sqrt{2}}{3}$ , the value under the square root is $x = 2$ , not $4$ .

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