The escape velocity of a body from Earth is 11.2 km/s. If the rad… — Solution | JEE Physics

MCQEasyJEE 2024Escape Velocity

The escape velocity of a body from Earth is $11.2 \, \text{km/s}$ . If the radius of a planet is one-third the radius of Earth and its mass is one-sixth that of Earth, the escape velocity from the planet is:

A
$11.2 \, \text{km/s}$
B
$8.4 \, \text{km/s}$
C
$4.2 \, \text{km/s}$
D
$7.9 \, \text{km/s}$

Answer

Correct answer:D

Step-by-step solution

the solution unavailable

Given: Escape velocity from Earth is $11.2 \, \text{km/s}$ . The planet has mass $M_p = \frac{M_E}{6}$ and radius $R_p = \frac{R_E}{3}$ .

Find: Escape velocity from the planet.

Working could not be extracted from the solution. Using the escape velocity relation

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

we compare the planet with Earth:

\frac{v_p}{v_E} = \sqrt{\frac{M_p/R_p}{M_E/R_E}} = \sqrt{\frac{(M_E/6)/(R_E/3)}{M_E/R_E}} = \sqrt{\frac{1}{2}}

Therefore,

v_p = 11.2 \times \frac{1}{\sqrt{2}} \approx 7.9 \, \text{km/s}

So, the correct option is D.

Common mistakes

Using a direct proportionality $v_e \propto \frac{M}{R}$ is wrong because escape velocity depends on the square root: $v_e \propto \sqrt{\frac{M}{R}}$ . Always apply the square root before comparing values.
Taking the planet radius ratio as $\frac{R_E}{3}$ in the denominator without simplifying carefully can cause ratio errors. First write $M_p = \frac{M_E}{6}$ and $R_p = \frac{R_E}{3}$ , then evaluate $\frac{M_p/R_p}{M_E/R_E}$ step by step.

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Answer

Step-by-step solution

Common mistakes

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