Question:
Escape velocity on a planet is \(10\ \text{km/s}\). If the radius remains same but the mass becomes \(4\) times, the new escape velocity is:
Escape velocity on a planet is \(10\ \text{km/s}\). If the radius remains same but the mass becomes \(4\) times, the new escape velocity is:
Show Hint
Escape velocity depends on both mass and radius:
\[
v_e \propto \sqrt{\frac{M}{R}}
\]
If mass becomes \(n\) times and radius remains constant, escape velocity becomes:
\[
\sqrt{n}\times \text{original escape velocity}
\]
Updated On: Jun 3, 2026
- \(10\ \text{km/s}\)
- \(20\ \text{km/s}\)
- \(5\ \text{km/s}\)
- \(40\ \text{km/s}\)
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The Correct Option is B
Solution and Explanation
Concept:
Escape velocity is the minimum velocity required for an object to escape completely from the gravitational field of a planet without further propulsion.
The formula for escape velocity is:
:contentReference[oaicite:0]{index=0}
where:
\[
v_e = \text{escape velocity}
\]
\[
G = \text{gravitational constant}
\]
\[
M = \text{mass of the planet}
\]
\[
R = \text{radius of the planet}
\]
From the formula:
\[
v_e \propto \sqrt{M}
\]
when radius remains constant.
Step 1: Write the relation between old and new escape velocity. Initial escape velocity: \[ v_1=10\ \text{km/s} \] New mass: \[ M_2=4M_1 \] Since radius remains same: \[ \frac{v_2}{v_1}=\sqrt{\frac{M_2}{M_1}} \] Substitute: \[ \frac{v_2}{10}=\sqrt{\frac{4M_1}{M_1}} \] \[ \frac{v_2}{10}=\sqrt{4} \] \[ \frac{v_2}{10}=2 \] \[ v_2=20\ \text{km/s} \]
Step 2: Write the final answer. Hence, the new escape velocity is: \[ \boxed{20\ \text{km/s}} \] Therefore, the correct option is: \[ \boxed{(B)\ 20\ \text{km/s}} \]
Step 1: Write the relation between old and new escape velocity. Initial escape velocity: \[ v_1=10\ \text{km/s} \] New mass: \[ M_2=4M_1 \] Since radius remains same: \[ \frac{v_2}{v_1}=\sqrt{\frac{M_2}{M_1}} \] Substitute: \[ \frac{v_2}{10}=\sqrt{\frac{4M_1}{M_1}} \] \[ \frac{v_2}{10}=\sqrt{4} \] \[ \frac{v_2}{10}=2 \] \[ v_2=20\ \text{km/s} \]
Step 2: Write the final answer. Hence, the new escape velocity is: \[ \boxed{20\ \text{km/s}} \] Therefore, the correct option is: \[ \boxed{(B)\ 20\ \text{km/s}} \]
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