Question:medium

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:

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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

Step 1: Understanding the Concept:
Escape velocity ($v_e$) is the speed needed for an object to break free from a planet's gravitational pull and reach infinity with zero kinetic energy.
It is determined by the gravitational potential energy at the surface.
Step 2: Key Formula or Approach:
The formula for escape velocity is:
\[ v_e = \sqrt{\frac{2GM}{R}} \] where $G$ is the Gravitational constant, $M$ is the mass of the planet, and $R$ is its radius.
This shows that $v_e \propto \sqrt{M}$ and $v_e \propto \frac{1}{\sqrt{R}}$.
Detailed Explanation:
Let the initial escape velocity be $v_{e1} = 10$ km/s for a planet of mass $M$ and radius $R$.
The new conditions for the second planet are:
New Mass $M' = 4M$.
New Radius $R' = R$.
The new escape velocity $v_{e2}$ is:
\[ v_{e2} = \sqrt{\frac{2GM'}{R'}} \] Substitute $M'$ and $R'$ into the formula:
\[ v_{e2} = \sqrt{\frac{2G(4M)}{R}} \] Factor out the constant 4:
\[ v_{e2} = \sqrt{4} \cdot \sqrt{\frac{2GM}{R}} \] \[ v_{e2} = 2 \cdot v_{e1} \] Substituting the value of $v_{e1} = 10$ km/s:
\[ v_{e2} = 2 \cdot 10 = 20 \text{ km/s} \] Step 3: Final Answer:
The new escape velocity is 20 km/s.
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