Question:medium

The escape velocity from the surface of a planet is 11.2 km/s. If the radius of the planet is doubled but the mass remains the same, what will be the new escape velocity?

Show Hint

Escape velocity is inversely proportional to the square root of the radius if the mass is constant: \[ v_e \propto \frac{1}{\sqrt{R}} \]
Updated On: Jan 13, 2026
  • 22.4 km/s
  • 7.9 km/s
  • 15.8 km/s
  • 5.6 km/s
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Recall the escape velocity formula. \[ v_e = \sqrt{\frac{2GM}{R}} \] Step 2: Consider the scenario where the radius is doubled (\( R \rightarrow 2R \)) while the mass remains constant: \[ v_e' = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{2GM}{R}} = \frac{v_e}{\sqrt{2}} \] Step 3: Calculate the new escape velocity: \[ v_e' = \frac{11.2}{\sqrt{2}} \approx \frac{11.2}{1.414} \approx 7.92 \, \text{km/s} \approx 7.9 \, \text{km/s} \]
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