Question:
Escape velocity at surface of earth is \(11.2\ \text{km/s}\). Escape velocity from a planet whose mass is the same as that of earth and radius \(1/4\) that of earth, is
Escape velocity at surface of earth is \(11.2\ \text{km/s}\). Escape velocity from a planet whose mass is the same as that of earth and radius \(1/4\) that of earth, is
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\(v_e \propto \frac{1}{\sqrt{R}}\) when \(M\) is constant.
Updated On: Apr 23, 2026
- \(2.8\ \text{km/s}\)
- \(15.6\ \text{km/s}\)
- \(22.4\ \text{km/s}\)
- \(44.8\ \text{km/s}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Escape velocity \(v_e = \sqrt{\frac{2GM}{R}}\).
Step 2: Detailed Explanation:
For earth: \(v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2\ \text{km/s}\).
For planet: \(M_p = M_e\), \(R_p = R_e/4\).
\(v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2GM_e}{R_e/4}} = \sqrt{4 \times \frac{2GM_e}{R_e}} = 2 \times \sqrt{\frac{2GM_e}{R_e}} = 2 \times 11.2 = 22.4\ \text{km/s}\).
Step 3: Final Answer:
Thus, escape velocity = \(22.4\ \text{km/s}\).
Escape velocity \(v_e = \sqrt{\frac{2GM}{R}}\).
Step 2: Detailed Explanation:
For earth: \(v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2\ \text{km/s}\).
For planet: \(M_p = M_e\), \(R_p = R_e/4\).
\(v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2GM_e}{R_e/4}} = \sqrt{4 \times \frac{2GM_e}{R_e}} = 2 \times \sqrt{\frac{2GM_e}{R_e}} = 2 \times 11.2 = 22.4\ \text{km/s}\).
Step 3: Final Answer:
Thus, escape velocity = \(22.4\ \text{km/s}\).
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