1. 1 : 2 2. 1 : 2√(2) 3. 1 : 4 4. 1 : √(2)

Answer: 2. 1 : 2√(2) Key idea: v_e=√(2GM/R)=R√(8π Gρ/3) , so v_e∝ Rsqrtρ . Planet: radius 2R , density 2ρ , so v_p=v_e×2×sqrt2=2sqrt2 v_e . Ratio v_e:v_p=1:2sqrt2 .

Topic: Escape velocity· Gravitation

The ratio of escape velocity at earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice as that of earth is :

1. 1 : 2

2. \(1 : 2\sqrt{2}\)

3. 1 : 4

4. \(1 : \sqrt{2}\)

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Answer: 2. \(1 : 2\sqrt{2}\)

Key idea: \(v_e=\sqrt{\dfrac{2GM}{R}}=R\sqrt{\dfrac{8\pi G\rho}{3}}\), so \(v_e\propto R\sqrt\rho\).

Planet: radius \(2R\), density \(2\rho\), so \(v_p=v_e\times2\times\sqrt2=2\sqrt2\,v_e\). Ratio \(v_e:v_p=1:2\sqrt2\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The ratio of escape velocity at earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice as that of earth is :

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