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Question:

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

1:2

1:2√2

1:4

1:√2

Solution:

We know that: ve = √(2GM/R)
Given that: R' = 2R and ρ' = 2ρ ⇒ M = (4/3)πR³ x ρ ⇒ M' = (4/3)π(2R)³ x 2ρ = 16M
⇒ vp = √(2G(16M)/2R) = √(16GM/R)
⇒ ve/vp = √(2GM/R) / √(16GM/R) = 1/2√2