Question

The escape velocity for a rocket from earth is 11.2 kms^–1 value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be = ………… kms^–1 

(A) 11.2 

(B) 22.4 

(C) 5.6 

(C) 53.6

Answer 1

Correct option: (B) 22.4

Explanation:

V_e = √(2gR)

[(V_(e)2 / V_(e)1] = √[(g₂R₂) / (g₁R₁)]

Given :  V_(e)1 = 11.2

g₂ = 2g₁ 

R₂ = 2R₁ 

V_(e)2 = 11.2 × √(2 × 2)

V_(e)2 = 22.4 km/s