Escape velocity on the surface of the earth is 11.2 kms-1. If the mass of the earth increases to twice its value and the radius of the earth become half the escape velocity is:
Correct answer: C. 22.4 kms-1
- A. 5.6 kms-1
- B. 11.2 kms-1
- C. 22.4 kms-1
- D. 33.6 kms-1
Explanation
escape velocity = √ (2GM / r)where: G= gravitation constantM = mass of the earthr = radius of earth2 and G are constant in this equation, sov ∝ √ M / rM' = 2M ; r' = 1/2 rv' ∝ √2M / (1/2 r) ==> v' ∝ √ 4 M / rv' ∝ 2 √ M / rv' = 2 v∴ new escape velocity = 2 x 11.2 = 22.4 kms-1
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