Escape velocity of projectile on a planet's surface is `11.2 kms^(-1)` . If a body is projected at double the speed its speed at an infinite distance from the planet is

Options

A. `11.2 km s^(-1)`
B. `1.2 km s^(-1)`
C. `7.4 km s^(-1)`
D. `19.4 km s^(-1)`

Answer

D

Explanation

From the principle of conservation energy
`(1)/(2) mv^(2) = (1)/(2) mv_(1)^(2) - (1)/(2) mv_(e)^(2) or v = sqrt(v_(1)^(2) - v_(e)^(2)) = sqrt((2 v_(e))^(2) - v_(e)^(2))`
`= sqrt(3) v_(e) = 1. 732 xx 11.2 = 19 . 4 km s^(-1)`