A rocket is launched striaght up from the surface of the earth the earth the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is `11.2 kms^(-1)`)

Options

A. `11.2 kms^(-1)`
B. `10.7 kms^(-1)`
C. `9.7 kms^(-1)`
D. `8.7kms^(-1)`

Answer

C

Explanation

Given escape velocity on the surface of the earth .
`v_(e) = sqrt(2gR_(e )) =11.2 km//s`
where R is the radius of earth
Height of the rocket from the surface of earth
`h = (R_(e ))/( 3) `. If the rocket has to escape from the
`[v_(e)` = escape velocity of rocket fromm `(R_(e))/(3)`]
`(g R_(e)^(2))/((4R_(3))/(3)) = (1)/(2) v_(e)^(2)`
`rArr (3)/(4) gR_(e) = (v_(e)^(2))/( 2)`
`rArr (sqrt(3))/(2) sqrt(2g R_(e) = v_(e)`
`rArr 9.7 = v_(e)`
`v_(e_(1)) = 9.7 km//s`