A planet in a distant solar systyem is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is `11kms^(-1)`, the escape velocity from the surface of the planet would be

Options

A. 1.1 km `s^(-1)`
B. 11 km `s^(-1)`
C. 110 km `s^(-1)`
D. 0.11 km `s^(-1)`

Answer

C

Explanation

`((v_(e))p)/((v_(e))e) = (sqrt((2GM_(p))/(R_(p))))/(sqrt((2GM_(e))/(R_(e)))) = sqrt((M_(p))/(M_(e)) xx (R_(e))/(R_(p)))`
` = sqrt((10 M_(e))/(M_(e)) xx (R_(e))/(R_(e)//10)) = 10 `
`therefore (v_(e))_(p) = 10 xx (v_(e))_(e) = 10 xx 11 = 110` km/s