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

Answer

D

Explanation

For earth , ` v_e = sqrt((2GM)/( R) ) = 11 kms^(-1)`
For plant , `v._e = sqrt((2G xx 10M)/(R//10))= 10 sqrt((2GM)/(R ) ) = 10 xx 11 = 110 kms^(-1)`