## In This post you will get answer of Q.

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is $ce{11 , km , s^{-1}}$, the escape velocity from the surface of the planet would be

Solution:

## Escape velocity from the surface of the earth is

$upsilon_e = sqrt{frac{2GM}{R}}$

$upsilon_e = 11 , km , s^{-1}$

Mass of the planet = $10M$.

Radius of the planet = $R/10$.

$ therefore , upsilon_p = sqrt{frac{2GM times 10}{R/10}} = 10 upsilon_e = 10 times 11$

$ = ce{ 110 , km , s^{-1}}$