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}}$
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}}$