Question

What is the value of escape velocity on a planet having radius 3 times that of earth and on which acceleration due to gravity is three times that on earth?

Answer 1

escape velocity $v_e = \sqrt{2gR}$....(1)

as $v_e = \sqrt{\dfrac{2GM}{R}}$

and $g = \dfrac{GM}{R^2}$
where,
G = Gravitational Constants
M= Mass of the planet
R= radius of the planet
g= gravitational acceleration of that planet

From (1),

$\dfrac{v_{e_{1}}}{v_{e_{2}}} = \sqrt{\dfrac{2g_1 R_1}{2g_2 R_2}}$

And, as $R_2 = 3R_1$ and $g_2 = 3g_1$

$\dfrac{v_{e_{1}}}{v_{e_{2}}} = \sqrt{\dfrac{g_1 R_1}{3g_1 3R_1}}$
or, $\dfrac{v_{e_{1}}}{v_{e_{2}}} = \sqrt{\dfrac{1}{9}}$
or, $\dfrac{v_{e_{1}}}{v_{e_{2}}} = \dfrac13$
or, $\dfrac{v_{e_{2}}}{v_{e_{1}}} = 3$
or, $v_{e_{2}} = 3 v_{e_{1}}$

The escape velocity of that planet is three times of earth.

as it is 11.2 km/s, it will be 3 × 11.2 = 33.6 km/s on that planet.

Answer 2

I hope this will help you
if not then comment me