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 11 km s⁻¹, the escape velocity from the surface of the planet would be

The escape velocity vₑ of a body depends on the gravitational pull of the celestial body it is escaping from. For Earth, the escape velocity is derived based on its mass M and radius R using the formula:

vₑ ∝ √((2 GM)/R)

On a different planet, if its mass is 10 times that of Earth and its radius is 1/10th of Earth’s, the escape velocity will increase significantly. Substituting the planet’s properties into the formula reveals that the escape velocity is proportional to:

vₑ ∝ 10 x √((2 GM)/R)

This results in an escape velocity that is 10 times that of Earth, making it approximately 110 km/s.

vₑ(earth) = √((2 GM)/R) = 11 km s⁻¹
vₑ(planet) = √((2 G x 10 M)/R/10) = 10√((2 GM)/R)
= 10 x 11 = 110 km s⁻¹