The planet Mars is 10 times less massive than Earth and its radius is half that of Earth.

Question

What is the ratio of the escape velocity from Mars’s surface compared to the escape velocity from Earth’s surface?

Answer 1

Escape velocity is the minimum velocity required to overcome the gravitational field. In other words, kinetic energy (which includes velocity) must be equal to or greater than gravitational potential energy, or "binding energy."

PE = KE

(GMm)/R = (1/2)mv²

v_E = √(2GM/R)

G is a gravitational constant which will reman the same for all planets. We can set M to be the mass of Earth and R for the radius of earth.

On mars, with (1/10)M and (R/2),

v = √((2G(M/10) / (R/2))

v_mars = √(2GM/5R)

V_mars/V_E= √(2GM/5R)/√(2GM/R) = √(1/5) : 1

(All similar terms cancel out)