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Verified answer
Let the mass and radius of Mars be m and r respectively.
According to the Question,
A generalized expression for escape velocity is,
[tex] \sf \: v = \sqrt{\dfrac{GM}{R} }[/tex]
For Mars,
[tex] \sf \: v_m = v = \sqrt{ \dfrac{Gm}{r} }[/tex]
For Earth,
[tex] \sf \: 11.2 = \sqrt{ \dfrac{9Gm}{2r} }[/tex]
Dividing both the equations, we get :
[tex] \sf \: \dfrac{v}{11.2} = \frac{ \sqrt{ \dfrac{Gm}{r} }}{ \sqrt{ \dfrac{9Gm}{2r} }} \\ \\ \implies \sf \: \dfrac{v}{11.2} = \sqrt{ \dfrac{2}{9} } \\ \\ \implies \sf \: v = 11.2 \times \sqrt{0.22} \\ \\ \implies \sf \: v = 0.46 \times 11.2 \\ \\ \implies \boxed{ \boxed{ \sf v_m \: = 5.25 \: kmh {}^{ - 1} }}[/tex]
Escape Velocity on Mars is 5.3 Km/h