Answer:
2.73 km/s
Explanation:
The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)
[tex]v=\sqrt{\frac{2GM}{r}}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the gravitational constant
[tex]M=7.34\cdot 10^{22} kg[/tex] is the mass of the Moon
[tex]r=1.74\cdot 10^6 m[/tex] is the radius of the Moon
As we can see, the escape velocity does not depend on the mass of the lunar module.
Substituting the numbers into the formula, we find
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(7.34\cdot 10^{22}kg)}{(1.74\cdot 10^6 m)}}=2732 m/s=2.73 km/s[/tex]
