The escape velocity on the moon is:

a) 5.64x10^6 m/s
b) 1.68x10^3
c) 160 m/s
d) 2.38x10^3 m/s

potential energy of mass m at surface = -GMmoon m/r

= - m (6.67*10^-11) (7.35*10^22) /(1.74*10^6)
= m (28.2*10^5)

Needs that many Joules of Ke so
(1/2) m v^2 = 2.82*10^6 m

v^2 = 5.64 *10^6
v = 2.37*10^3 = 2370 m/s
by the way:
http://www.1728.org/escvel.htm

thanks! and thanks for the link, its quite useful!

To find the escape velocity on the moon, we need to use the formula for escape velocity:

Ve = √(2 * G * M / R)

where Ve is the escape velocity, G is the gravitational constant, M is the mass of the moon, and R is the radius of the moon.

The mass of the moon (M) is approximately 7.35 × 10^22 kg, and the radius of the moon (R) is approximately 1.74 × 10^6 meters.

Plugging in these values into the formula and calculating:

Ve = √(2 * (6.67430 × 10^-11 N m^2/kg^2) * (7.35 × 10^22 kg) / (1.74 × 10^6 meters))

Ve ≈ 2380 m/s

Therefore, the correct answer is:

d) 2.38 × 10^3 m/s