What Is the Gravitational Potential on the Moon's Surface?

AI Thread Summary
The discussion revolves around calculating the gravitational potential on the Moon's surface using the formula V = -G[Mm/Rm + Me/(D - Rm)]. The user initially attempted to use V = -GMm/Rm but was confused about why the Earth's mass is added rather than subtracted. It was clarified that the Earth's gravitational potential also contributes positively to the total potential experienced on the Moon's surface. The assumption is made that the calculation pertains to the Moon's light side, where the Earth's influence is more significant. Understanding these contributions is crucial for accurately determining the gravitational potential in such scenarios.
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Homework Statement



Given:
G = 6.67*10^ -11 Nm^2kg^-2
Radius of Moon = Rm = 1.73*10^6 m
Radius of Earth = Re = 6.37*10^6 m
acceleration due to gravity = g = 9.8m/s
Mass of moon = Mm
Mass of Earth = Me
Distance between the centre of the Earth and the centre of the moon = D = 3.8*10^8 m

Find the gravitational potentials on the moon's surface.

The Attempt at a Solution



V = - GMm/Rm
V = - g * Re^2 / Rm
V = - 9.8 * (6.37*10^6)^2 / 1.73*10^6
V = - 2.30*10^8 J/kg

The solution said that V should be
-G* [ Mm/Rm + Me/(D - Rm) ]
I don't understand why it should not be "Mm/Rm - Me/(D - Rm)" :confused:

Thank you very much!
 
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Because the Earth contributes a potential too. I guess you are supposed to assume that the problem wants the potential on the light side of the moon.
 
Thank you! In the past I thought that some potential of Earth will be canceled out by potential of the moon :-p
 
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