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

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SUMMARY

The gravitational potential on the Moon's surface is calculated using the formula V = -G * [Mm/Rm + Me/(D - Rm)], where G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), Rm is the Moon's radius (1.73 x 10^6 m), and D is the distance between the centers of the Earth and the Moon (3.8 x 10^8 m). The gravitational potential is determined to be approximately -2.30 x 10^8 J/kg. The discussion clarifies that the potential from Earth must be added rather than subtracted, as both celestial bodies contribute to the gravitational potential experienced on the Moon's surface.

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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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