The gravitational potential energy of a mass $x$ meters away from the center of the Earth is calculated from the formula $$G\frac {Mm}{x}$$ where $M$ is the mass of the Earth and $m$ is the mass of the other mass and $x$ is the distance it is away from the center. In this formula, why isn't the gravitational potential energy between the Earth and the Moon found by $$G\frac{M_{\text{Earth}}M_{\text{Moon}}}{r}$$ affect the result found?
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3 Answers
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The mass of the Moon is about 1.2% the mass of Earth. Unless the mass $m$ is very close to the Moon, its effect on the Earth gravitational field can be neglected. Indeed,
$$U=m\left(\frac{M_e}{x} + \frac{M_m}{R-x}\right)$$
where $R$ is the distance between the Earth and the Moon. With $x\approx R_e$, that is, the Earth radius, $R-R_e\approx R$, so that
$$U\approx\frac{mM_e}{x}\left(1 + \frac{M_m}{M_e}\frac {R_e}R\right).$$
The second term inside the parenthesis is negligible, so a good approximation is
$$U\approx\frac{mM_e}{x}$$
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$\begingroup$ RE: "Unless the mass m is very close to the Moon, its effect on the Earth gravitational field can be neglected." Unless, for example, you want to know when the tide is going out so you can get your sailing ship out of port. $\endgroup$ Commented Apr 17, 2020 at 15:46
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$\begingroup$ Compared to $R$, the tide is also negligible $\endgroup$ Commented Apr 17, 2020 at 16:14
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It is not the gravitational potential energy of the mass alone rather it is the gravitational potential energy of the system which comprises the Earth and the mass.
$ {\Large \color{red}{\mathbf-}}\dfrac{GMm}{x}$ is the gravitational potential energy of a system of two masses, $M$ and $m$, with a separation of $x$ assuming that the zero of gravitational potential energy is when the two masses have an infinite separation.
If you now add the Moon then the system now contains three objects, Earth, Moon and mass.
You now have to include two extra terms to the gravitational potential energy of the system which also depend on the separation between the Earth and the Moon, $y$, and the separation between the Moon and the mass, $z$.
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Your question could be worded better. Are you asking if the Earth moon potential is part of the Earth mass system or if the Moon affects the potential energy that the mass will have near the earth? The answer is yes to both.
If you bring the Moon into the problem then there are three bodies in the problem and one speaks of the potential energy of the system, that is proportional to the amount of work required to bring the masses into their configuration from infinity. In this case you will have three contributions, Earth-mass, Earth-moon, and Moon-mass.
If you consider the Earth and Moon to be fixed for a moment and want to calculate the potential energy of the mass that you brought in to a fixed location then you have two terms to calculate, the potential from Earth-mass (which you have in the first formula with $M = M_{Earth}$) and the potential from Moon=mass (which you get by replacing $M = M_{Moon}$, and x with the distance from the Moon to the mass). Adding the results gives the total potential energy of the mass in the presence of the Earth and Moon.
If you are asking why the Moon is not considered as part of the calculation I think this has been answered by another post. In a nutshell you could ask why the Sun, Jupiter, etc are not included. In reality the presence of all the matter in the universe creates a Net gravitational field at any location. The "complete" potential energy of a mass placed at a location is $m\;U$, where U is the potential at that point due to all other mass. If the test mass in question is very close to one object and very far from all others then we are justified in neglecting everything else. Near the Earth (near the surface) it is the effect of the Earth's gravity that we feel the most. The Moon can affect the tides, so perhaps it is not negligible, but for most applications we can ignore it. If you are solving basic physic problems you are probably taught to ignore it to make things easy. In fact everything affects the result, just to varying degrees.