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See additional information below added after the answer to the question.

When the moon is between the earth and the sun, the gravitational force on the moon from the sun is greater than the gravitational force from the earth, hence the moon should continue its path towards the sun instead of staying in its orbit around the earth.

When the moon is at 90 degrees to the earth relative to the sun, the moon is moving almost directly towards the sun and then the gravitational force on the moon from the sun is roughly double the gravitational force from the earth, hence there is nothing to stop the moon's trajectory towards the sun and it should continue towards the sun away from the earth. What makes the moon "turn" back towards the earth and away from the sun at the point where it is right between the earth and the sun when at this point right when it turns, the gravitational force from the sun is roughly double the gravitational force from the earth?

I am aware of the movement of the sun and also of the fact that the moon is moving on a different plane to that of the earth and I don't think that can explain my question above.

Thank you for your time, I would like to understand this situation.


Additional information:

The answer does not address the question, let me explain. Although it is true that the moon orbits the sun as described in the answer, this does not explain why the sun does not pull the moon out of its orbit around the earth.

For an object to change direction a net force has to be applied to it according to Newtons laws. a=F/m applied to the moon from the point in its orbit where the moon gets closer to the sun than the earth is to the sun, shows a force vector on the moon towards the sun applying twice the force as another force vector on the moon towards the earth. This results in a combined net force vector on the moon pointing away from the earth and this net force vector will progressively point further away from the earth and more towards the sun as the moon moves closer to the sun and therefore should pull the moon out of its orbit around the earth and pull it towards the sun.

Newtons laws must be true on the three separate objects; the sun, the earth and the moon in the example above.

I have a Master of Science in Engineering and know my physics. I have some friends who are convinced they can prove that the earth is actually flat and that a search on "flat earth" will prove it, and they have asked me the question in this thread to prove to me that the heliocentric model is not true. I took up the challenge, however when looking at this simple example I have not been able to find an answer that is in accordance with normal physical laws.

I hope somebody can come up with an explanation that is in accordance with physical laws.

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    $\begingroup$ "When the moon is between the earth and the sun, the gravitational force on the moon from the sun is greater than the gravitational force from the earth" that is not true. $\endgroup$ – A. C. A. C. Dec 29 '17 at 20:44
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    $\begingroup$ But it is true...By about a factor of 2 times $\endgroup$ – DJohnM Dec 29 '17 at 22:22
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    $\begingroup$ @Chappo, and A.C.A.C: Do the math. The gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth on the Moon. The answer is that force is a lousy metric. $\endgroup$ – David Hammen Dec 29 '17 at 22:58
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    $\begingroup$ @Chappo -- You are the one who has overlooked the question. The question is quite simple: Since the Moon's gravitational acceleration toward the Sun is about twice that of it's acceleration toward the Earth, why doesn't the Sun pull the Moon out of Earth's orbit? This is not a dumb question. Some otherwise very intelligent people have used this same line of reasoning to argue that the Moon does not orbit the Earth. (It does.) $\endgroup$ – David Hammen Dec 29 '17 at 23:48
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    $\begingroup$ Maybe it will help you to understand what's going on if you turn the question around: why doesn't the Sun pull the Earth away from the Moon? $\endgroup$ – PM 2Ring Dec 31 '17 at 6:41
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  • The speed of the Earth (and Moon) around the Sun is $\approx 29.8 \frac{km}{s}$.
  • The speed of the Moon around the Earth is $\approx 1.05 \frac{km}{s}$.

The result is that the Moon has a nearly perfect circular orbit around the Sun, the effect of the Earth is only a perturbation in it:

enter image description here

In fact, the "waves" are much smaller (400.000km Moon orbital radius to 150.000.000km Earth orbital radius)

This is mainly because $R_{Earth-Moon} \ll R_{Earth-Sun}$, and $M_{Moon} \ll M_{Earth} \ll M_{Sun}$. In general, 3-body gravitationally bound systems tend to be unstable, i.e. one of the components tends to fly away with time. But this system seems stable, because the ratio of the masses are big.

The same can't be said from the whole Solar System, (it is a notoriously unsolved problem) although also this seems stable in human time scales.

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    $\begingroup$ This is a great approach for an answer. It's not exactly true that "in general, three-body gravitationally bound systems tend to be unstable." That instability applies to third body orbits with sufficiently large "energy" (sufficiently negative Jacobi constant). The energy associated with the Moon's orbit is much lower than this threshold (which is why it's so close to Earth) and so it is certainly in a stable orbit when viewing the system as a CR3BP. $\endgroup$ – uhoh Dec 30 '17 at 16:26
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    $\begingroup$ Only if you tried to raise the Moon's orbit out to say 1.5 million kilometers (instead of its current ~350,000 km) near where the Lagrange points are would you start to worry that the orbit would become unstable. See @DavidHammen 's answer for more clarification. Also, I've just realized that he's already said what I've said here in this comment above. $\endgroup$ – uhoh Dec 30 '17 at 16:33
  • $\begingroup$ I have added more information to the original question/post as the answer did not directly address the question $\endgroup$ – John Bentsey Dec 30 '17 at 20:21

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