<ul>
<li>The speed of the Earth (and Moon) around the Sun is $\approx 29.8 \frac{km}{s}$.</li>
<li>The speed of the Moon around the Earth is $\approx 1.05 \frac{km}{s}$.</li>
</ul>

<p>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:</p>

<p><a href="https://i.sstatic.net/nKadC.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nKadC.gif" alt="enter image description here"></a></p>

<p><sup>In fact, the "waves" are much smaller (400.000km Moon orbital radius to  150.000.000km Earth orbital radius)</sup></p>

<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2017-12-30T16:26:36.000Z
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](https://en.wikipedia.org/wiki/Jacobi_integral)). 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](https://en.wikipedia.org/wiki/Three-body_problem).
uhoh
CC BY-SA 3.0
2017-12-30T16:33:07.197Z
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](https://astronomy.stackexchange.com/a/10950/7982) for more clarification. Also, I've just realized that he's already said what I've said here in [this comment](https://astronomy.stackexchange.com/questions/24243/why-does-the-moon-not-leave-its-orbit-around-the-earth-due-to-the-gravitational?noredirect=1#comment41563_24243) above.
uhoh
CC BY-SA 3.0
2017-12-30T20:21:10.330Z
I have added more information to the original question/post as the answer did not directly address the question
John Bentsey
CC BY-SA 3.0
2017-12-30T01:11:16.450Z
2017-12-30T01:25:43.453Z
2017-12-30T01:25:43.453Z
peterh
<p>See additional information below added after the answer to the question.</p>

<p>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.</p>

<p>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?</p>

<p>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.</p>

<p>Thank you for your time, I would like to understand this situation.</p>

<hr>

<p>Additional information:</p>

<p>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.</p>

<p>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.</p>

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

<p>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. </p>

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

CC BY-SA 3.0
2017-12-29T20:44:20.463Z
"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.
A. C. A. C.
CC BY-SA 3.0
2017-12-29T22:22:52.950Z
But it is true...By about a factor of 2 times
DJohnM
CC BY-SA 3.0
2017-12-29T22:52:13.883Z
@DJohnM Are you saying that getting <1% closer to the Sun doubles the gravitational force? This new theory will certainly rewrite our understanding of physics.
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-12-29T22:58:21.260Z
@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.
David Hammen
CC BY-SA 3.0
2017-12-29T23:22:56.093Z
@DavidHammen You seem to have overlooked the question. The *change* in the Sun’s gravitational attraction from the position of full Moon to new Moon is minuscule, and nowhere near enough to disrupt the Moon’s orbit around the Earth-Moon barycentre. The math is indeed simple but not on my modest smartphone.
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-12-29T23:48:03.620Z
@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.)
David Hammen
CC BY-SA 3.0
2017-12-29T23:56:14.467Z
Because the Moon and the Earth are quite close to one another (compared to the distance to the Sun), the Earth's and Moon's sunward accelerations are nearly equal to one another. The difference between the two is very small, a couple of orders of magnitude smaller than the Moon's gravitational acceleration toward the Earth. This would not be the case if the Moon's orbit about the Earth was four times or so wider than it is. In that case, the perturbing acceleration from the Sun would be equal to the Moon's acceleration towards the Earth, and the Sun would pull the Moon out of orbit.
David Hammen
CC BY-SA 3.0
2017-12-30T00:56:53.593Z
@chappo:  mass of earth over earth-moon distance squared is *half* mass of sun over moon-sun distance squared, isn't it?  Start re-writing...
DJohnM
CC BY-SA 3.0
2017-12-31T00:01:23.967Z
Re your edit: You are still asking the exact same question as the duplicate from 2 1/2 years ago. Re "I hope somebody can come up with an explanation that is in accordance with physical laws." You are the one stuck on a false metric. This is a three body gravitational problem, not the [problem of two fixed centers](https://en.wikipedia.org/wiki/Euler%27s_three-body_problem), where your metric of gravitational acceleration toward the Sun would be quite applicable. Unlike the problem of two fixed centers, the Earth is not at a fixed position. Instead, it, like the Moon, is falling sunward.
David Hammen
CC BY-SA 3.0
2017-12-31T06:41:44.063Z
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?
PM 2Ring
CC BY-SA 3.0
2018-01-01T09:50:00.110Z
The answer is here: http://mathpages.com/home/kmath405/kmath405.htm
Victor Storm
CC BY-SA 3.0
2018-01-01T17:22:32.340Z
David Hammen how does the moon move further away from the sun without a force to change its direction? E.g. when the moon moves from the outer syzygy to the inner syzygy (or at a point near to this), it is moving more or less straight towards the sun at the tanget to its orbit arount the earth.
John Bentsey
CC BY-SA 3.0
2018-01-01T17:22:41.460Z
At this point the force from the sun is twice the force from the earth so it should accelerate towards the sun and this acceleration should continue as it gets closer to the sun. What force cancels out this acceleration and changes the moon's direction, so it stays in its orbit around the earth and therefore later on ends up further away from the sun?
John Bentsey
CC BY-SA 3.0
2018-01-01T23:56:41.573Z
@JohnBentsey - I don't know what to say other than to stop using the wrong metric. It is only serving to confuse you. Suggesting research material for you: the three body problem (but not Euler's three body problem, aka the problem of two fixed centers), the Jacobi integral, Laplace's gravitational sphere of influence, the Hill sphere.
David Hammen
CC BY-SA 3.0
2018-01-04T11:52:31.003Z
@DavidHammen - I am familiar with the material you are referencing. The problem is that the material does not explain how the model can work, but rather describes attributes of the model from the assumption that the model is true in the first place.
John Bentsey
CC BY-SA 3.0
2018-01-04T11:57:35.127Z
I still need an answer to the question; "what force makes the moon move further away from the sun when on the one "side" where it is moving closer to the sun, the force from the sun is about twice the force from the earth?" The fact that all three bodies; the sun, the earth and the moon are all moving through the universe "together" does not explain why the distance between the moon and the sun can increase without a net force acting upon the moon in a direction away from the sun.
John Bentsey
CC BY-SA 3.0
2018-01-04T11:58:51.700Z
What force makes this happen? Are you suggesting that Newtons laws do not apply here or is there another force at work or does the moon not move further away from the sun at some point in its orbit? These are the only possibilities.
John Bentsey
CC BY-SA 3.0
2018-01-04T17:46:46.927Z
@JohnBentsey - They are not the only possibilities. The Earth's gravitational field is all that is needed. No other force is needed, no deviation from Newtonian mechanics is needed.
David Hammen
CC BY-SA 3.0
2018-01-08T22:05:38.943Z
The heliocentric model of the universe does not have an answer for the question. It is simply not possible for an orbit to exist within a larger orbit as the moon in this case would be pulled out of its orbit of the earth by the larger force from the sun.
John Bentsey
CC BY-SA 3.0
2018-01-08T22:05:46.717Z
An orbit is a very unstable path as the smallest acceleration or the smallest change in direction would make the moon (in this case) leave the orbit or collide with the earth. An orbit cannot exist if other forces exist as they would pull differently at different parts of the orbit, hence making the orbit unstable and therefore impossible over a relatively short period of time.
John Bentsey
<the-sun><orbit><the-moon>
Why does the moon not leave its orbit around the earth due to the gravitational force from the sun?
John Bentsey
# Why does the moon not leave its orbit around the earth due to the gravitational force from the sun?

(Asked by: username_0 on 2017-12-29)

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.

**Tags:** the-sun, orbit, the-moon
> **Comments:**
> **username_1**: "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.  
> **username_2**: But it is true...By about a factor of 2 times  
> **username_3**: @DJohnM Are you saying that getting <1% closer to the Sun doubles the gravitational force? This new theory will certainly rewrite our understanding of physics.  
> **username_4**: @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.  
> **username_3**: @DavidHammen You seem to have overlooked the question. The *change* in the Sun’s gravitational attraction from the position of full Moon to new Moon is minuscule, and nowhere near enough to disrupt the Moon’s orbit around the Earth-Moon barycentre. The math is indeed simple but not on my modest smartphone.  
> **username_4**: @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.)  
> **username_4**: Because the Moon and the Earth are quite close to one another (compared to the distance to the Sun), the Earth's and Moon's sunward accelerations are nearly equal to one another. The difference between the two is very small, a couple of orders of magnitude smaller than the Moon's gravitational acceleration toward the Earth. This would not be the case if the Moon's orbit about the Earth was four times or so wider than it is. In that case, the perturbing acceleration from the Sun would be equal to the Moon's acceleration towards the Earth, and the Sun would pull the Moon out of orbit.  
> **username_2**: @chappo:  mass of earth over earth-moon distance squared is *half* mass of sun over moon-sun distance squared, isn't it?  Start re-writing...  
> **username_4**: Re your edit: You are still asking the exact same question as the duplicate from 2 1/2 years ago. Re "I hope somebody can come up with an explanation that is in accordance with physical laws." You are the one stuck on a false metric. This is a three body gravitational problem, not the [problem of two fixed centers](https://en.wikipedia.org/wiki/Euler%27s_three-body_problem), where your metric of gravitational acceleration toward the Sun would be quite applicable. Unlike the problem of two fixed centers, the Earth is not at a fixed position. Instead, it, like the Moon, is falling sunward.  
> **username_5**: 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?  
> **username_6**: The answer is here: http://mathpages.com/home/kmath405/kmath405.htm  
> **username_0**: David Hammen how does the moon move further away from the sun without a force to change its direction? E.g. when the moon moves from the outer syzygy to the inner syzygy (or at a point near to this), it is moving more or less straight towards the sun at the tanget to its orbit arount the earth.  
> **username_0**: At this point the force from the sun is twice the force from the earth so it should accelerate towards the sun and this acceleration should continue as it gets closer to the sun. What force cancels out this acceleration and changes the moon's direction, so it stays in its orbit around the earth and therefore later on ends up further away from the sun?  
> **username_4**: @JohnBentsey - I don't know what to say other than to stop using the wrong metric. It is only serving to confuse you. Suggesting research material for you: the three body problem (but not Euler's three body problem, aka the problem of two fixed centers), the Jacobi integral, Laplace's gravitational sphere of influence, the Hill sphere.  
> **username_0**: @DavidHammen - I am familiar with the material you are referencing. The problem is that the material does not explain how the model can work, but rather describes attributes of the model from the assumption that the model is true in the first place.  
> **username_0**: I still need an answer to the question; "what force makes the moon move further away from the sun when on the one "side" where it is moving closer to the sun, the force from the sun is about twice the force from the earth?" The fact that all three bodies; the sun, the earth and the moon are all moving through the universe "together" does not explain why the distance between the moon and the sun can increase without a net force acting upon the moon in a direction away from the sun.  
> **username_0**: What force makes this happen? Are you suggesting that Newtons laws do not apply here or is there another force at work or does the moon not move further away from the sun at some point in its orbit? These are the only possibilities.  
> **username_4**: @JohnBentsey - They are not the only possibilities. The Earth's gravitational field is all that is needed. No other force is needed, no deviation from Newtonian mechanics is needed.  
> **username_0**: The heliocentric model of the universe does not have an answer for the question. It is simply not possible for an orbit to exist within a larger orbit as the moon in this case would be pulled out of its orbit of the earth by the larger force from the sun.  
> **username_0**: An orbit is a very unstable path as the smallest acceleration or the smallest change in direction would make the moon (in this case) leave the orbit or collide with the earth. An orbit cannot exist if other forces exist as they would pull differently at different parts of the orbit, hence making the orbit unstable and therefore impossible over a relatively short period of time.  

---

## 1 answers

### Answer by username_7 on 2017-12-30. Score: 4

* 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](https://i.sstatic.net/nKadC.gif)](https://i.sstatic.net/nKadC.gif)

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.

> **Comments:**
> **username_8**: 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](https://en.wikipedia.org/wiki/Jacobi_integral)). 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](https://en.wikipedia.org/wiki/Three-body_problem).  
> **username_8**: 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](https://astronomy.stackexchange.com/a/10950/7982) for more clarification. Also, I've just realized that he's already said what I've said here in [this comment](https://astronomy.stackexchange.com/questions/24243/why-does-the-moon-not-leave-its-orbit-around-the-earth-due-to-the-gravitational?noredirect=1#comment41563_24243) above.  
> **username_0**: I have added more information to the original question/post as the answer did not directly address the question  

---

