If the suns gravitational pull is strong enough to hold much larger masses in place (all the planets) and at much greater distances (all planets further away from the sun then earth) why does it not pull the moon away from earth?
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4$\begingroup$ Short answer: Earth is much, much closer to the Moon than the Sun is. $\endgroup$– HDE 226868 ♦Jun 8, 2015 at 17:43
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3$\begingroup$ But the path of the Moon is always concave towards the Sun; the gravitational force exerted by the Sun on the Moon is always greater than the pull of the Earth on the Moon... $\endgroup$– DJohnMJun 10, 2015 at 23:03
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$\begingroup$ the sun's gravity is demonstrated by the sun's effect on tides. $\endgroup$– bandybabboonJan 1, 2018 at 21:44
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$\begingroup$ Related (probably not dupe). $\endgroup$– peterhFeb 22, 2018 at 19:44
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6 Answers
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Why doesn't the sun pull the moon away from earth?
Short answer: Because the Moon is much closer to the Earth than it is to the Sun. This means the gravitational acceleration of the Earth toward the Sun is almost the same as is the gravitational acceleration of the Moon toward the Sun.
The Moon's acceleration toward the Sun, $-GM_\odot\frac{\boldsymbol R+\boldsymbol r}{||\boldsymbol R+\boldsymbol r||^3}$ is indeed about twice that of the Moon toward the Earth, $-GM_\oplus\frac{\boldsymbol r}{||\boldsymbol r||^3}$. This is irrelevant. What is relevant is the Moon's earthward acceleration due to gravitation compared to the difference between the Moon's and Earth's sunward gravitational acceleration, $$\boldsymbol a_{\odot,\text{rel}} = -GM_{\odot}\left(\frac{\boldsymbol R + \boldsymbol r}{||\boldsymbol R + \boldsymbol r||^3} - \frac{\boldsymbol R}{||\boldsymbol R||^3}\right)$$ This relative acceleration toward the Sun is a small perturbation (less than 1/87th in magnitude) on the Moon's gravitational acceleration toward the Earth. Given the current circumstances, the Sun can't pull the Moon away from the Earth.
Longer answer:
The gravitational force exerted by the Sun on the Moon is more twice that exerted by the Earth on the Moon. So why do we say the Moon orbits the Earth? This has two answers. One is that "orbit" is not a mutually exclusive term. Just because Moon orbits the Earth (and it does) does not mean that it doesn't also orbit the Sun (or the Milky Way, for that matter). It does.
The other answer is that gravitational force as-is is not a good metric. The gravitational force from the Sun and Earth are equal at a distance of about 260000 km from the Earth. The short-term and long-term behaviors of an object orbiting the Earth at 270000 km are essentially the same as those of an object orbiting the Earth at 250000 km. That 260000 km where the gravitational forces from the Sun and Earth are equal in magnitude is effectively meaningless.
A better metric is the distance at which an orbit remain stable for a long, long, long time. In the two body problem, orbits at any distance are stable so long as the total mechanical energy is negative. This is no longer the case in the multi-body problem. The Hill sphere is a somewhat reasonable metric in the three body problem.
The Hill sphere is an approximation of a much more complex shape, and this complex shape doesn't capture long-term dynamics. An object that is orbiting circularly at (for example) 2/3 of the Hill sphere radius won't remain in a circular orbit for long. Its orbit will instead become rather convoluted, sometimes dipping as close to 1/3 of the Hill sphere radius from the planet, other times moving slightly outside the Hill sphere. The object escapes the gravitational clutches of the planet if one of those excursions beyond the Hill sphere occurs near the L1 or L2 Lagrange point.
In the N-body problem (for example, the Sun plus the Earth plus Venus, Jupiter, and all of the other planets), the Hill sphere remains a reasonably good metric, but it needs to be scaled down a bit. For an object in a prograde orbit such as the Moon, the object's orbit remains stable for a very long period of time so long as the orbital radius is less than 1/2 (and maybe 1/3) of the Hill sphere radius.
The Moon's orbit about the Earth is currently about 1/4 of the Earth's Hill sphere radius. That's well within even the most conservative bound. The Moon has been orbiting the Earth for 4.5 billion years, and will continue to do so for a few more billions of years into the future.
answered Jun 9, 2015 at 13:02
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1$\begingroup$ I’m too sh|tf@ced to understand any/all of this. But I will still upvote cus it sounds correct. Good night. $\endgroup$ Sep 23, 2017 at 0:02
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$\begingroup$ This answer has potential but doesn't address the apparent paradox by 1.) clearly stating what the difference between Hill-sphere and gravitational balance is. I think the key here is that most of the Sun's acceleration is compensated by the centrifugal acceleration of the Earth-Moon system around the sun. Same game then for the orbit around Earth. $\endgroup$ Jan 5, 2018 at 13:29
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1$\begingroup$ @AtmosphericPrisonEscape - What paradox? This is an apparent paradox only. I clearly addressed this with my latest update, showing that the gravitational acceleration of the Moon relative to the Earth is always earthward, even after including the acceleration of the Sun. No need to invoke a fictitious centrifugal force. ... (continued) $\endgroup$ Jan 5, 2018 at 15:36
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$\begingroup$ Suppose the Earth and Moon were falling together in a 600 micro-g uniform gravitational field. Rhetorical question: Would the Moon be pulled away from the Earth because the Moon's gravitational acceleration toward the Earth is a mere 270 micro-g? The answer is no. There's no distinguishing freefall in that uniform gravitational field from no gravitational field whatsoever. The Sun's gravitational field at one AU is very close to a 600 micro-g uniform gravitational field. The gravity gradient, the local deviation from uniformity, is very small. $\endgroup$ Jan 5, 2018 at 15:39
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$\begingroup$ The point is a valid and correct one, one that is much clearer than your long answer. That's why I'm confused why you shy away from 'fictious' forces, as those 1.) give intuition 2.) are necessary in calculating the Hill-surface. $\endgroup$ Jan 5, 2018 at 15:46
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The Moon is in orbit about the Sun, much as the Earth is. Although this is not the usual perspective from the Earth, a plot of the Moon's trajectory shows the Moon in an elliptic orbit about the Sun. Essentially the Earth, Moon, Sun system is (meta) stable, like that of other planets orbiting the Sun.
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$\begingroup$ The Moon certainly does orbit the Earth, unlike an object such as 2016 HO3. So I don't think this answers the question, and may serve only to confuse. $\endgroup$– James KJul 6, 2016 at 16:31
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2$\begingroup$ Where did I say the Moon didn't orbit the Earth. My point comes from V.A. Firsoff's classic "The Old Moon and the New" - the Moon orbits both the Earth and the Sun. $\endgroup$ Jan 5, 2018 at 17:02
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If we "hold" the Earth and "move" the Sun away, the Moon wouldn't stay with the Earth, but would follow the Sun. It is the only satellite in the Solar System that is attracted to the Sun stronger than to its own host planet:
our Moon is unique among all the satellites of the planets, is so far as it is the only planetary satellite whose orbital radius exceeds the threshold value, which means it is the one satellite on which the Sun's gravitational acceleration exceeds the host planet's gravitational acceleration. Consequently, it is the only moon in the solar system that is always falling toward the Sun.
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1$\begingroup$ This is correct, but it doesn't answer the question, which is "why doesn't the sun pull the moon away from the Earth". $\endgroup$– James KSep 23, 2017 at 20:54
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1$\begingroup$ @JamesK: Yes, but the question is trivial and answered already several times while this point is largely unknown and unique.Why aren't our GPS satellites pulled away by the Sun? Why isn't the Earth pulled away from the Sun by the Milky Way? (Yawn) They are all in a free fall, there are no forces in General Relativity. $\endgroup$ Sep 23, 2017 at 23:23
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1$\begingroup$ Re If we "hold" the Earth and "move" the Sun away, the Moon wouldn't stay with the Earth, but would follow the Sun: This is nonsense. Re It is the only satellite in the Solar System that is attracted to the Sun stronger than to its own host planet This is not the case. Jupiter, Saturn, and Uranus have several moons for which the gravitational force due to the Sun is stronger than that of the host planet. $\endgroup$ Feb 22, 2018 at 9:04
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$\begingroup$ @DavidHammen Have you checked the quoted link, Mr. "ex-ex-rocket scientist"? Perhaps nonsense is what you say rather than Kevin Brown of MathPages.com. $\endgroup$ Feb 22, 2018 at 10:07
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I agree with Adrian's answer. If you look at the moons orbit, in a very real sense it orbits the sun maybe more than it orbits the earth. The Earth/Moon system orbits the sun at 30 KM/s, the Moon orbits the earth at about 1 KM per second. Both orbits are reasonably eliptical.
The entire solarsystem orbits around the center of the Milky-way, so orbiting more than one center of mass isn't unusual. Orbits can exist within other orbits, within limits. The orbital limit is sometimes referred to as the Sphere of Influence http://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29
If the moon was a bit more than twice as far as it is from the Earth as it is now, the Earth might lose it.
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2$\begingroup$ The article to which you linked describes the Lagrange sphere of influence. An arguably better metric is the Hill sphere. In the case of the Earth orbiting the Sun, the diameter Earth's Hill sphere is about 60% larger than that of the Earth's sphere of influence. The Moon is currently at about 1/4 of the Earth's Hill sphere radius, so safely ensconced within it. $\endgroup$ Jun 9, 2015 at 10:03
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$\begingroup$ According to Wiki, only about 1/2 to 1/3rd of the Hill Sphere is actually a stable orbit. en.wikipedia.org/wiki/Hill_sphere I agree, the moon is safe and secure, but the entire Hill Sphere isn't stable. I may have been too generous with my "bit more than twice", estimate. Might be a bit less than twice it's current distance and the earth could lose the moon. But I think we both agree, the moon is stable where it is and would be stable for a fair distance further. $\endgroup$– userLTKJun 9, 2015 at 11:53
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Now, if the Moon needs to escape the Earth and go for Sun, it needs more speed to do so. It cannot escape Earth until it's speed is enough for escaping. It needs more velocity.
The orbit of the Moon around the Sun is essentially a circle with a radius of 150million km. Its orbit around the Earth has only a 400 000 km radius, thus the effect of the Earth is only a minor perturbation of it.
Looking from the Sun, the Moon has a circular orbit around it, just like Earth, and their effect to eachother is nearly negligible.
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Newton law: https://en.m.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
F=G*(m1*m2)/d² is the gravitationnal force between 2 things of mass m1 and m2, separated by a distance d. G is the gravitational constant (I don't remember the value).
--> F_earth/moon=F_moon/earth=G*(m_moon*m_earth)/d²
Same thing for F_sun/moon
You'll notice that F_earth/moon is greater than the other force, so the Moon is more attracted by Earth than Sun.
