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The sun pulls on the moon with a force that is more than twice the magnitude of the force with which the earth attracts the moon. Why, then, doesn’t the sun take the moon away from the earth?

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The sun exerts a force on the moon larger than the earth's times 2. –  Hckr Jan 5 at 16:56
    
what exactly do you mean by "revolves around the sun directly", for me it makes perfectly good sense to say it is revolving the sun directly, but with a non-conic sectional orbit, why is it not conic sectional? Because earth is perturbing it. –  Jia Yiyang Jan 5 at 17:09
    
    
The link helps a lot, I edited the question with the original one. Could you write an answer on it? –  Hckr Jan 5 at 17:26
    
If you like this question you may also enjoy reading this Phys.SE post. –  Qmechanic Jan 5 at 23:51

6 Answers 6

up vote 9 down vote accepted

The simple answer is that the sun's gravity produces the same acceleration on both the Earth and the Moon. The Sun is pulling both of them along, but they are falling together.

You may imagine two skydiver jumping out of a plane at the same time (and we'd better ignore air resistance). They are subjected to gravitational forces from the Earth that vastly larger than the forces between them, but that doesn't rip them away from each other because they both experience the same acceleration.

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If you're a skydiver and you ignore air resistance, you're going to have a bad time. –  rob Aug 10 at 23:37

Let's do a back-of-the-envelope calculation:

Let $M\approx 2.0\cdot 10^{30}\text{kg}$ be the mass of the sun and $m\approx 6.0\cdot 10^{24}\text{kg}$ the mass of the earth, $R\approx 1.5\cdot 10^{11}\text{m}$ the distance between earth and sun and $r\approx 3.8\cdot 10^8\text{m}$ the distance between earth and moon.

The relative acceleration of the moon respective to earth due to the difference in gravity of the sun can be approximated by $$ \Delta a = \frac{GM}{R^2} - \frac{GM}{(R+r)^2} = \frac {GM}{R^2}\left( 1 - \frac 1{(1+\frac rR)^2} \right)\approx \frac{2GMr}{R^3} $$ via Taylor expansion.

The moon's acceleration due to earth's gravity is $$ a = \frac {Gm}{r^2} $$ and we end up with $$ \frac a{\Delta a} \approx \frac{mR^3}{2Mr^3}\approx 92 $$ Personally, I'd have expected some more powers of ten here, but of course this should still be more than enough to keep the moon from wandering off...

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This is the correct way to analyze when my simplistic argument fails. –  dmckee Jan 5 at 21:34

There are 5 so called Lagrangian points, where the gravitational forces and the centrifugal forces of the rotating system balance out. This link gives a nice picture of it http://upload.wikimedia.org/wikipedia/commons/thumb/8/88/Lagrange_points.jpg/685px-Lagrange_points.jpg.

Take the point L1 from the picture and you can imagine that at least theoretically an equilibrium is achieved at this point. If an object is closer to the Sun from this point, it will fall towards the Sun and if nearer the Earth it would fall towards the Earth.

Now the distance from Earth to Lagrangian point L1 is about 1 500 000 km so anything closer than that will fall towards Earth. The moon orbits the Earth about 384 000km from Earth so Moon will fall towards Earth and therefore not orbiting the Sun.

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Sun is greater 330,000 / 93 mil squared / 1(earth mass) x 250,000 mil sqared = ratio of g sun to g earth towards the moon at new moon. = 2.38 The ratio of sun to earth gravity is 2.38 , so the sun is more influential. However the moon is at a velocity that precludes it from fallimg into the sun.
Just as the earth has has a velocity that allows it to revolve around the Sun and not fall into it. in fact the Earth and Moon are almost considered a binary planet system. The binary system has a barycenter which basically is the center of rotation of the two masses about a thousand kilometers into the surface of the earth. This makes the earth appear to wobble as the moon rotates around it.

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Hi Greg, can you please format your answer better and maybe make use of MathJax? See meta.math.stackexchange.com/questions/5020/… –  Brandon Enright Aug 10 at 22:06

Moon is attracted to both Earth and Sun, but Earth gravitational force exerted on the moon is larger than the Sun's, because Sun distance from the moon is enormously larger, even if Sun mass is more than 300 thousands times the Earth's.

If Sun gravitational force was larger than Earth's, the Moon would no longer revolve around the Earth, but only around the Sun.

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As it is specified in the question and as I know in reality, the sun's is larger. –  Hckr Jan 5 at 16:54
    
But force does not depend only on the mass, but also on the distance (squared). The law written by Newton is: $F=G\frac{m_1m_2}{r^2}$ –  Ale Jan 5 at 16:59
    
I did not say the sun's mass is bigger. I said the sun's force on the moon is bigger. –  Hckr Jan 5 at 17:00
    
-1. the force of sun is bigger. –  Jia Yiyang Jan 5 at 17:21
    
Sorry, you're right. I got confused. Sun tidal force is smaller, but its gravitational force is greater. This is because tidal force varies on the cube of $r$. However, moon revolves also around the Sun... –  Ale Jan 5 at 17:33

The force of gravity mainly depends upon mass and distance (gravitational force of objects directly depends upon product of masses of objects and inversely proportional to square of distance) Although Sun is bigger than earth ,moon is closer to the earth.the force the gravitational force of earth is more than that of sun on moon.so moon revolves around earth not sun.

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The question says and in reality ( as I know) the sun's force is larger than the earth's. –  Hckr Jan 5 at 16:55
    
the gravitational force of earth is more than sun.(by newton's law of gravity). –  Katz Jan 5 at 17:06
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-1, if you just take a few data from wikipedia and calculate the force you will see the sun's force is greater. –  Jia Yiyang Jan 5 at 17:20

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