Why does the Moon not revolve around the Sun directly?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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...

Answer 4

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.

Answer 5

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 ,the moon is closer to the earth.The force of the gravitational force of earth is more than the sun on moon.So the moon revolves around the earth not the sun.