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We know that Gravitational Force that is applied by the sun on the moon is greater than that by the earth on the moon.

My Question is then why does not the moon escape from the earth and go towards sun?

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marked as duplicate by John Rennie newtonian-mechanics Jul 2 '18 at 9:20

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    $\begingroup$ Duplicate. Https://astronomy.stackexchange.com/q/10946 $\endgroup$ – user198207 Jul 2 '18 at 8:42
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    $\begingroup$ @Mauricio The force exerted by the Sun on the Moon is about 2.38 times larger than the force exerted by the Earth on the Moon. See physics.stackexchange.com/q/92465 and Countto10's link. $\endgroup$ – PM 2Ring Jul 2 '18 at 9:17
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    $\begingroup$ Possible duplicate of Why does the Moon not revolve around the Sun directly? $\endgroup$ – John Rennie Jul 2 '18 at 9:20
  • $\begingroup$ Why should the Moon go closer to the Sun? It's got the right speed to orbit the Sun at its current distance, just like you do. The Moon's gravitational attraction to the Earth merely perturbs its solar orbit slightly. If you plot that orbit on your computer screen, the wobble induced by the Earth is only a pixel or two. $\endgroup$ – PM 2Ring Jul 2 '18 at 9:31
  • $\begingroup$ The o/p asked the wrong question. What he should have asked is why do both the Earth and the Moon not get pulled into the Sun -- since the Earth and Moon are a conjoined system. If he had asked that, he would have immediately realised his mistake. Plainly, the Moon does not fall into the Sun for the same reason the Earth does not. Both have identical orbital velocity, and that element is the key element in the existence of orbital equilibrium. The fact that the Earth has a stable orbit thus acts as a proof that the Moon, which has the same velocity, must be in a stable orbit too. $\endgroup$ – Ed999 May 23 at 16:22
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The Sun's pull in the Moon is indeed larger; a bit more than twice that of the Earth's pull in the Moon. Depending on where in the orbit you calculate it, you'll get numbers such as:

$$F_\text{Earth-Moon}=1.8\times 10^{20}\;\mathrm N\qquad F_\text{Sun-Moon}=4.3\times 10^{20}\;\mathrm N$$

These forces cause a gravitational acceleration towards Earth or the Sun, respectively, of:

$$g_\text{Moon-towards-Earth}=2.5\;\mathrm{mm/s^2}\qquad g_\text{Moon-towards-Sun}=5.9\;\mathrm{mm/s^2}$$

But remember that motions superimpose. Both of these two centripetal accelerations will cause circular motions. Both can take place at the same time. Think of the Earth's attraction as causing a small "wobble" in the Moon's Sun-orbit - that "wobble" is the Earth-orbit, which indeed is a much smaller orbit.

The Sun pulls both the Moon and the Earth towards it. This causes

  • the Moon to experience a centripetal acceleration of $g=5.9\;\mathrm{mm/s^2}$ towards the Sun;
  • but Earth also experiences a centripetal acceleration of $g=5.9\;\mathrm{mm/s^2}$ towards the Sun!

Were there no Earth-Moon attraction, then the Moon would have stayed in a fixed place compared to the Moon, since they "fall" towards the Sun (circulate it) equally fast. When Earth then on top of that motion causes an extra pull in the Moon, the Moon will - during the "fall" towards or circulation around the Sun - have a superimposed motion around the Earth.

(And the Earth will likewise wobble a little towards the Moon - it is just too massive in comparison for any circular motion to be seen.)

You can compare it with a skydiver letting go of a bowling ball; it will stay next to him (more or less due to air resistance) because they both fall equally fast. He can easily grab the ball and move it to the other side or above or below him - or start moving it in a circle around him. That is, he can easily superimpose that motion on top of the falling motion of the ball.

Compare such circulating motion of the ball due to the force from his hands (which is much smaller than the Earth's pull in them) with the Earth's pull in the Moon (which is smaller than the Sun's pull). Such motion is possible due to the superposition principle of motions.

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