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I need help finding the gravitational null point point between the Earth and the Moon, the point beyond which you start falling towards Moon. I am having difficulty in taking into account the orbiting of the moon around Earth.

If the Moon is stationary with respect to Earth:

  • Mass of Earth : $M$
  • Mass of moon : m
  • Distance between Earth and Moon: d
  • Distance between Earth and null point: x

At null point, forces cancel out: $$\frac{GM}{x^2} = \frac{Gm}{(d-x)^2}$$

$$\frac{d-x}{x}=\sqrt{\frac{m}{M}}$$

$$\frac{d}{x}-1=\sqrt{\frac{m}{M}}$$

$$x = \frac{d}{1+\sqrt{\frac{m}{M}}}$$

How would this change if Moon is not assumed to be stationary?

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    $\begingroup$ @HritikNarayan Thanks for formatting the equations $\endgroup$ – farizrahman4u Sep 19 '15 at 6:38
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    $\begingroup$ You can check this out for future reference! meta.math.stackexchange.com/q/5020 $\endgroup$ – Hritik Narayan Sep 19 '15 at 6:39
  • $\begingroup$ I guess if it is not stationary then d is a function of time. $\endgroup$ – Horus Sep 19 '15 at 8:56
  • $\begingroup$ @Hours d is the distance between earth and moon. Moon is orbiting earth in an almost circular orbit. So d should be constant. $\endgroup$ – farizrahman4u Sep 19 '15 at 9:02
  • $\begingroup$ It appears the question was answered in some detail here: physics.stackexchange.com/q/147908 $\endgroup$ – userLTK Sep 19 '15 at 9:19
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If you assume both the Earth and Moon to be moving (because, you know, they are) then you actually have five different points where an object would be in a stable configuration, and appear to be stationary with respect to the two massive bodies.

Note that this answer is slightly different than the question that you're asking; the net force at each of the Lagrangian points is not necessary zero, and you're not doing physics in an inertial frame anymore. However, practically speaking it's a more meaningful answer. If you were to put a test mass at the location you calculated in your question, it would be in unstable equilibrium and soon have a net force directed at either the Earth or the Moon. This is not very helpful when trying to place satellites in a stable location. Instead, we put satellites as Lagrangian points because the satellites are much more stable there. The Earth and Moon are moving, and we don't want our satellites to fall out of orbit, so instead of placing them where there's no net force, we place them where they'll stay stationary w.r.t. the Earth and Moon.

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How would this change if Moon is not assumed to be stationary?

Nothing in space is stationary. Everything is falling through space based on the gravitational shape of the space they are in. In theory, if the Moon was held by some force, stationary with respect to the Earth, you're math is correct, the distance ratio is 9 to 1. Essentially the square root of the mass difference (81 to 1) or 1/10th if you measure the Moon's region to the distance between earth and moon.

Now if (using this impossible scenario) you're moving and the Earth and Moon are stationary to each other, the easiest way to calculate is your velocity vs the center of mass of the Earth-Moon combination and if you're velocity exceeds escape velocity, you'd land on neither, if less than, you either orbit for a while or land on one or the other. But all that seems rather clumsy to me because objects don't stay in one place.

It's much more interesting with moving objects. The Moon orbits the Earth at about 2,290 miles per hour or 1.023 kilometers per second. If you're just "still" in reference to the Earth, you'd be moving 2,290 miles per hour in relation to the moon, or, perhaps a more useful number about 43% of the Moon's escape velocity. With that relative velocity you'd need to be much closer to the moon to have a chance of landing on it, either conveniently in the Moon's orbital path or within (square root of 1/0.43), within 1.52 moon radii from the moon's surface to be close enough where the Moon's gravity could hold onto you . . . kinda/sorta. Too close to the escape velocity and you'd fall too far from the moon and then the earth wins, so, lets say 1.5 to 1.51 moon radii in this scenario.

This means, the Moon's range of keeping you shrinks from about 24,000 miles to about 1,500 miles.

The most interesting scenario, to me, is if you're orbiting the Earth. Earth orbits are a little more tricky. Certainly low earth orbits tend to fall towards the earth due to atmospheric drag, but higher earth orbits (I think - somebody verify this if they like), but high enough so that atmospheric drag is less of an issue, the tidal forces will cause you to gradually raise your orbit and you'd drift, over a significantly long period of time, towards the moon. Because of the sun, the Lagrange points aren't stable either, so if you're in earth's orbit, you're arguably more likely to eventually fall onto the moon than you are to fall to earth, unless the atmospheric drag is strong enough to slow you down, which would depend in part on your mass, for example, were you in a heavy capsule or just a space suit? Mass is irrelevant to orbit but very relevant to atmospheric drag. You could also fly just right and get a gravity assist around the moon, adding 2 times the moon's orbital velocity to your velocity and in this scenario, fly away from the Earth-moon system and hit neither.

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If your test mass is being placed in an orbit at same angular speed as the lunar orbit i guess you'd need to super impose its centripetal force with the gravitational force.
The new null orbit will probably be much lower because the null point is already near the orbit altitude.

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