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If the moon was close in orbit that it's surface was like 100 km away from the earth's surface. And it had a large enough angular velocity will it be able to hold orbit?

If this was possible, is something similar possible to exist in the universe (the very close orbit)?

If this was hypothetically in effect (ignoring the damage that would cause to life on earth), would a person on earth, or a person on the moon tend to levitate, or gain weight on the far sides?

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  • $\begingroup$ How fast does the moon have to be in order to maintain such an orbit? $\endgroup$
    – Force
    Commented Apr 11, 2013 at 17:59
  • $\begingroup$ You can work it out with Newton's gravitational law: $F = \frac{GMm}{r^2}$ and the expresion of the normal force (centripedal): $F = m \frac{V^2}{r}$ $\endgroup$
    – iiqof
    Commented Apr 12, 2013 at 10:53

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You should do some hard assumptions, as no atmosphere, but the real problem is the so called Roche Limit, that states that at this limit, the tidal forces , difference of the gravitational potential between the face facing the Earth and the one opposite, is so large that rips the body (Moon) apart.Wiki

With the Wikipedia formula:

$ d = 2.44 R_{Earth} \left(\frac{\rho_{Earth}}{\rho_{Moon}}\right)^{\frac{1}{3}} \sim 2.44\cdot R_{Earth}km \left(\frac{5.5}{3.3}\right)^{\frac{1}{3}} \sim 2.89R_{Earth} $

That would be well beyond the Moon in your scenario.

But despite everything, the Moon and the Earth survives, and people also, then you should simply equate the gravitational pulls of each body.

$Earth_{Surface} = m_{individual}g = 70 \cdot 9.8m/s^2$.

$Moon_{100km above} = G\frac{M_{Moon}m_{individual}}{(d+R_{Moon})^2} \sim m_{individual}1.44 m/s^2$

where $m_{individual}$ is your mass. Earth pull would still be $6.8$ times stronger. You would not levitate, but you wouldn't need to follow a diet plan.

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  • $\begingroup$ If the moon doesn't collide with Earth we wold end up with nice rings due the tidal forces. $\endgroup$
    – iiqof
    Commented Apr 11, 2013 at 9:49
  • $\begingroup$ To answer the other part of the question: a person on the near side of the moon would levitate. The Roche limit quoted in this answer is, in fact, the balance point, where a person on the surface of the moon would be exactly weightless (assuming the moon stayed spherical). $\endgroup$
    – Retarded Potential
    Commented Apr 11, 2013 at 17:44
  • $\begingroup$ @Angel is there something similar in the universe, or will just rip off according to the above formula? and how about people in the moon will they fly away? how fast does the moon need to be to maintain such an orbit? $\endgroup$
    – Force
    Commented Apr 11, 2013 at 17:53
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    $\begingroup$ There could not be souch places, as they get destroyed. Although there are places with similar sights, Jupiter's moons have almost the whole sky filled by Jupiter. As for the gravitational pull, try working the third fórmula in extreme places, close binary stars, or systems with blackholes. But not planetary. And as Lubos Motl says, the system is really unstable. $\endgroup$
    – iiqof
    Commented Apr 11, 2013 at 21:04
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Yes, it is possible for objects to orbit 100 km away from the Earth. After all, lots of man-made satellites are doing so. It's also possible for a moon to be much closer than it is now. However, you must realize that the radius of the Moon is 1,700 km so 100 km can surely not be the distance between the Moon's center and Earth's surface.

However, if the Moon were this close, it would have severe consequences. Yes, a person would feel lighter on the side closer to the Moon and heavier on the opposite side but despite the Moon's substantial size, it would still change the weight roughly by 1 percent in both directions (because the Moon is substantially lighter than the Earth).

This difference would primarily manifest itself as the tidal forces. They already exist but they're much weaker than in your hypothetical world.

In your hypothetical world, the Moon's center is about 8,000 km from the Earth's center – about 50 times closer than it is in the real world. Because the tidal forces scale like $1/r^3$, this amplification would translate to $50^3=125,000$ times stronger tidal forces. Instead of 1-meter tides, one would get 100-kilometer megatsunami on both sides of the Earth. Moreover, the tides wouldn't have a cycle of 12 hours. They would get repeated every 45 minutes or so.

Needless to say, 100-kilometer tides would splash everything on the surface of the globe. Lots of energy and angular momentum would be lost in this way, too. As a result, the Moon would soon either collapse to the Earth or flew away from the Earth.

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  • $\begingroup$ I apologize what I meant by the 100 km is the surface of the moon not the center of gravity. I will edit my question to clarify that. $\endgroup$
    – Force
    Commented Apr 11, 2013 at 17:58
  • $\begingroup$ No megatsunamis--the tides deliver so much energy the oceans would have boiled off. $\endgroup$
    – Loren Pechtel
    Commented Jan 28, 2015 at 1:21
  • $\begingroup$ A good point if it's right. $\endgroup$
    – Luboš Motl
    Commented Jan 28, 2015 at 14:04
  • $\begingroup$ You would be lighter both on the side closest to the Moon (due to stronger gravity when closer) and on the side farthest from the Moon (due to weaker gravity when farther). At the Earth's center of gravity (or on the other sides of the Earth) the Moon's gravity and the centripetal acceleration of that point around the Earth-Moon barycenter (center of their mutual revolution) balance out and leave you the heaviest (at the surface, that is). The tide stretches both towards and away from the other body. $\endgroup$
    – Rob Parker
    Commented Dec 2, 2016 at 18:52

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