Let's say we somehow speed up the moon's rotation. (I'm thinking about a big asteroid smashing into it, but I'm open to better ideas.) Here are some questions:

How fast could we get the moon to spin before gravity stopped holding it together? (An answer in revolutions per minute would be nice.)

At half that speed, what would the consequences be on the moon, on space missions to the moon, and here on earth?

How many asteroids would it take to get it to spin at that speed?


1 Answer 1


"before gravity stopped holding it together" is the same (pretty much) as "so that apparent gravity at the surface is zero".

This means that

$$m \omega^2 r = \frac{GM_{moon}m}{r^2}$$

with $G=6.7\cdot 10^{-11}$, $M_{moon}=7.3\cdot 10^{22} kg$, $r_{moon}=1740 km$, we find

$$\omega=\sqrt{\frac{GM_{moon}}{r^3}}=0.0092 rev/min = 0.55 rev/hour$$

It is interesting to see that the result has the ratio of mass and the third power of the radius - in other words it depends on the density and not the size. Whether the moon consists of loose dust that will just fly apart when you reach this speed is another question - not one you asked.

You asked for rev/minute but that's a hard number to grok. Note that one lap per two hours is what you would have to do in order to remain "in orbit" at the surface of the moon. Interestingly, the Apollo 11 command module (piloted by Michael Collins, the least known of the three astronauts that formed the crew of Apollo 11) was orbiting at about 60 nautical miles, and going around once in just about 2 hours. Pretty similar... And that was no coincidence.

Also note that this speed would start the moon ripping apart at the equator - where the centrifugal force appears strongest. At other points along the surface, you have a force of gravity pointing towards the center; the component of gravity that counters the centrifugal force (normal to the axis of rotation) scales with $\cos(\text{latitude})$ which is also how the centrifugal force scales. This means that matter will start drifting from higher latitudes towards the equator.

I haven't even begun addressing the other questions you had - about the impact on earth if the moon was spinning at half that speed (4 hours / revolution). I guess we would get to see the back of it, and watching the moon at night would be a less peaceful experience - but I can't think of any real physical effects (tides etc) on earth that would affect life. Chances are great that if the moon is being hit by asteroids to make it spin faster, that this would affect its orbit around the earth: that would impact tides, and that would really wreck coastal ecology and possibly climates; but I'm assuming that the moon would accelerate from glancing impacts with "twin asteroids" coming from opposite directions, leaving no net linear momentum, and no net mass increase / decrease - just make it spin faster.

The impact on a moon base would be more substantial. Every point on the moon would experience days and nights on a four hour cycle, with the temperatures cycling violently. Communication with earth would be disturbed - you would want to place relay transmitters on the poles to maintain contact. And of course gravity would be halved again - instead of 1/6th of the earth's pull, you would appear to have 1/12th. Coriolis forces would be wicked too - although you could drive a golf ball a very long way with such low gravity, you'd get a horrible hook / slice, depending on whether you're on the Northern or Southern hemisphere of the moon (and whether you're left- or right-handed).

Finally a word on the energy of the moon if it spun that fast. We know that

$$E = \frac12 I \omega^2$$


$$I = 2/5 M R^2 = 8.3\cdot 10^{28} J$$

If you took all the power of the sunlight that hits the earth (assuming 1 kW/m^2 over the side that is lit) for 20,000 years - you would have the amount of energy needed to give the moon that kind of energy. I imagine that if a sufficient density of meteorites came near enough to the moon to impart that kind of energy, we would not be worrying on earth about building moon bases...

Footnote about Collins' orbit.

It was said that while he was on the far side of the moon, he was the "loneliest human since Adam" - because he was literally further from any human beings (and with no means of contacting them) than any other human, anywhere on earth (this was before Major Tom, of course). The NASA log of the flight contains the tidbits needed to reconstruct his orbital speed:

  • hidden by the moon for 47 minutes per orbit (July 21, 9:44 am)
  • Altitude of 60 nautical miles
  • orbital velocity was 5329 ft/sec (July 22, 12:56 am)

After converting to sensible units (really - they put a man on the moon with feet and nautical miles? What are we doing, trying to force SI units on a nation that was capable of such a feat, with those units…) and drawing a little diagram, I convince myself that the fraction of the orbit when the command module was "dark" is computed as

$$f = \frac{\pi - 2 \cos^{-1}\left(\frac{R_m}{R_m + h}\right)}{2\pi} \approx 0.35$$ And if that took 47 minutes, then the orbit took 121 minutes.

  • 2
    $\begingroup$ "Pretty similar..." and not just a coincidence! $\endgroup$
    – BMS
    Sep 11, 2014 at 21:49
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    $\begingroup$ +1. But note this answer tells you how fast the moon would need to rotate such that lose items on the surface begin to float up. I suspect this is a lower-bound on the actual answer since there are some internal structure/binding mechanisms present. $\endgroup$
    – BMS
    Sep 11, 2014 at 21:51
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    $\begingroup$ Why thank you! This was the answer I was looking for! Also I read about Collins and what he was feeling like and it was beautiful! $\endgroup$ Sep 12, 2014 at 11:14

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