Someday, the Moon will be colonized. Before this happens, it would be nice if we could speed up the Moon's rotation, since we won't be able to grow crops with a 2-Earth-week day and a 2-Earth-week night unless we use artificial light.
Here's some calculations on speeding up rotation of a moon or planet that I thought were interesting.
Both the Moon (28 days) and Venus (~243 days, retrograde) currently spin
very slowly, and we would want to speed them up to something close to 24
hours. (although perhaps we could stop at 48 hours, for 1/4 the energy)
The energy required to do this is enormous, on the order of 10^26 (Moon)
to 10^29 (Venus) Joules. That is a *lot* of energy, the only ways I can
think of to generate this much is via impacts (where a slight nudge now
could deliver a huge amount of
energy later), and some kind of matter/anti-matter nuclear device that allows us to get .5MC^2 of energy from M mass.
Let's try impacts first, since if this would work, it could actually be done with something like today's technology. I'll make some vigorous-hand-waving assumptions:
* All impactors have a density of 5 (I know, it's their mass that counts, not their density, but bear with me for just a bit)
* All impactors have a impact velocity of 5 km/second.
* 10% of the kinetic energy from an impact goes into rotational energy (obviously, we have to target the impacts at the right spot). The remaining 90% goes to translational energy, heating, throwing of ejecta, etc.
* The Moon's density is not a function of distance from the center. This is obviously false, but it does give us a worst-case calculation, any increase in density toward the center just makes the Moon easier to spin.
For the Moon, I find that a single impact from a 380 km-diameter asteroid,
given the above assumptions, would deliver the
energy required. Of course, there are a host of other concerns: even if we targeted the impact for the far side, we'd get some
meteorite-rain here on Earth, and that could be dangerous. Also, targeting anything that big to come close to Earth is potentially scary. If you thought the protests over Cassini were bad, just think of this! Finally, hitting the Moon with an asteroid of this size would impart a lot of damage, perhaps rendering large areas of the surface unstable for some time.
So, we could consider smaller impacts. Here's a quick table of impactor
size, resultant Lunar day, and number of asteroids
of this size required to get the Lunar day close to one (note that we don't get anything for free here, we need the mass of a 370 km diameter asteroid, whether it is in a single impact, or spread out over multiple impacts)
Well, maybe this is better. With 400 hits from 50-km asteroids, we could
have a 24 hour Lunar day. Even if danger to Earth
from the impacts is ignored, we have another significant problem: these asteroids don't exist.
Here are some links to the known Earth-crossing asteroids:
Including the Amors is cheating a bit, these asteroids do not cross Earth's orbit. A couple of weeks ago I was looking at a web site (that I now am unable to find) that gave diameters for these asteroids, but the above sites give absolute magnitudes and hints for relating magnitude to size.
Note that there are only 6 or 7 that are over 10 km in diameter, and
the largest is 40 km. If all of these are targeted for Lunar impact, I find
the resultant lunar day is about 19 Earth days - and that's assuming a density
of 5, which is maximal. On one hand,
this is a improvement over 28 days - plus, we get to actually see the far side from Earth with a telescope! On the other hand, we're still very far away from a 24-hour Lunar day. There may be an additional complication that the Moon must first be knocked out of its tidal lock with Earth, and I'm not sure what energy this would require (I'll make a guess later in this page).
Changing some of the parameters helps some; if the impact velocity is
10 km/sec, and *all* the impact energy goes into increasing the Moon's rotation
speed, then we "only" need 11 50-km asteroids (none of which are
known to exist!) hitting in the
So, unless we find a "nice" source of large Earth-crossing asteroids (or ones that can be easily perturbed into an Earth-crossing orbit), we're not going to speed up the rotation of the Moon via impacts.
And, here's an additional complication: If there really is water ice at the Lunar poles, we don't want to lose it by changing the rotation axis, which increases the required precision of the hit.
Since these asteroids don't currently exist in Earth-crossing orbits, step 1 would be to get them there. Perhaps some celestial billiards in the Kupier belt could throw a big chunk toward Jupiter, and it could then be deflected to the Moon.
Knocking the Moon out of tidal lock:
I have no idea how to compute this exactly, but we can bound it: Suppose the Moon was replaced with a mass-dipole, each pole with mass .5(MoonMass), how close can two poles be while still giving us confidence that we're overestimating the tidal-lock energy? Obviously, a separation of 2Rmoon is maximal, but I think we can wave our hands and get it down to a tenth of that, or .2Rmoon.
So, let's compute the potential energy difference between this mass-dipole and the Earth, at the two critical orientations of the dipole:
Rm=radius of Moon, Dm=Earth-Moon distance, Mm=mass of Moon, Me=mass of Earth:
-G(MmMe/Dm - .5MmMe/(Dm-.2Rm) - .5MmMe/(Dm+.2Rm))
= -GMmMe(1/Dm - 1/(2(Dm-.2Rm)) - 1/(2(Dm+.2Rm)))
Crank this, set it equal to 10% of .5mv^2 and I find that a single asteroid of roughly 25km diameter (with a density of 5), impacting at 5 km/sec, imparting 10% of its kinetic energy to the Moon's rotational energy, would knock the Moon out of tidal lock. Hey, this would be fun, and it would allow us to see the far side from Earth! We'd have to hit has the Moon approached Perigee, for a little extra boost.
I read somewhere once (oooh, there's a good reference!) that the Moon is about 5km "off center". If this is true, and we crank the above equation with a delta of 10km, then:
.05 m v^2 = -GMmMe(1/Dm - 1/(2(Dm-10 km) - 1/(2(Dm+ 10km)))
The equations come from:
(.5MV^2)*.1 = .5 I W^2
10% of K.E. of impact = Rotational energy after impact
W (the rotational velocity in rad/sec)
= sqrt(.1*5mV^2 / 2MR^2)
with V = velocity of impactor
m = mass of impactor
M = mass of moon
R = radius of moom
Now, 2PI/W gives us the length of a rotation due to this
impact (in seconds), and we then divide by 86,400 to convert
it to days.
So, Days = 2PI / (86,400 W)
= (2PI/86,400) sqrt(.1*2MR^2/5mV^2)
= (2PI/86,400) (R/V) sqrt(.1*2M/5m)
Of course, the moon is not starting from a standstill, it has
a rotation period of 28 days. In the general case, this gives us:
.5 I W2^2 - .5 I W1^2 = .1 E
Or, the difference in the rotational energies before and after
impact is equal to a tenth of the impact kinetic energy.