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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: spinning up the Moon*From*: Steve VanDevender <stevev@efn.org>*Date*: Wed, 7 Mar 2001 00:18:14 -0800*Reply-To*: Steve VanDevender <stevev@efn.org>*Sender*: owner-starship-design@lists.uoregon.edu

I did a little research into what it would really take to spin up the Moon to a higher rotation rate. The angular momentum of a body is in units of kg * m^2 / s, which you can think of as kg (mass) * m (moment arm) * m/s (rotational velocity). The Moon has a mass of some 7.35 * 10^22 kg and a radius of 1740 km. Assuming it's a uniform-density sphere (not quite right, but sufficient for approximation), its angular momentum for a given rotation rate w (in 1/s) will be m * 2/5 r^2 * w. Its rotation rate is about 1 / 2.55 * 10^6 s. The Moon's current angular momentum is therefore about 8.72 * 10^28 kg*m^2/s. Spinning it up to have a rotation rate of 24 hours would increase its angular momentum to 1.03 * 10^30 kg*m^2/s, for a total change in angular momentum of 9.43 * 10^29 kg*m^2/s. So let's suppose, just for the sake of argument, that we have a way to smack comets or asteroids tangentially into the Moon's surface in such a way as to perfectly transfer their linear momentum to the Moon's angular momentum. That would mean a mass M striking at a radius of 1740 km with a velocity of v m/s, transferring a total angular momentum of M * v * 1.74 * 10^6 kg*m^2/s. So to spin up the Moon, we need a quantity of momentum M * v = 5.42 kg*m/s. So you can determine how much mass M would be required for a given impact velocity v, or given mass M what velocity v it would have to be traveling at to achieve the intended angular momentum change. Let's say you can manage to smack those objects into the Moon at 50 km/s (rather untenable, since that velocity's probably much too high to expect that the momentum would be efficently transferred). That means you'd need a mass of comets/asteroids of 1.08 * 10^19 kg, or that you'd need about 1/6780 the mass of the Moon in comets/asteroids impacting at that velocity _and transferring ther momentum entirely_ to get the Moon spun up. With lower impact velocities the mass ratio goes up proportionally; i.e. for impactors moving at 10 km/s you'd need five times as much mass as for impactors moving at 50 km/s. Even 1/10000 the mass of the Moon might not seem like much, but that translates into a pretty huge quantity of comets.

**Follow-Ups**:**Re: starship-design: spinning up the Moon***From:*"bugzapper" <bugzappr@bellsouth.net>

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