[extropy-chat] Astronomical question

[spike]

> -----Original Message-----
> From: extropy-chat-bounces at lists.extropy.org [mailto:extropy-chat-
> bounces at lists.extropy.org] On Behalf Of Dan Clemmensen
> Subject: Re: [extropy-chat] Astronomical question
> 
> Mike Lorrey wrote:
> 
> >It is a well known phenomenon that the moon is draining angular
> >momentum from the earth...
> >
> I seem to recall that the earth slows to a stop before the moon escapes...


Don't look it up Mike, figure it out.  We know the mass of the earth is
about 6e24 kg and the mass of the moon is about 7e22 kg.  The moment
of inertia of a sphere is 2/5MR^2 and the radius of the earth is
about 6E6 meters.  So the moment of inertia of the earth is about
1e38 kg-m^2 and the earth goes thru 2 pi radians a day so the angular
momentum of the earth is about 7e33 kg-m^2/sec.  The moon goes 2 pi
radians at a distance of about 4e8 meters in about 30 days.  
(It really doesn't work exactly that way, but this calc will 
get us the answer to one significant digit, which is all you 
often need.)  

So the angular momentum of the moon is about 2.7e34 kg-m^2/sec, 
so now you know the total angular momentum of the moon at its 
new distance and new month length must carry a total angular 
momentum equal to the sum of the earths angular momentum and
the moon's or about 3.4e34 kg-m^2/sec.

Set up the equation Mike.  You can see that the moon already
carries most of the angular momentum of the earth-moon system
and you know how to calc the orbit time as a function of distance,
so set that up, set the earth's day equal to the new month
length, calc the distance and set the result equal to 3.4e34 kg-m^2/sec.

Before I calc that, I can see the moon won't be getting away,
but that our day length will be longer than 30 days.

spike