Well Ken, using the figures over 100 years of the moon receding 4 m and the Earth slowing 1 milli second may be just too approximate, as you say. Because using them, it appears the Earth would have to lose 1.823 times as much angular momentum as I calculated, in order to conserve angular...
...Back to Ken's angular momentum...
Yes thanks Jenk, I came up with the similar formulas and derivation to show root 2 once Janus explained what Wiki meant, and so the quandary referred to in the post title is solved for me.
However, the thread has progressed to a disussion of conservation of...
Ken, I understand that in the end angular momentum is conserved (somehow), however I’m finding my calculations are NOT showing this. For the situation where the primary (the Earth) slows 1 msec over 100 years, I’m finding the change in the rotating Earth’s angular momentum L = 2/5 * Me * Re *...
Janus, I did previously calculate the ratio of the sun's force of gravity on the moon compared to the Earth's, in its present orbit, and came up with 2.21
Should there be a theoretical maximum for the ratio of forces of gravity at the Hill sphere for different primary and secondary objects?
So the ratio of the energy the Earth loses compared to the moon gains is
6.00 E21 / 3.96 E20 = 15.14
Total energy of the moon after receiving 1/15th of the Earth’s rotational energy = -2.11 E28 joules.
Radius for that energy = 694,000 km, compared to 384,400 present radius of moon’s...
Here’s what I calculated:
The Earth’s loss in rotational energy due to a change of one millisecond (happens over a century) is 6.00 E21 joules
The moon’s gain in energy receding 4 m, (receding 4 cm per year ) is 3.96 E20 joules
So the fraction of the energy the Earth loses compared to the...
So Janus, using your formula for the moon’s total energy (other than some small rotational energy about it’s center), I get:
E = -6.6726 E-11 * 5.974 E24 * 7.348 E22 /(2*3.844 E8) = -3.81 E28 Joules
Now for the rotation energy of the Earth, I used Ek = mv^2/5
Where m is the mass of a...
DH, just trying to get my head around this quote of yours. If the spinning results in a minimizing of the total energy, then there must have been more energy at an earlier time. If so, where does this energy go?
Secondly, isn't any change in the shape of a spinning object just the result of...
DM, as you know the Earth spins around once every 24 hours or so. The moon circles the Earth about every 28 days. The tidal bulge created by the moon's gravity on the Earth is subjected to the Earth's rotation, and hence this bulge ends up ** ahead ** of the Earth-moon centre line. The effect...
thanks Janus, that explains it. I used Vesc = sq root(2*G*M of moon/R) and calculated the escape vel for an object on the surface of the moon to be 2.375 x 10E3 m/sec, almost exactly what wikipedia states. Further, I used the same equation to calculate the escape velocity of the moon from the...
I had thought that for an object in a simple circular orbit, the escape velocity would be root 2 times it's orbital speed. And that this should approximately apply for the Moon too. However Wikipedia here
http://en.wikipedia.org/wiki/Moon
gives the moon's escape velocity as 2.38 km/sec...