Quote:
On 2002-04-01 16:30, Gary Redmond wrote:
..xxxy............
.xxxx.............o
zxxx..............
If you let the x's, the y, and the z represent Earth, and the o the Moon. You will find that gravitational force between y and o can not affect the force between y and z. Inversely the force on z by o will not affect y.
If the parallelogram xyxz were to rotate it would need to be due to something other than o.
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I inserted dots to be empty space, otherwise the parallelogram seems to turn into a square on my display, and the moon rushes up to earth adjacency...
Note that y is closer to o; by the inverse square law, it is more attracted. z is farther away, and by the isl, is less attracted -- and, in fact, is so much less attracted that it does not compensate for the additional attraction felt by y.
Thus, the force drawing y toward o is unbalanced, overall. Assuming that, due to orbital velocity, the overall p-gram does not actuall draw nearer the moon, it will, at least, rotate until y is closest to the moon.
Do the math: Fyo^2 :: 170 Fz0^2 :: 383
Is anyone good enough to do a simple BASIC sim? (I think I am, but I'm not wholly sure...)
Silas