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Re: ZetaTalk and Spaceguard UK (D6)

Nancy was rescued by M.C. Harrison who inserted the right figures.
  Article <6kfst2$>
  27 May 1998 02:13:54 GMT
Copied below and at


In article <> M.C.Harrison writes:
> What is more, the acceleration of the moon will be equal to
> (-W.W.r.cosWt,-W.W.r.sinWt, 0) in a cartesian system, where
> W is angular velocity, r is radius of orbit and t is time. Radians
> and meters and seconds. This corresponds to
> (-W.r.sinWt,W.r.cosWt,0) for velocity, as rate of change of
> velocity is acceleration, and (r.cosWt,r.sinWt,0) is the position,
> as rate of change of position is velocity. The rate of change of
> something with time is d/dt exp. This goes like the following,
> d/dx of A.sin(Bx) is -B.A.cos(Bx) d2/dx2 is -B.B.A.sin(Bx),
> d3/dx3 is B.B.B.A.cos(Bx) d4/dx4 is B.B.B.B.A.sin(Bx), and
> so forth.  To evaluate the d/dx of a matrix, (A,B,C) merely
> d/dx each of the terms.

Dear M.C. Harrison, you lost me (not the Zetas, who can only speak
though this lowly instrument which plays the math notes poorly) when
you put commas between the cos and sin.  Can you put this into numbers
I can relate to?  Did I calculate the inverse square thingie for the
gravity pull between Moon and Earth correctly, as in equating to 88
metric tons on the surface of the Earth (so the common man can relate,
um, including the common woman).

If so, how to deal with the V=v+At thingie?

The issue is, if the Moon is pulling toward Earth at 88 trillion metric
tons, and only moving at 32 km/minute, then what is it's acceleration
as it FALLS toward Earth?  How does this balance out against the push
AWAY that the straight line momentum encompases?  Please note, my
calculator only does square root, no cos or sin, and I'm a bit boggled.
Thanks in advance for your help!