Anisotropy and Mercury (2)
On May 29, 11:41 pm, Koobee Wublee wrote:
Max Keon wrote:
[...]
Centrifigal forces change at the rate of orbital speed squared,
so if Mercury was traveling at an average of 24000 m/sec instead
of the average 48000 m/sec, it would be restrained from falling
at the full rate by 24000^2 / 48000^2 = .25 of the .0395 m/sec^2
gravity rate. The fall rate is .25 * .0395 = 9.875e-3 m/sec^2.
[...]
The gravity force is pointing directly at the Sun, so unless
Mercury falls closer to the Sun on average its orbital speed
cannot be increased. Adding a new force does not change the pull
direction, so orbital speed cannot change from the normal unless
the average radial length changes. Can you now see that?
Have you asked yourself why the centrifugal force is (m v^2 / r)?
I wonder if you have ever studied Newton's second law in a rotating
coordinate system....
In spacetime with the Schwarzschild metric, the Euler-Lagrange
equation associated with r can simply be derived as the following
where the orbital motion is confined to the equatorial plane of the
gravitating body.
It isn't "spacetime", chuckles. It is an affine parameter.
d^2r/ds^2 - (1 - 3 U) r (dO/ds)^2 = - U / r
Where
** O = Longitude
** s = Spacetime
** U = G M / c^2 / r
[snip all]
Wrong.
The actual radial equation of motion is:
d^2r/dl^2 + GM/r^3 * (r - 2GM) * (dt/dl)^2 - GM/r(r-2GM) *(dr/dl)^2 -
r(r-2GM)[(d\theta/dl)^2 + sin(\theta)^2*(d\phi/dl)^2] = 0, where l is
an arbitrary affine parameter [none of this 'spacetime' bull****].
Notice this isn't anywhere even close to your idiotic equation. As
usual you get even the simplest mathematics totally and completely
wrong.
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