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The most simple and reasonable solution about mercury precession
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The most simple and reasonable solution about mercury precession
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ZHUYH
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The most simple and reasonable solution about mercury precession
«
Reply #20 on:
04/11/2008 01:37:22 »
Google:The Complement of Newton Gravitation Law
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lyner
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The most simple and reasonable solution about mercury precession
«
Reply #21 on:
13/11/2008 12:39:13 »
Give me the bottom line on this.
How big is this new force, compared with the force due to 'normal' Gravity?
It would need to be less than the spread of errors involved in all the established measurements or it would already have been revealed as a constant error, wouldn't it?
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ZHUYH
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The most simple and reasonable solution about mercury precession
«
Reply #22 on:
14/11/2008 03:27:54 »
F = Fn + Ft
To the weight of Fn = GMm / r ^ 2 -
Tangential component [vortex force] Ft = kGMm ω Cos α / r ^ 2
K = 0.4 for one factor,the unit: sec/cycle ; ω is the angular velocity rotating ball, the unit: cycle / sec; α to the orbital inclination.
By the formula, we can see that:[earth]
Fn =216000 Ft
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lyner
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The most simple and reasonable solution about mercury precession
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Reply #23 on:
14/11/2008 11:09:28 »
One point:
angular
velocity has the units of Radians per second (as opposed to
frequency
). Is your expression out by a factor of Pi?
Have you any measurements which reveal and confirm this factor of 1/2000,000? It's a tiny factor. If you obtained your value from the Moon's behaviour, is there any other data to support it?
As you know, there is an alternative mechanism which needs to be included. This is a real effect and needs to be factored into your sums.
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ZHUYH
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The most simple and reasonable solution about mercury precession
«
Reply #24 on:
17/11/2008 06:48:07 »
Fn = GMm / r ^ 2
Tangential component [vortex force] Ft = kGMm ω Cos α / r ^ 2
By the formula, we can see that:[earth]
Ft = k ω Cos α Fn
Cos α=1,ω = 1/ 24×3600[cycle / sec] , K = 0.4 for one factor,the unit: sec/cycle
Ft = [0.4/ 24×3600]Fn
=Fn/216,000
is not: 1/2000,000
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lyner
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The most simple and reasonable solution about mercury precession
«
Reply #25 on:
17/11/2008 10:33:26 »
Same comment as in the other thread - I slipped another zero into my post.
You still really need to put this force in context with the other forces in operation. Is it greater? Is it less? Is it a major factor or a tiny modification?
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