Re: Perihelion of Mercury with classical mechanics ?

carlip-nospam_at_physics.ucdavis.edu
Date: 02/14/05 

Date: Mon, 14 Feb 2005 22:02:07 +0000 (UTC)

Nicolaas Vroom <nicolaas.vroom@pandora.be> wrote:

> <carlip-nospam@physics.ucdavis.edu> schreef in bericht
> news:cu3mcl$ma4$1@skeeter.ucdavis.edu...
>> Nicolaas Vroom <nicolaas.vroom@pandora.be> wrote:
 
>> To a very good approximation, for nearly circular orbits the radius
>> at time t will satisfy
>>
>> r^2 - (r_0)^2 = (4GM/c_g)(t-t_0)
>>
>> where r_0 is the radius at time t_0 , M is the mass of the Sun, and
>> c_g is the speed of (Newtonian) gravity. Take r_0 to be the radius of
>> the Sun and r the present radius of the Earth's orbit, and you can
>> use this to compute t-t_0, the time in the past that the Earth must
>> have been at r_0. If c_g=c, this comes out to about 400 years. The
>> time is directly proportional to c_g, so for c_g=300c, this becomes
>> about 120,000 years.
 
>> Again, the computation is fairly simple; see the Lightman reference
>> I gave before. All you really have to do is to note that the effect
>> of propagation delay in Newtonian gravity is to impart a tangential
>> acceleration equal to v/c_g times the radial acceleration, and
>> compute the change in energy.

> I have studied the book from a library but still I have a couple
> of unsettled questions.

> In the above equation the stability is a function of the mass
> of the Sun.
[...]

> The reason IMO comes from the equation at page 350
> that the earth's energie increases at a function of
> v * theta = v * v / c.
> IMO that is wrong.
> IMO the sun's energie is a function of angle v / c
> i.e. v+/c with v+ being the speed of the earth.
> IMO the earth's energie is a function of angle v0/c
> with v0 being the speed of the sun.
> In total earth's energie is a function of v+*v0/c
> (which is a factor M+/M0 smaller)

Ah... You're partly right. There is an additional ambiguity here in
what one means by "Newtonian gravity with a finite propagation speed."

The most common model assumes, explicitly or implicitly, that something
is traveling between the Earth and the Sun at a speed c_g, and then
imparting a force in the direction of its motion. In that case, the
relevant speed is the Earth's speed, and the tangential acceleration
of the Earth is proportional to v+/c_g. (Think of the usual analogy of
"walking in the rain" -- it's your speed that determines the angle the
rain hits you.) This is the assumption of, for example, Van Flandern,
and is the starting point of Lightman et al.

But one could divorce oneself from this picture, and simply postulate
that the Sun's gravitational force at time t points to the position
of the Sun at time t - r/c_g. In that case, you would be right in
saying the tangential acceleration of the Earth would be proportional
to v0/c_g. This would give you your factor of M+/M0.

This still won't agree with observation, though. Probably the most
clear-cut case is the Moon (this is the example Laplace looked at).
You should check my arithmetic, but even with a tangential acceleration
proportional to the Moon's velocity, I get a change of about 2500 m/yr
for gravity propagating at c. Thanks to Lunar laser ranging, we've
known the position of the Moon to an accuracy of a centimeter or so for
more than 30 years. An anomaly that size -- or even one of 8 m/yr, for
c_g = 300c -- would be impossible to miss. If you assume, generously,
that a steady increase of 1 cm/yr could have been missed for 35 years,
you need c_g to be at least 250,000c.

Steve Carlip

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