Perihelion of Mercury with classical mechanics ?

wrote:

{snip}

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.


The effect of aberration delay in Newtonian physics is a tangential
acceleration roughly proportional to (m / M) (v / v_g). You have left
out the first term. Quite simply, you (and Lightman) forgot that the
planet and the Sun orbit the center-of-mass -- not the midpoint of the
orbit. It is only when the masses are equal that the aberration angle
is proportional to v / v_g, alone.

Hence, your equation that resulted in 400 years for aberration, alone
(if you did the calculation correctly), would result in a 400 million
year time when v_g = c. (Since m / M is about 1 / 1,000,000 for the
Earth and Sun.)


greywolf42
ubi dubium ibi libertas

Nicolaas Vroom wrote:

schreef in bericht
...
Nicolaas Vroom 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

Dear Nicolaas Vroom:

"Nicolaas Vroom" wrote in message
...

schreef in bericht
...
Nicolaas Vroom wrote:

schreef in bericht
...
Nicolaas Vroom 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


Accordingly to this url the increase is 3.8 cm / year.
http://curious.astro.cornell.edu/que...php?number=124

For more info about McDonald Laser Ranging Station see:
http://www.csr.utexas.edu/mlrs/

I have performed a simulation for an Earth Moon system
with a speed of gravity cg equal to 3000km/sec
The increases are in km per rev: 10 (0.32), 20(0.64) 30 (0.95)
The values in brackets is the time in years.

That means there is an increase of 30 km / year.
For cg = 300000 we get 30000/100 = 300 m / year
For cg = 300*c we get 1 m /year = 100 cm / yr
which is a factor 25 higher than observed.

For cg = 8000*c we get 3.8 cm / yr.

I doubt if that value of cg is small enough to simulate
the precession of the perihelion of Mercury.

What if about half that value (3.8 cm/year) is due to actual angular
momentum transfer (via tides) between the Earth and the Moon? What will
that do to cg?

David A. Smith

"Nicolaas Vroom" schreef in bericht
news
Is there any one who can give me more informartion
how based on observations:
first the observed distance d1 Earth Moon,
(Using McDonald Laser Ranging Station ?)
secondly the actual distance
and finally this increase in distance of 3.8 cm / year
are calculated ?
How many observatories are involved as part of
these observations / calculations ?
Is the observed distance d1 at t1 calculated as:
(t2 - t0) * c / 2
or is a more complex algorithm used ?
With t0 = moment of emission of light from laser
With t2 = moment of receiving of light from laser
With t1 = (t0+t2) / 2

Nicolaas Vroom
http://users.pandora.be/nicvroom/


In order to get some idea about the Earth Moon
distance I have studied the program MOON supplied
as part of the book "Astronomy with your personal
computer" by Peter Duffett-Smith.
This program calculates the Earth-Moon distance
at a certain time and place, but with some modifications
you can also use that to calculate the maximum
distances for a certain period.

Following is the result for 1984:
Column 1 = distance in km
Column 2-4 = date Column 5-7 = time
405608.3849943659 7 1 1984 10 1 46.99
406430.5054588925 3 2 1984 16 25 17.56
406714.4260363030 1 3 1984 17 25 55.86
406350.3714502475 28 3 1984 19 57 7.43
405395.6816477149 25 4 1984 15 20 49.34
404468.1197446492 24 5 1984 0 30 48.53
404243.9745394128 19 6 1984 21 49 21.88
404852.1254151030 17 7 1984 18 47 23.51
405781.0821973901 14 8 1984 14 35 26.28
406359.6689450809 10 9 1984 18 47 0.53
406322.6275803399 7 10 1984 19 16 32.25
405690.9634341389 3 11 1984 23 25 22.60
404813.0296634666 1 12 1984 19 46 35.07
404338.7797164128 29 12 1984 18 10 7.13
404628.8869445491 26 1 1985 16 46 53.47

What this tells you is that the maximum distance
highly flexible over a period of one year
but what is more important that it is very difficult
to explain that suppose that all the distance value
are measured values that the value of 29 12 1984
should not be 404338 km and 77972 cm
but 77968 cm if this 3.8 cm increase is not taken
into account.

I expect that one important object that influences
these distances is the planet Jupiter.
That means you have to remove the influence of
Jupiter out of these observations.
In order to do that you have to know the positions
of the Earth and Jupiter.
Accurately ?
And if you do not know them accurately I expect
that makes any claim about an increase of the
Earth-Moon distance rather controversial.

(in the near future I will give more information about
a simulation with the four objects:
Sun, Earth, Moon and Jupiter)

With the same program I also calculated the
maximum distances over a much longer period:

406703.9617799576 28 12 1902 21 25 47.68
406707.6002670664 8 1 1921 16 35 35.06
406672.4259527441 27 1 1930 20 2 30.48
406696.1499911513 8 2 1948 15 8 22.16
406687.4182780707 30 1 1957 18 51 42.11
406710.1184460392 18 2 1966 22 16 29.22
406669.1009584760 8 10 1980 19 32 19.80
406714.4260363030 1 3 1984 17 25 55.86
406672.3202105450 8 11 2007 18 18 29.29
406656.9669761913 1 4 2011 16 28 15.11
406694.5184556800 23 3 2020 19 44 22.79
406705.0145049790 30 11 2043 20 37 8.66
406671.6101895833 23 4 2047 18 37 35.73
406707.5434456542 11 12 2061 15 46 49.81
406675.2184529630 30 12 2070 19 7 47.45
406698.4571414733 10 1 2089 14 13 52.00
406684.7218982943 2 1 2098 18 1 32.62
406712.3839595916 22 1 2107 21 21 57.43
406717.0200725148 2 2 2125 16 31 6.44
406667.1142930359 11 10 2148 17 21 43.40
406683.4550249315 4 3 2152 15 17 20.52
406698.3902277122 24 2 2161 18 48 20.10
406696.3255528670 15 3 2170 22 23 7.87
406701.2674920212 2 11 2184 19 39 35.86
406699.5379324314 26 3 2188 17 30 37.53
406704.4021879547 15 11 2202 14 48 44.01

Accordingly to my simulation the distance R
increases with 100 cm / year = 1 m / year.
That means with 1 km per 1000 years.

Does the above information, again assuming
that they are actual measured/observed values
invalidates such a claim ?

Nicolaas Vroom
http://users.pandora.be/nicvroom/

schreef in bericht
...

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


In my posting of 24 Feb 2005 I discussed the longest
distance between the Earth and the Sun with a programme
by Peter Duffett-Smith

406707.6002670664 8 Jan 1921 16 35 35.06
406714.4260363030 1 Mar 1984 17 25 55.86
406712.3839595916 22 Jan 2107 21 21 57.43
406717.0200725148 2 Feb 2125 16 31 6.44

406667.1142930359 11 10 2148 17 21 43.40
406683.4550249315 4 3 2152 15 17 20.52
406698.3902277122 24 2 2161 18 48 20.10
406696.3255528670 15 3 2170 22 23 7.87
406701.2674920212 2 11 2184 19 39 35.86
406699.5379324314 26 3 2188 17 30 37.53
406704.4021879547 15 11 2202 14 48 44.01

Jean Meeus in his book Astronomical Algorithms
at page 332 writes for apogee 406710 km
1921 jan 9 406710
1984 mar 2 406712
2107 jan 23 406716
2125 Feb 3 406720
2143 feb 14 406713
2247 dec 27 406715

The first 4 are close but the 5th one by Jean Meeus
is missing in the table by Peter Duffett-Smith

At page 331 JM writes : mainly by reason of the perturbing
action of the Sun, the actual time interval between
consecutive perigees varies greatly
At page 332 JM writes: It is evident that the variable Earth-Sun
distance somewhat affects the Earth-Moon distance

IMO also the distance (position) of Jupiter is important.

Accordingly to this url the increase between Earth-Moon
distance is 3.8 cm / year.
http://curious.astro.cornell.edu/que...php?number=124
that is roughly 1m in the last 30 years.

IMO there is no observational evidence to support this
claim (I' am not saying that the above value is wrong
but the value could be wrong and could be larger)
In order to do that it is not enough to know the Earth-Moon
distance accurately.
You must also know the Earth-Sun, Earth-Jupiter
and Sun-Jupiter distances accurately.

Nicolaas Vroom
http://users.pandora.be/nicvroom/