# The effect of the Doppler effect on planetary orbits

Discussion in 'Alternative Theories' started by TonyYuan, Apr 2, 2020.

1. ### HalcRegistered Senior Member

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350
Exactly. Thus a magnetic field is not a force field, for the same reason that a gravitational field is not a force field.
Maybe so, but a link would help me know what you mean by this. Without units, this is nonsense. What is the spatial density where you are now?
I don't think I mentioned a difference of acceleration in any of that. I'm comparing gravity to gravitational waves, only one of which produces acceleration.
I also made no mention of different locations.

I don't think they give out Nobel prizes for applying a principle known for at least 400 years.

A long rod generates gravitational waves if rotated. At some distance X from its center of gravity, it the gravity is greater if the rod is aligned with X than if it is perpendicular to a line drawn from it to X. The gravity from the rod is not 'conveyed' by gravitational waves or anything else, but changes to the gravitational field are conveyed by such waves. By this argument, no waves at all are sent along the axis of rotation of the rod since the field doesn't change in that direction. Nevertheless, an object there will be drawn by gravity to the rod.

3. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

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852
This is the concept in GR, please tell me your understanding, how to define space bending in GR. Hope it can help us correct understanding. Please note that our space is three-dimensional, not two-dimensional.

5. ### DaveC426913Valued Senior Member

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18,892
There are no waves. There is nothing for Doppler to affect.

Stop this and move on to something more productive.

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7. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

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852
https://photos.app.goo.gl/FGb6LqhESERuZ6zg7
Assume that the earth speed v0 = 0. Star began to approach the earth at the speed of v. Star came to position B, and the elapsed time was T seconds. The gravitational changes experienced by the earth are as follows:
T = 0s: F = G * M * m / R^2
T = 1s: F = G * M * m / (R + 1000)^2 ≈ G * M * m / R^2
According to the law of conservation of momentum:
F * T = mv1-mv0 = mv1, then v1 = F * T / m = T * G * M / R^2
T = 0s: v1 = 2 * G * M / R^2
T = 1s: v1 = 1 * G * M / R^2
Because of the speed v of the star, the earth obtained a different speed v1. The larger v is, the larger v1 will be, but the gravitational force on the planet from the star has hardly changed. If the star's v = X * cos(wt), then the speed change of earth will also show volatility.
The relative speed between them really affects the speed of the earth. Do you still think there is no Doppler effect between them? If you understand the Doppler effect, you will know that it exists anywhere.

8. ### HalcRegistered Senior Member

Messages:
350
My understanding is that there is nothing in GR called spatial density. You're making things up.

GR doesn't even have space bending. It has non-Euclidean (bent) spacetime, not bent space. Spacetime is 4D, not 3D. GR equations are not derived from deformed 3D space.

This is a good point.
Not sure why I ever joined in.

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9. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

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852
Okay, as you said, the curved space-time. I have already said that no matter what it is in the end (gravitational field or space-time bending ...), what matters is the change in the state of motion of the object.
"You are no longer suitable for discussion on this topic." This sentence was not written for you, but for the exchemist.
There is no problem in the discussion, but we all must pay attention to our words and deeds, and we should not talk nonsense.

https://photos.app.goo.gl/axeDPZggPZAjComR7
Assume that the earth speed v0 = 0. Star began to approach the earth at the speed of v. Star came to position B, and the elapsed time was T seconds. The gravitational changes experienced by the earth are as follows:
A: F = G * M * m / R^2
B: F = G * M * m / (R + 1000)^2 ≈ G * M * m / R^2

According to the law of conservation of momentum:
F * T = mv1-mv0 = mv1, then v1 = F * T / m = T * G * M / R^2
T = 2s: v1 = 2 * G * M / R^2
T = 1s: v1 = 1 * G * M / R^2

Because of the speed v of the star, the earth obtained a different speed v1. The larger v is, the larger v1 will be, but the gravitational force on the planet from the star has hardly changed. If the star's v = X * cos(wt), then the speed change of earth will also show volatility.
The relative speed between them really affects the speed of the earth. Do you still think there is no Doppler effect between them? If you understand the Doppler effect, you will know that it exists anywhere.
T: elapsed time

Last edited: Apr 6, 2020
10. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
Let us return to this topic, we would like to know why the planetary precession calculated by GR is accurate? We ca n’t really see where it ’s accurate.
https://photos.app.goo.gl/PWRJBQkNtNf7c2Mm6
Planet-------------observed-----------------theoretical-------------delt( "per year)-------------------( "per century)
Mercury--------------5.75----------------------5.50.....................................0.25.......................................25
Venus-----------------2.04----------------------10.75...............................-8.71.......................................-871
Earth------------------11.45---------------------11.87..............................-0.42.......................................-42
Mars------------------16.28----------------------17.60........................... -1.32........................................-132
Jupiter----------------6.55------------------------7.42.............................-0.87.......................................-87
Saturn----------------19.50-----------------------18.36..........................-1.14.......................................-114
Uranus----------------3.34------------------------2.72.............................. 0.62.......................................62
Neptune--------------0.36-----------------------0.65................................-0.29....................................-29

......R ......................e...........................Mine................GR( "per century)
46001200..........0.2056......Mercury: 40.4 "......GR: 42.93"
107476259.........0.0068......Venus: ...0.85 "......GR: ..8.64"
147098074.........0.0167......Earth: .....1.90 "......GR: ..3.85"
227936637 ........0.0934......Mars: ..........8 "......GR: ...1.34"
740573600........0.0483 .....Jupiter:......2.3 "......GR: 0.078"

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html
I can't see where the Einstein GR data is correct?
How do you find that the precession results obtained by GR calculation are very close to the observed data?
What do you think？ The data is the best proof. In addition to the correctness of GR in calculating Mercury's precession, other planets' precession calculations performed poorly.
GR didn't even consider the eccentricity e, just calculating the precession, this is ridiculous. GR data is more like a piece of data.

Last edited: Apr 6, 2020
11. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
Table 2:The observed perihelion precession rates of the planets compared with the theoretical precession rates calculated from Equation (1024) and Table 1. The precession rates are in arc seconds per year.
https://photos.app.goo.gl/d4AYsA6u9mLuGQW86
Mercury

Earth

I want to know why the earth's precession is bigger than Mercury's every year? From the perspective of the universe sand table game, the precession of Mercury should be much larger than that of the earth.

Who can help me? Thanks very much.

12. ### DaveC426913Valued Senior Member

Messages:
18,892
Since this discussion is based on an egregious lack of understanding of the subject matter by the OP - and he is entirely deaf to hearing about it - he is going down an imaginary rabbit hole to nowhere.

Since I'm not one for nonsense, I too am no longer suitable for discussion on this topic.

... click...

13. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
I will generally respond to your questions and queries, and will explain my point of view. I also did a detailed analysis of the Doppler effect of the gravitational field #106, but you turned a blind eye. So I do n’t know what else to say. I will continue my work. good luck.

14. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
Planetary precession data under the Doppler effect:
Mercury: e = 0.205608, Precession per century = 568.075883 "
Venus: e = 0.006811, Precession per century = 277.589395 "
Earth: e = 0.016780, Precession per century = 240.844614 "
Mars: e = 0.093332, Precession per century = 216.645123 "
Jupiter: e = 0.048700, Precession per century = 112.550073 "

Observed precession data:
Mercury -------------- 5.75 ------------- 575 "per century
Venus ---------------- 2.04 ------------- 204 "per century
Earth ----------------- 11.45 ----------- 1145 "per century
Mars ----------------- 16.28 ----------- 1628 "per century
Jupiter --------------- 6.55 ------------- 655 "per century
Saturn ------------- 19.50 ------------ 1950 "per century
Uranus --------------- 3.34 ------------- 334 "per century
Neptune ------------- 0.36 ------------- 36 "per century

Except for Mercury 575 "PK 568" and Venus 204 "PK 277", the precessions of the other planets differ greatly. But we all know that Mercury's precession is the largest, so I suspect there is a problem with the observation data.

15. ### phytiRegistered Senior Member

Messages:
732
DaveC;

Is this what you cited?
The 2013 critique of D.J.Sadhu's video on a vortex/helix solar system only corrected the flaws, but did not eliminate the idea.
The angle of inclination to the ecliptic plane was reduced from 90 to 60 deg.
With the sun orbiting the galactic center at 220 km/sec or .007c, and gravitational influence at c, a cone centered on the sun would be approx. flat. This configuration allows the earth to maintain an approx. constant speed. The motion of the sun is not in the ecliptic plane as shown in the Tony graphics.

16. ### phytiRegistered Senior Member

Messages:
732
Halc;

How can a fictitious force kill someone who falls from a great height?

Einstein, 1920 lecture at U of Leiden, ether and general relativity:
"The next position which it was possible to take up in face of this state of things appeared to be the following. The ether does not exist at all. The electromagnetic fields are not states of a medium, and are not bound down to any bearer, but they are independent realities which are not reducible to anything else, exactly like the atoms of ponderable matter."
He kept the concept of a redefined ether in the course of forming a unified field theory, for the remainder of his life.

17. ### phytiRegistered Senior Member

Messages:
732
Tony;

This will elaborate on what has been posted.

A gravitational field is static relative to the mass that generates it (via an unknown process).
Motion is relative per Relativity. The g-field varies 1/x^2 with changing distance x.
Doppler shift is the perceived change in the frequency of a process resulting from relative motion. This means you need a gravitational source that varies. The varying distances involved in any sun-planet system are too weak. You need large masses and high frequencies. That was the success of the LIGO experiment.

This may interest you.
If Newton had considered a finite light speed (discovered in the 1670's), he could have predicted gravity waves.
In the graphic, a pair of unit masses rotate around a common center. A test object is 3 units distant from the center.
Where would a single 2 unit mass be located to produce the same gravitational effect, when aligned in the x or y directions?
The equivalent center of gravity would oscillate between the 2 tick marks in a very eccentric ellipse, at a frequency twice the orbital period.

18. ### HalcRegistered Senior Member

Messages:
350
Are you just asking me, or are you replying to some comment I've made? I've no context here. The ether quote you included with your post seems irrelevant to the question asked.

Gravity, interpreted as a force, is not fictitious. Falling from a great height doesn't kill you. High acceleration from collision with the ground does. A sufficiently non-uniform gravitational field can kill you directly via tidal forces without help from the ground.

19. ### Write4UValued Senior Member

Messages:
19,970
From a layman's perspective that is an interesting choice of terms.
Is a collision a form of relative "acceleration" or is it direct "deceleration"?

Last edited: Apr 8, 2020
20. ### HalcRegistered Senior Member

Messages:
350
Acceleration is not a relative term. It means change in velocity. The rate of acceleration (which is what I'm talking about here) is the change in velocity over time. Phyti's description involves being hit by the ground causing perhaps 200 g's or more of acceleration rate, which is fatal.

exchemist likes this.
21. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
I don't understand the model you designed too much. If you can give some hints, I might understand.

22. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
https://photos.app.goo.gl/axeDPZggPZAjComR7
Assume that the earth speed v0 = 0. Star began to approach the earth at the speed of v. Star came to position B, and the elapsed time was T seconds. The gravitational changes experienced by the earth are as follows:
A: F = G * M * m / R^2
B: F = G * M * m / (R + 1000)^2 ≈ G * M * m / R^2

According to the law of conservation of momentum:
F * T = mv1-mv0 = mv1, then v1 = F * T / m = T * G * M / R^2
T = 2s: v1 = 2 * G * M / R^2
T = 1s: v1 = 1 * G * M / R^2

Because of the speed v of the star, the earth obtained a different speed v1. The larger v is, the larger v1 will be, but the gravitational force on the planet from the star has hardly changed. If the star's v = X * cos(wt), then the speed change of earth will also show volatility.
The relative speed between them really affects the speed of the earth. Do you still think there is no Doppler effect between them? If you understand the Doppler effect, you will know that it exists anywhere.

23. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

Messages:
852
https://photos.app.goo.gl/FNohkKDepHhgx2b29
To simplify the calculation, we do not consider the displacement in the y direction.
F = G*M*m/L^2=G*M*m/(R - r*cos(w*t))^2
F * t = mv1-mv0, assume v0=0 then F * t = mv1, then v1 = F * t / m = t*G*M/(R - r*cos(w*t))^2, then v1 = g(t)*t, g(t) = G*M/(R - r*cos(w*t))^2.
if R>100r, v1 ≈ t*G*M/R^2 = g*t , g = G*M/R^2.

The volatility of the gravitational field is obvious. When the celestial body is relatively far away, the g between them is close to a constant, but when they are relatively close, g is a function of relative speed and time.

post # 119 has shown that when relative motions between celestial bodies are on the same straight line, the speed change of celestial bodies with mass m is related to the relative speed between them. If the relative speed change shows volatility, then the speed change value v1 will also show volatility. Then the gravitational acceleration g (t) will also show volatility.

No matter what gravity is, the affected object will show volatility, it also has a Doppler effect.