# The effect of the Doppler effect on planetary orbits

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

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3. ### DaveC426913Valued Senior Member

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Tony, what produces this Doppler Effect of yours?

The sun's gravitational field is constant. There are no ripples in it, thus nothing for the Doppler Effect to affect.

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

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Mercury's precession deviation, my data is consistent with GR. I do n’t know what the deviation of the earth is, can you help me see it?

Is this the formula for all planets? 0.383 / (R / Rearth * days / 365)

7. ### DaveC426913Valued Senior Member

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Where does this Doppler Effect come from? What gravitational waves are you talking about? The sun doesn't emit any; its field is constant.

Gravitational waves occur when there is a change in gravity - such as merging black holes.

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

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The sun has a gravitational field, and moving in this field will produce a Doppler effect.

Is this the formula for all planets? 0.383 / (R / Rearth * days / 365)
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node116.html

9. ### exchemistValued Senior Member

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There are no waves in a static gravitational field.

No waves -> no Doppler effect.

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

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I have calculated that the precession of Mercury is 40.3 " per century and the precession of the earth is 1.85 " per century under the Doppler effect.

I checked the information, the calculation result of GR is 44.1 "and 3.85"

11. ### exchemistValued Senior Member

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12,235
No waves, no Doppler effect.

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

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Let me improve the accuracy of the earth's radius and see if the result is consistent with GR.

the precession of the earth is 1.88 "
Venus is 0.85"

Last edited: Apr 3, 2020
13. ### exchemistValued Senior Member

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12,235
No waves, no Doppler effect.

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

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My calculated data:
Mercury: 40.2 "......................GR: 41.0" .....................Observed: 43 "
Venus: 0.85 ".........................GR: 8.64"
Earth: 1.92 "..........................GR: 3.85" ......................Observed: 5.0 "
Mars: 8 "...............................GR:

15. ### DaveC426913Valued Senior Member

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18,689
No waves, no Doppler effect.

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

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852
I have given the data, and use the data to speak.

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

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This data will determine whether my theory and calculations are correct, because it is very different from the GR calculation results.

18. ### exchemistValued Senior Member

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12,235
No waves, no Doppler effect.

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

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852
Yesterday I learned how to use GR to calculate the precession of a planet: GR: 0.0383 / (days / 365 * R_planet / R_earth)
Compare my data with GR data
......R ......................e...........................Mine................GR
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"

If the eccentricity is small, the Doppler effect will be relatively small without considering the rotation of the celestial body. I haven't studied how much rotation has contributed to precession. I don't want to delve into this. Hahaha, my purpose here seems not to be for this GR, but for SR.
But it is true that this time I learned a lot of astrophysics. Thanks for the scientific forum. And those good brothers who continue to give me motivation.

The light is relatively simple, without considering its rotation and its quality. We only need to consider the strength of the gravitational field, including the strength brought by R and the strength brought by relative motion. I need to focus on the bending angle of the light. After considering the Doppler effect, the calculated bending angle is compared with the bending angle without considering the Doppler effect.

Last edited: Apr 4, 2020
20. ### exchemistValued Senior Member

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You are quite plainly incapable of explaining your supposed idea about Doppler effects. You have been given apple opportunity to do so on two threads now. The fact you can't address this, even though it is the very title of this thread, speaks volumes.

I think it's time to report this now and suggest the thread be moved to Pseudoscience.

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

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852
I think I have made a lot of explanations. My program also verified my assumption. However, the calculated results of the celestial bodies do deviate from the observed values. I just didn't consider their rotation. I disagree with you.

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

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https://photos.app.goo.gl/LXPNvN1ANViDVpoE8
In this running competition, I got two different answers from other special relativists.
1. It is always unclear.
2. They reach the platform at the same time.
He said: "As A.T. points out, observers at rest or only moving parallel to y in the Earth frame (such as Earth and your platform) will regard this race as a draw. Observers with any motion in the x direction in the Earth frame (such as Newton and Einstein in your example) will get different results for who won. It isn't clear from what you've said which frame your astronomer was using. He may have made a mistake, or interpreted your question differently, or he may be using a different frame from the one you think he is. "

I have got at least 5 different answers. Is there a standard answer for special relativity?

According to your ideas, all scholars on the physicsforum all must be moved to Pseudoscience. You can try to express your opinion about SR there. Good luck.

exchemist, if you can build a physical model to make the celestial body precession close to the observation result. I will admire your work. But I have never seen any your research in this area.

Mercury:
Mine:40.4 "......GR: 42.93"
In the calculation of the precession of celestial motion under a large eccentricity, my theory has already shown correctness. This is a research direction, and it is worth our in-depth study.

Last edited: Apr 4, 2020
23. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

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I do n’t know if you have ever thought about it, the precession calculation formula given by Einstein:
GR: 0.0383 / (days / 365 * R_planet / R_earth)
From this formula, the precession is only related to the planet's orbital radius and period. The orbit radius and eccentricity determine the period. So the last thing that has to do with precession is R and e.
The equation given by Einstein is a linear relational equation, which looks very simple, in fact, it is also very simple when calculating.

My calculation focused on e, and Einstein's equation focused on cycle, which is why our final results are different. According to my derivation, if e = 0, then there will be no precession, the greater the e, the greater the precession. However, GR, e = 0 or e is very large, and can only slightly change the planet's cycle, which has no effect on precession. According to the formula given by GR, even if e = 0, precession still exists. I don't know if this is the case, why will there be precession if e = 0. This is confusing.

Last edited: Apr 4, 2020