# The effect of the Doppler effect on planetary orbits

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

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

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852
According to the law of conservation of energy, I have improved the accuracy of planetary speed at any point on the elliptical orbit.

Planet -------- Perihelion precession ------------ Afar precession (per century)
------------------------------------------------------------------------------------------
Mercury********** 58.404"***************** 32.938"
Venus********** 38.004"****************** 38.466"
Earth********** 33.073"****************** 32.392"
Mars********** 27.609"****************** 24.686"
Jupiter********** 14.524"****************** 13.042"
Saturn********** 10.831"****************** 9.519"
Uranus********** 7.446"****************** 6.992"
Neptune********** 6.064"***************** 5.929"

( 58.404" + 32.938" )/2 = 45.671" , I still don't know what this average data means.

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

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852
We study a physical force situation, often by observing its speed change. Then under the Doppler effect of the gravitational field, the speed change of the object will also be different from the change under the ordinary gravitational field (non-Doppler effect).

I wrote the planetary orbit precession under the action of the Doppler effect of the gravitational field. I found that if I remove this effect, there will be no precession in the planetary orbit, and with the Doppler effect, you can see that there is precession.

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

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852
Precession under the Doppler effect of gravitational field:
Planet -------- Perihelion precession ------------ Afar precession (per century)
------------------------------------------------------------------------------------------
Mercury********** 58.404"***************** 32.938"
Venus********** 38.004"****************** 38.466"
Earth********** 33.073"****************** 32.392"
Mars********** 27.609"****************** 24.686"
Jupiter********** 14.524"****************** 13.042"
Saturn********** 10.831"****************** 9.519"
Uranus********** 7.446"****************** 6.992"
Neptune********** 6.064"***************** 5.929"

Precession under the normal gravitational field: (non-Doppler effect)
Planet -------- Perihelion precession ---- Afar precession per century
-----------------------------------------------------------------------
Mercury********** 0.000"****************** 0.000"
Venus********** 0.000"****************** 0.000"
Earth********** 0.000"****************** 0.000"
Mars********** 0.000"****************** 0.000"
Jupiter********** 0.000"****************** 0.000"
Saturn********** 0.000"****************** 0.000"
Uranus********** 0.000"****************** 0.000"
Neptune********** 0.000"****************** 0.000"

7. ### HalcRegistered Senior Member

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350
Mass cannot be either 'removed' or 'placed' in a g-field. It must be transported in from elsewhere.
Gravitational potential energy is negative, so a mass transported into the g-field of another object acquires negative GPE by doing so.
Energy conservation doesn't allow destruction of energy, so typically it gathers positive kinetic energy in balance.
It didn't accumulate energy. It just traded PE for KE, or as I worded it above, it acquired negative PE in equal quantity with positive KE. If there is impact, much of the KE is lost to heat, resulting in a negative overall energy state. Earth for instance has negative overall mechanical energy. That's why it's in orbit. If it had enough KE to balance its negative GPE, it would be moving at escape velocity.

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

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https://photos.app.goo.gl/Ck4zXdiVFiKzQQrV9
Assume M makes a linear oscillating motion between A and B. In order to simplify the calculation, assume the initial speed of m is 0.
M moves back and forth between A and B, then
delta E = m*G*M/(R - v(t)) - m*G*M/R = ½*m*v*v, v=sqrt(2*G*M)*sqrt(1/(R - vx(t)) - 1/R)
Delta g(t) = sqrt(2*G*M)*sqrt(1/(R - vx(t)) - 1/R) /t, t=r/vx(t)
Delta g(t) = vx(t)/r*sqrt(2*G*M)*sqrt(1/(R - vx(t)) - 1/R)

The amount of gravitational acceleration change delta g(t) of m will increase as vx(t) increases. When the distance between M and m becomes smaller, the direction of delta g(t) is toward M; when the distance becomes larger, the direction of g(t) is away from M.

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

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852
Planet -------- Perihelion precession ---- Afar precession per century under the Doppler effect
---------------------------------------------------------------------------------------------------
Mercury********** 58.404"****************** 32.938"*******meanArc= 43.053"
Venus********** 38.004"****************** 38.466"*******meanArc= 38.236"
Earth********** 33.073"****************** 32.392"*******meanArc= 32.726"
Mars********** 27.609"****************** 24.686"*******meanArc= 26.075"
Jupiter********** 14.524"****************** 13.042"*******meanArc= 13.747"
Saturn********** 10.831"****************** 9.519"*******meanArc= 10.138"
Uranus********** 7.446"****************** 6.992"*******meanArc= 7.209"
Neptune********** 6.064"****************** 5.929"*******meanArc= 5.996"

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

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852
Mercury********** 58.404"****************** 32.938"*******meanArc= 43.053"
Mercury Precession Deviation 43" comes from the Doppler effect of the gravitational field.

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

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852
https://photos.app.goo.gl/D4kgZ8dyzBmpjrux5

12. ### BellsStaff Member

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24,247
Mod Note

TonyYuan, please stop spamming the same links across several sub-forums and threads. If you persist, you will face further moderation.

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

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852
Ok, I see, thank you for reminding me.

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

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852
I have finished the paper on the Doppler effect of the gravitational field, and will submit the paper to nature, I hope good luck.

Thank you very much to every scholar who has followed me. The discussion with you has given me a lot of motivation and passion. I hope that we can continue. Thanks.

15. ### exchemistValued Senior Member

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That should be funny.

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16. ### phytiRegistered Senior Member

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732
halc#144
You are interpreting my words too literally.If a small test mass m is on the surface of a dominant mass M, there is a continuous acceleration mg of m, detected with a scale.If m is transported to a region of space empty of any M's, there is no force mg on m. The source of the energy accelerating m in the first case would appear to be M.
When the same mass m is released from a height, its KE=0.On impact its KE=mv^2/2 >0. It gained energy from the g-field. Thus the energy of the system of m and M remains constant.I don't interpret the g-field as negative energy.The boost in energy for space probes that pass near a large mass M is positive, and reduces the g-field of M.

17. ### HalcRegistered Senior Member

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350
Weight is force, not acceleration. Take away the m to get acceleration. Yes, the gravitational field of M is an acceleration field, which means anything in that field will accelerated at some fixed rate independent of its mass m.

Careful because energy is not a function of acceleration. Yes, m defines an amount of energy in terms of gravitational potential between various distances from M. Such energy is negative, so zero energy is defined as the energy of m with M not there at all, or at least indefinitely far away. As a object gains negative gravitational potential energy, per energy conservation, the system must gain some positive energy as well. A rock sitting on a surface of a planet does not gain or lose energy. It just sits there on the scale.

It gained no energy since the KE it gained are offset by the negative PE gained. Net gain is zero because gravity doesn't create energy.

Right. Just making sure we're on the same page with that.

But the probe goes into and out of the field of M when getting a boost from it. The energy gained by the probe didn't come from gravity, but rather KE stolen from M. Jupiter falls a little closer to the sun each time we slingshot a probe past it.

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18. ### phytiRegistered Senior Member

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732
halc#154
weight=f=ma=mgAcceleration is proportional to mass, thus the same speed for all objects.
It can be expressed as such with: ke=.5mv^2=.5m(at)^2It's expressed as negative potential energy, only because gravity accelerates toward the earth center, or downward. A pilot doing aerobatics may be inverted while pulling 3g, with acceleration directed upward. Direction is not absolute, but relative.The g-field disappears much closer than 'infinitely distant', as in an isolated two body system. Within a small sphere in remote space, there is em noise and gravitational noise that would obscure any effect from Earth. There is a longer history of isolating experiments from extraneous em noise.Gravity is always on. The mass on the surface is continuously receiving energy from the g-field, which is dispersed into the ground as heat.If it didn't, a scale between it and the ground would indicate it was weightless!
The field concept replaced 'action at a distance'. The dominant mass M could be considered a bank and the field a group of agents who hand out money to whoever passes by.

19. ### HalcRegistered Senior Member

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350
According to the equation you've chosen to illustrate your statement (f=ma), mass in inversely proportional to acceleration. For a given force, the larger the mass is, the lower its acceleration, which is why its easier to throw (accelerate) a pebble than it is to throw a car.
Then you say acceleration is proportional to mass, which would imply that a large rock dropped from a height would outrun a small one. It does not.

Kinetic energy can be expressed that way. Gravitational potential energy is not kinetic energy since it is a function of neither motion nor acceleration.

You make it sound like gravitational energy is taken from Jupiter during a flyby. If the gravity of Jupiter is the same after the probe has left the system, then this is a violation of energy conservation. If you claim the gravity 'bank' is reduced by the flyby, then the gravity of Jupiter is a function of how much 'gravity fuel' it has remaining in its 'bank', and not a function of its mass. One could exhaust the gravity of any object by having enough objects fly by it, leaving a massive planet with no gravity left, having handed it all out to these passing objects that are no longer present.

20. ### phytiRegistered Senior Member

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732
halc#156
If a composite object contains 10 units of mass, and each unit is accelerated simultaneously at g, then the object is accelerated at g, with a speed of gt. If a composite object contains 5 units of mass, and each unit is accelerated simultaneously at g, then the object is accelerated at g, with a speed of gt.All objects fall at the same rate (ignoring atmosphere) independent of total mass.

Agree it is potential energy, a function of position. When an object is in the g-field it receives energy from it, and acquires KE (no action at a distance).

Yes, if it leaves with more KE than when it entered (energy conservation). "The 'gravity fuel' is its mass. Considering the weakness of gravity, it would require an enormous number of slingshot masses to reduce the dominant mass.

21. ### HalcRegistered Senior Member

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350
I'm glad we agree on that, since before you said otherwise: "Acceleration is proportional to mass" (155)

gt (where g I presume is acceleration specifically due to gravity) computes delta-V, or change in velocity. It does not work out to speed.
My example, as always, is the ISS, which after one second of acceleration at 1g is moving not at speed 8.9 m/sec, but rather a significant 7666 m/sec, same speed as it was moving one second ago. But its velocity has changed by 8.9 m/sec, not its speed.
Acceleration, velocity, delta-V and momentum are all vectors. Speed and energy are scalars. Acceleration and delta-V are absolute, whereas speed, velocity, momentum and kinetic energy are all frame dependent. Be careful mixing terms on different sides of those lists, because you end up with ambiguous or meaningless statements like that.
For instance, speed of light is constant in any frame, but velocity of light is not, not even the velocity of a given photon.

It does no such thing. The ISS (again) is in Earth's g-field yet it gets no energy from that. Even falling straight down doesn't gain energy. It just adds to both negative and positive energy, but the sum of the two is still the same as it was before. Gravity is not energy. Gravitational waves are, but that's something else, and I know of no way to harvest it.

A probe flying past Jupiter briefly acquires KE, but it loses every bit of that as it moves away again. The KE it gains permanently does not come from gravity, but rather KE (not gravitational energy) taken from Jupiter, and it only takes this energy in some frames of reference. It actually passes net KE to Jupiter in different frames, as indicated above where energy is listed on the frame-dependent list.

You seem to assert here that Jupiter loses mass whenever something goes by it like that. This is wrong. Jupiter loses only kinetic energy, not mass, and it only loses that energy in some frames, and gains it in others, so the energy isn't even lost in any absolute sense.

22. ### phytiRegistered Senior Member

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Halc#158;
It seems difficult to form a statement that someone will misinterpret or read things into it that aren't there.

The example of the two objects has no initial velocity components. When they are released at a designated height, the acceleration vector is pointing toward earth center.

A different scenario, with the ISS velocity having 2 components, 9.8 m/s vertical and 7666 m/s horizontal, which balances out to stable orbit.

How does it gain speed as NASA claims? How does the KE transfer from Jupiter to the probe?
Energy and mass are transformable, E=mc^2.

23. ### Janus58Valued Senior Member

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2,388
Simplified scenario:
Probe is launched from Earth into elliptical orbit that reaches Jupiter orbit. It arrives moving at a velocity less than Jupiter's but ahead of it. Let's say that it is moving at 8 km/sec or 5km/sec slower than Jupiter relative to the Sun.* This means it starts with a 5/km sec velocity in Jupiter's direction.

Jupiter's gravity pulls it towards it even faster, and it goes into a parabolic orbit around Jupiter.** This means it falls in towards Jupiter, whips around it and heads back the way it came from. It returns to its starting distance from Jupiter, with the same difference in speed 5 km/sec, but heading away from Jupiter in the same direction as Jupiter is moving away relative to the Sun. So before it started falling towards Jupiter, it was moving at 13-5 = 8km/sec relative to the Sun, and after whipping around Jupiter, it is moving at 13+5 = 18 km/sec relative to the Sun. The flip side of this is that Jupiter also "falls'' toward the probe, and performs a "mirrored" trajectory, which ends up subtracting a bit of its orbital velocity relative to the Sun. There is a momentum exchange between Jupiter and the Probe. But since Jupiter out masses the probe by so much, the change in Jupiter's velocity is infinitesimal.

* These values are just for illustration purposes and are not meant to represent the real numbers involved in an actual encounter.

** Again, over simplified. The actual trajectory would be hyperbolic, but this doesn't change the basic principle.

In essence, it is like an elastic collision where one object comes from behind another slower object, bumps into it, increasing the other object's velocity while giving up some of its own. Just instead of physical contact, gravitational interaction provides the means of momentum exchange.

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