The effect of the Doppler effect on planetary orbits

[exchemist]

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.

Nice explanation.

 

[phyti]

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.

The question is how the probe acquires the energy to speed up. It can only get it from the g-field of Jupiter, since action at a distance is not acceptable in current physics. If the g-field results from the mass of Jupiter, then Jupiter loses the energy that the probe gains.

 

[Halc]

The question is how the probe acquires the energy to speed up. It can only get it from the g-field of Jupiter, since action at a distance is not acceptable in current physics.

Yes, the gravity of Jupiter is the means of momentum transfer in this case. It does not provide energy since it isn't an energy field. Momentum is transferred between the two objects, which may or may not involve an energy transfer. This is no different than a collision between a very small and very large billiard ball with no gravity involved at all. Assuming no energy is wasted in friction, the end numbers are the same.

If the g-field results from the mass of Jupiter, then Jupiter loses the energy that the probe gains.

Yes, in a frame where the probe gains energy, Jupiter must lose it equally. So in the frame of the solar system for instance, this is what happens. Momentum is transferred to the probe, and Jupiter loses some, and loses corresponding orbital energy (a combination of negative gravitational potential and kinetic energy). Thus Jupiter is slowed by the exchange and drops into a lower orbit.

For an example with no energy transfer, consider the same exchange from the point of view of Jupiter's frame. The probe comes in a 5 km/sec and leaves at 5 km/sec in a different direction. No difference in speed, and thus no net energy transfer, just a momentum transfer. This illustrates that kinetic energy is frame dependent, not an absolute amount of energy that an object has. In the frame of the solar system, the probe has speed perhaps 10 km/sec before and 18 km/sec after, a gain of energy, and Jupiter loses energy in that frame. In the exit frame of the probe, the probe has speed 8 km/sec before and is stationary after the encounter, and thus the probe loses energy and Jupiter gains it in that frame.

Regardless of frame, the encounter has no effect on Jupiter's gravitational field, which is based on its mass, which doesn't change due to the encounter. Its field is the same both before and after the encounter.

 

[TonyYuan]

I have created a new thread and uploaded my paper to the forum. I made a detailed description and proof of the theory I proposed.
http://www.sciforums.com/threads/doppler-effect-of-gravitational-field.163203/

 

[TonyYuan]

The 43" we know is the precession of Mercury's perihelion or the precession of Mercury's elliptical orbital axis?

According to my theory, Mercury’s perihelion precession is 58.404", aphelion precession is 32.938", and the ellipse axis precession is 43.05"
Does anyone know? This will determine whether my theory is correct or GR is correct.

https://photos.app.goo.gl/ziioYb7pAFx72tNH6

 

[TonyYuan]

Planet precession----- Perihelion ------- Aphelion ------- Ellipse axis (per century) under the Doppler effect
------------------------------------------------------------------------------------------------------------------------------
Mercury 58.404" | 32.938" | 43.053"
Venus 38.004" | 38.466" | 38.236"
Earth 33.073" | 32.392" | 32.726"
Mars 27.609" | 24.686" | 26.075"
Jupiter 14.524" | 13.042" | 13.747"
Saturn 10.831" | 9.519" | 10.138"
Uranus 7.446" | 6.992" | 7.209"
Neptune 6.064" | 5.929" | 5.996"

 

[James R]

TonyYuan:

Are the perihelion precession figures from your simulation the total precession angle per century, or only the amount of precession due to a single effect?

 

[TonyYuan]

TonyYuan:

Are the perihelion precession figures from your simulation the total precession angle per century, or only the amount of precession due to a single effect?

only the amount of precession due to a single effect.
43.05" is only the contribution of the Doppler effect to Mercury's precession.

From Google:
The calculation of Mercury's perihelion precession is actually based on Newton's law. The result is a precession of 5,557.62" per century, 90% of which is caused by the precession of the coordinate system, and the rest are caused by other planets, especially Venus. It is caused by the perturbation of the Earth and Jupiter; and the actual observation value is =5,600.73", the subtraction of the two is 43.11" per century.

GR calculated this 43". My theory also calculated this value, which comes from the contribution of the Doppler effect of the gravitational field.

 

[Write4U]

GR calculated this 43". My theory also calculated this value, which comes from the contribution of the Doppler effect of the gravitational field.

question: are you saying that spacetime itself is compressed or expanded inside a gravitational field, depending on its proximity to the gravity well? Does a wave-length shrink or expand depending on it's coordinate in a gravitational well?

Is this relevant to the conversation?
Measuring the Tidal Force on a Particle’s Matter Wave
Matt Jaffe , Department of Physics, University of California, Berkeley, CA 94720, USA

Figure 1:Kasevich and colleagues [1] have measured the effect of the tidal force, and thus of spacetime curvature (illustrated in blue), on the wave function of individual particles using a “sensor” atom interferometer (green) and a “reference” atom interfero...Show more

The tidal force of gravity, the focus of Kasevich and colleagues’ study, is the second spatial derivative of the gravitational potential. It is closely linked to the curvature of spacetime, the quantity that in general relativity describes the gravitational field.

The gravitational potential itself is unmeasurable, while its first spatial derivative, the acceleration of free fall, is as much a function of the observer’s state of acceleration as it is a function of the gravitational field. For example, operating an atom interferometer in an accelerated frame of reference is the same as operating it in a gravitational field.

The curvature, however, is a function only of the gravitational field and cannot be eliminated by going to a different frame of reference. Therefore, the curvature describes the gravitational field itself. Unlike phase shifts on a particle’s wave function arising from local acceleration, curvature-induced phase shifts may be described as an effect resulting purely from the influence of gravitation on a quantum system.

https://physics.aps.org/articles/v10/47

 

[TonyYuan]

are you saying that spacetime itself is compressed or expanded inside a gravitational field, depending on its proximity to the gravity well? Does a wave-length shrink or expand depending on it's coordinate in a gravitational well?

My physical model is the Newtonian model of universal gravitation, but Newton believed that the speed of the gravitational field is infinite, and the force on the object is instantaneous. We can think that the gravity described by Newton is a static gravity.
In fact, the speed of the gravitational field is equal to the speed of light c, and so there will be a chasing effect between the gravitational field and the object.
For example: (Gravity source as a reference)
When the velocity of the object in the direction of the gravitational field is zero, the gravitational force on the object is Newtonian gravity.
When the velocity of the object in the direction of the gravitational field is c, the gravitational force on the object is 0.

GR describes gravity from spacetime curvature, which is a different gravity model. As we know, the planetary orbit calculated by GR is more accurate. This is because the GR model considers the influence of the object's speed on gravity, and Newton ignores this influence, because the speed of the object is generally much smaller than c, so this influence can always be ignored. But it is reflected when accurate orbit calculations are required.

I just improved the Newtonian gravity equation and let it consider the influence of speed on gravity. And it has been verified on the Mercury precession.
F(v) = G*M*m/R^2 * (c-v)/c. From the formal point of view, it is a Doppler effect, and its essence is the chasing effect between objects.

 

[TonyYuan]

I have a question: How did the result 0.41" obtained by the GR calculation become 0.43"? This requires precise calculations, but the document does not give such important instructions. Does anyone know how 0.02" is calculated? Is this an estimate? So how is it estimated? Why is it not 0.025" or 0.03"?

I hope a GR scholar can help us answer this question, thank you.
@Jamas, can you invite GR scholars on the scientific forum to answer this question?

Reference link:
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node116.html

 

[TonyYuan]

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html
Table 1:Data for the major planets in the Solar System, giving the planetary mass relative to that of the Sun, the orbital period in years, and the mean orbital radius relative to that of the Earth.

Obviously, we cannot see the "perihelion" information from the calculation of GR. Why is the precession of Mercury's perihelion calculated by GR?
I want to know how GR calculates the aphelion precession of Mercury. Very interesting idea.

 

[TonyYuan]

From Table 2 in http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node115.html

I noticed such a problem: Except for Mercury's precession deviation: 5.75-5.50, which is closer to the GR calculation of 0.41, the precession deviations of other planets are very different from the GR calculation results. such as:
planet *****obs *****th ***obs-th ***GR: 0.0383/(RT)
Mercury ***5.75 ***5.50 ***0.25 ***0.41
Venus ***2.04 ***10.75 ***-10.71 ***0.086
Earth ***11.45 ***11.87 ***-0.42 ***0.0383
Mars ***16.28 ***17.60 *** -1.32 ***0.0134
Jupiter ***6.55 *** 7.42 ***-0.78 ***0.00062
Saturn ***19.50 ***18.36 ***1.14 ***0.000136
Uranus ***3.34 *** 2.72 ***0.62 ***0.0000237
Neptune ***0.36 *** 0.65 ***0.29 ***0.0000077

GR analyzed the data of Mercury, th 5.50 needs to be revised to 5.32, and 0.41 needs to be revised to 0.43. Does that mean that other planets also need to be revised? Or does it mean that other planets cannot be calculated by GR's formula 0.0383/(RT)?

Looking forward to the reply of GR scholars. Thanks.

 

[James R]

My theory also calculated this value,

It didn't, really, though, did it? What you have done is to do a simulation using finite time steps in a computer, and you have tweaked the settings until you got close to the "required" value.

 

[James R]

I have a question: How did the result 0.41" obtained by the GR calculation become 0.43"? This requires precise calculations, but the document does not give such important instructions. Does anyone know how 0.02" is calculated? Is this an estimate? So how is it estimated? Why is it not 0.025" or 0.03"?

Didn't you read your own link?

If the above calculation is carried out sightly more accurately, taking the eccentricity of Mercury's orbit into account, then the general relativistic contribution to

becomes

arc seconds per year.​

I hope a GR scholar can help us answer this question, thank you.

No need. It has already been answered.

@Jamas, can you invite GR scholars on the scientific forum to answer this question?

We already have several members who have studied GR. Others are very welcome to join, if they wish.

 

[TonyYuan]

It didn't, really, though, did it? What you have done is to do a simulation using finite time steps in a computer, and you have tweaked the settings until you got close to the "required" value.

My calculation does not have the process you mentioned. It would be meaningless if the conditions were gradually added to the result.

My calculation is based on this gravity equation: F = G*M*m/R^2 * (c-v) /c , and mathematical geometry. In my paper, I have given detailed calculation steps.

 

[TonyYuan]

Didn't you read your own link?
If the above calculation is carried out sightly more accurately, taking the eccentricity of Mercury's orbit into account, then the general relativistic contribution to

becomes

arc seconds per year.

This is just an explanation, and no detailed calculation steps are given.

planet *****obs *****th ***obs-th ***GR: 0.0383/(RT)
Mercury ***5.75 ***5.50 ***0.25 ***0.41
Venus ***2.04 ***10.75 ***-10.71 ***0.086
Earth ***11.45 ***11.87 ***-0.42 ***0.0383
Mars ***16.28 ***17.60 *** -1.32 ***0.0134
Jupiter ***6.55 *** 7.42 ***-0.78 ***0.00062
Saturn ***19.50 ***18.36 ***1.14 ***0.000136
Uranus ***3.34 *** 2.72 ***0.62 ***0.0000237
Neptune ***0.36 *** 0.65 ***0.29 ***0.0000077

Venus, Earth, Mars. . . GR's data is unconvincing and the deviation is huge.

Welcome more GR scholars to join.

 

[TonyYuan]

This is just an explanation, and no detailed calculation steps are given.

The accurate general relativistic precession formula is given in Exercise 2 of http://farside.ph.utexas.edu/teaching/celestial/Celestial/node96.html
For Mercury :
a = 57909711000;
P = 365.25*24*60*60;
T = 88*24*60*60;
e = 0.20563069;
det_w = 24*Pi*Pi*Pi*a/c*a/c/(T*T*(1-e*e))*(P*100/T)*(180/Pi);
printf("w=%5.10lf \n",3600*w);

The final result is:
1.Mercury det_w=42.9363728609" , we get it 43" by GR Accurate calculation，but you can see a = 57909711000, it is perihelion.
2.Venus det_w=8.5083106705"
3.Earth det_w=3.8389197296"
4.Mars det_w=1.1099908860"
5.Jupiter det_w=0.0564879393"

But this data is still very different from obs-th, how to understand this.

planet *****obs *****th ***obs-th ***GR: 0.0383/(RT)
Mercury ***5.75 ***5.50 ***0.25 ***0.41
Venus ***2.04 ***10.75 ***-10.71 ***0.086
Earth ***11.45 ***11.87 ***-0.42 ***0.0383
Mars ***16.28 ***17.60 *** -1.32 ***0.0134
Jupiter ***6.55 *** 7.42 ***-0.78 ***0.00062
Saturn ***19.50 ***18.36 ***1.14 ***0.000136
Uranus ***3.34 *** 2.72 ***0.62 ***0.0000237
Neptune ***0.36 *** 0.65 ***0.29 ***0.0000077

 

[TonyYuan]

we get it 43" by GR Accurate calculation，but you can see a = 57909711000, it is perihelion.

sorry, a is the length of semi-major axis, so here we can not say "it is perihelion".
I don't understand why GR considers the calculated result to be Mercury's perihelion precession rather than aphelion precession. Is it more appropriate to call axis precession?

 

[TonyYuan]

A schematic diagram of the influence of the Doppler effect on the planet's orbit just drawn for a professor. Share with scholars in the sciforums.
https://photos.app.goo.gl/4VVCDFXFrKA82RwU8

From the figure below, we can see that in different orbital regions, such as A, B, C, D, the change of gravity is different.
Area A: Mercury's gravity will increase.
Area B: Mercury's gravity will reduce.
Area C: Mercury's gravity will increase.
Area D: Mercury's gravity will reduce.