Discussion in 'Alternative Theories' started by TonyYuan, Apr 2, 2020.
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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.
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.
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.
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.
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.
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"
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.
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.
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
Please Register or Log in to view the hidden image! Figure 1:Kasevich and colleagues  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
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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.
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?
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.
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.
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.
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 Please Register or Log in to view the hidden image! becomes Please Register or Log in to view the hidden image! arc seconds per year.
No need. It has already been answered.
We already have several members who have studied GR. Others are very welcome to join, if they wish.
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.
This is just an explanation, and no detailed calculation steps are given.
Venus, Earth, Mars. . . GR's data is unconvincing and the deviation is huge.
Welcome more GR scholars to join.
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);
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.
But this data is still very different from obs-th, how to understand this.
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?
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.
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.
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