An interesting gravitational problem

Hypothesis 1: The speed of the gravitational fieldis equal to c;
Hypothesis 2: There is a chasing effect between the gravitational field and an object.

Derive the relationship between gravity and velocity based on the hypothesis
In a very short time slice dt, we can suppose that m is static, with a constant gravity on it. Then we accumulate the impulse generated by the gravity on each time slice and average them over the entire period to obtain an equivalent constant gravity to observe the relationship between equivalent gravity and velocity.

Let us study the effect of velocity on gravity when the velocity of an object m relative to M is not zero.
M<---------------------------------R---------------------------------------------->m
Figure 1 Gravity Model

As shown above, there are two objects of mass M and mass m and the distance between the two is R. mhas a velocity v relative to M. The velocity direction is shown on the straight lineconnected to each other. We can use F(t) = G*M*m/(R+v*t)^2 to represent the force on m at time t. It is obvious that Newton’s gravity equation is used here. In any short time dt, m can be regarded as stationary. The sum of the gravitational impulse on m over a certain period can be obtained by adding up the impulse dp which is the gravity on these very small time slices multiplied by the time.

Suppose after the time T, m will obtain a gravitational impulse p, so the gravity is subject to an integral in the time domain:
https://photos.app.goo.gl/s7jYQy1q9oMQ6wsV6

For the object m with a velocity v, the impulse p accumulated in the time T can be represented by an equivalent constant force multiplied by the time T. To represent briefly, we can use F(v) to represent this equivalent force.

Suppose there is a situation where: m has two different velocities v1 and v2, and their equivalent forces in T are respectively:
F(v1) = p1/T = (G*M*m/R^2)/(1+v1/R*T),
F(v2) = p2/T = (G*M*m/R^2)/(1+v2/R*T),
F(v1)/F(v2) = (1+v2/R*T)/(1+v1/R*T) = (R+v2*T)/ (R+v1*T).
Suppose F(v1) =K* (R+v2*T), F(v2) =K* (R+v1*T),
So reach: F(v1) - F(v2) = K*(v2-v1)*T.

It shows that the difference in equivalent forces F(v1) - F(v2) is directly proportional to the difference in velocities v2-v1. F(v) and v have a linear relationship and it is observed that this linear relationship is unrelated to the velocity magnitude of the gravitational field. That is, regardless of whether the real speed of the gravitational field is c or c-v, this linear relationship will always hold.

The above process can be explained by a visual example. Suppose a gun fires at a speed of n bullets per second, the bullet with a velocity V generates an impact force F on a static car. Now use this gun to shoot a car with a constant speed V and seek the bullet’s average impact force on the car in T.
Step1: Calculate the total displacement of the car s = T*V;
Step2: Calculate the total bullet number Nb= T*n;
Step3: Calculate the time when the car cannot be hit T'= s/Vb= T*V/ Vb;
Step4: Calculate the number of bullets that can hit the car Ns = n*(T – T') = T*n* (Vb– V)/ Vb;
Step5: Calculate the average impact force F(V) = F*[(Vb-V)/Vb].

We can also get the equation from the chasing effect between the bullet and the car and find the average impact force of a bullet is a function of its velocity V.
When V = 0, the car is static and the average impact force of a bullet is F.
When V = bullet’s velocity Vb, the average impact force of a bullet is 0.
The relative velocity between the bullet and the car will lead to different average bullet impact forces. Similarly, we suppose there is a chasing effect in the gravitational field and we can deduce the equivalent gravity by a similar method.

Now we can use velocity boundary conditions v=0 and v=c to obtain the expression of K to find the gravitational equation with v as the parameter.
Velocity boundary condition 1: When v2 = 0, F(v1) - F(0) = K*(0-v1)*T = -K*v1*T. Then F(0)=G*M*m/R^2，So reach:
F(v1) = G*M*m/R^2- K*v1*T， (a)

Velocity boundary condition 2: When v2 = c, according to the chasing effect between the two objects, when they are at the same velocity, effects of forces cannot occur. So F(c) = 0，
So F(v1) - F(c) = K*(c-v1)*T
So F(v1) = K*(c-v1)*T ， (b)
From equations (a), (b), we can obtain:
F(v1) = G*M*m/R^2- K*v1*T = K*(c-v1)*T, so K*c*T = G*M*m/R^2 and K =(G*M*m/R^2)/(c*T), substitute K into,
F(v1) = K*(c-v1)*T = [(G*M*m/R^2)/(c*T) ]* (c-v1) * T = (G*M*m/R^2)* ((c-v1)/c).
Sorting out to obtain: F(v) =( G*M*m/R^2) * [(c-v)/c]. What it shows is the relationship between gravity and mass, distance and velocity.

We can also use another thought to analyze it. After we know that F(v) is a linear relationship to v, as shown below, F(v) can be easily calculated based on boundary conditions F(0)=G*M*m/R^2 and F(c) = 0:
F(v) =F(0) + v* ( F(c)-F(0) )/c = F(0)*(c-v)/c = (G*M*m/R^2) * [(c-v)/c]。

https://photos.app.goo.gl/vYxfdXsucCRPn6Ju8
Figure 2 Linear Relationship between Force and Velocity

Under the above analysis, we can write out a universal gravitation formula with the velocity parameter v: F(v)= (G*M*m/R^2) * f(v) , f(v) = [(c-v)/c], Here G is a gravitation constant measured when M and m are relatively static. From the f(v), it is consistent with our hypothetical chasing effect. If we have to keep the form of Newton’s equation of gravity, we can write it in the form of:
F(v)= G(v)*M*m/R^2 and G(v) =G*[(c-v)/c], G= 6.67259×10N·m²/kg².

That is to say that the gravitation constant will turn into a function G(v) of v, and you can understand it as the gravitation constant is different when the relative velocity of the gravitational field to m is different.

 

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Tony;

I thought your point was a delay in adjustment of the g-field, resulting from the mass M moving. After reading this latest post, I'm confused.
The ocean waves occur in a 2D environment vs a 3D environment for space. The nature of water waves is transverse oscillation of the medium. With gravity, an hypothesized medium (structured space or field) is aware of a mass via a messenger entity (graviton?) radiated at light speed (all force carriers move at c).

There is no complete theory of gravity which explains how the energy of a mass is transferred to the surrounding space (field), which then alters the path of a test object as it passes by.

Curved spacetime is another figure of speech. In the case of gravitational lensing, the field remains unchanged while the light is redirected, in the form of a curved path.

(x-v)/x does not have the dimensions of velocity. What does (distance-speed) mean?

 

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Phyti：

There are two situations that must be distinguished:
1. The disturbance of the gravitational source causes gravitational waves.
2. The movement of the object leads to a chasing effect with the gravitational field.

These are two different scenarios. Just like ships in voyage, ships and ocean currents have a chasing effect. The waves coming from other places will also have a chasing effect with ships

"I thought your point was a delay in adjustment of the g-field, resulting from the mass M moving"
===>This is the first scenario I pointed out，there is a chasing effect between gravitational waves and objects.

 

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Tony;

In a real world problem: m is moving away from M at v.

The distance x between M and m is

x=d + vt - .5gt^2 (green path)

The g is based on M just as the current g is based on earth's M.

This is substituted for x in GMm/x^2.

When m is at the point corresponding to time t, the field is charged from a signal (blue) sent earlier at t=0. This presents the appearance/perception of instantaneous interaction.

A rough analogy is a mouse caught in a trap that was set the day before.
or
The covid19 vaccination is not from the manufacturer directly to the patient. The serum is dispersed from the source (M) to injection clinics (field) and given to the patient (m) when they arrive.

Gravity is NOT a direct interaction by contact.

You can also rotate the graphic 90° CCW and see m at the top of its trajectory near t, as it descends back to M.

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I probably know what you mean. We have the same view "I thought your point was a delay in adjustment of the g-field, resulting from the mass M moving. "

 

Tony;

I was attempting to back you up on the delay idea. I could not follow the 'chasing' idea.


I respect anyone who analyzes the advancement of Mercury's orbit!

That and gravitational time dilation never got my interest. Too much advanced math with results only in nano seconds.

 

I agree with your point of view, but the form we describe is different. The g-field delay and the chase effect describe the same thing.
But the speed of gravitational waves can never be regarded as the propagation speed of gravitational fields. Just like the speed of waves cannot be regarded as the speed of ocean currents.

 

James, you are right.

phyti, no " g-field delay".
If there is a g-field delay, it only takes a few thousand years for the earth to escape from the solar system.

The speed of gravitational waves cannot represent the speed of action of the gravitational field. But gravitational waves can cause planetary precession.

If the sun disappears, the gravity on the planet will disappear instantly.

 

Yet there will be light for approx. 8 min. No change of state can propagate faster than light, thus gravity will cease when the light goes out! The g-field is updated as the sun changes positions

 

phyti,

I have derived the exact gravity equation: https://photos.app.goo.gl/HAPFKhAGhyizbQvo6

The planetary precession data was calculated using this formula. https://photos.app.goo.gl/FxgrdAssiB15BTyK8

You can apply my equation to planetary orbit calculations, and you will find that it is so accurate.

 

Tony;

When M is at position 1, the g-field (black line) at position p is directed toward M.
When M is at position 2, the g-field at position p must be adjusted for direction by M. The graphic is not to scale. The motion of M would be much smaller than the light motion from M to p. I.e., like a weather vane, the elements of the field point toward M (with a tiny time lag). If they didn’t the acceleration would be toward a phantom position!

Newton thought light speed was instantaneous, thus gravitational effects were also. With the knowledge that light speed is finite, 'action at a distance' is obsolete.

I don’t question the accuracy of your calculations for orbits. My point is the g-field is conditioned by the dominant mass M before any object enters the field.

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Correction: r0 is approximately equal to 0.00042.

 

I have submitted Nature and I believe I have opened the door to science. Starting in 2021, physics will enter a new era.

 

If my article is published, I will go to your country to explain my theory in detail. It will surpass Newton and Einstein.
A very famous physicist has already read my paper. He said that as long as nature's editors don't reject my manuscript because of formatting issues, it will cause a sensation in the world.

 

I wish you success in this effort.

 

It is possible and i would be very happy if this could help the understanding of gravity.
I dident intervein i this discussion since, not because i think you are wrong, but because i am trying myself to find out how your idea is connected to the representation of gravity i have myself and that match what you say (altought it is very speculativ).

It is a very difficult work because it is some mixing of particular situation (that permit to say that instantaneous force can be take in acount) and some other speculation and i dont like to say wrong things (that could be confusing).

You remember i proposed you to add some "more" physical explaination that could reinforce your proposition, hopping you could eventualy come to the same conclusion.
So, here is my point of view and the physical explaination that can be eventualy added to your calculations as a physical explaination (that you can use freely, it is only some imagination tought and i dont need to be cited) :

In my opinion, gravity is not an attractiv force but exactly the opposit, it is the result of the expansion that occur at every scale, in addition with the inertia phenomenon.
Per example, for Earth, every atom (in exact proportion of mass, ) is expanding, so anything that is not attached to earth with covalent links is pushed away by this expansion.
But inertia (in exact proportion of mass again) hinder the expansion of this non attached matter => there is an illusion of attraction because matter inertia tend to let matter staying at place.
So the fundamental phenomenon in the universe is the expansion of the space-time.

So, if you say that gravity decrease a little as matter get away at some speed, i agree !

If this could help.

Furthemore, with this mecanismus, the matter itself appear because of the differential speed of a space-time portion relativ to an other space-time portion.

 

More accurate program simulation results: (The latter is Einstein's)
Mercury 43" ----------------------- 41"
Venus 240"------------------------ 8.6"
Earth 3" ----------------------------3.8"
Mars 1" -----------------------------1.3"
Jupiter 0.8" -----------------------0.06"
Saturn 0.1" ------------------------0.01"

Mercury ,Venus ,Earth ,Mars : delta t = 0.1s
Jupiter,Saturn:delta t = 1s

 

Which scholar can help me endorse account on arVix? (General Physics)
I want to submit my paper to arVix and share it with everyone.

Thanks.

phyti, can you help me?

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You must be bloody joking. Nobody is going to endorse this. They would look like idiots if they did.

 

GR calculated that the precession of Venus per century is 8.6" is wrong. The correct result is about 240".