# An interesting gravitational problem

Discussion in 'Physics & Math' started by TonyYuan, Feb 28, 2021.

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

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1. Two objects with masses M and m are located at positions A and B. The distance between them is 100 ls. What is the gravitational force between them?
--------------------A ------------------100 ls--------------------B
--------------------M---------------------------------------------m

Answer: F = GMm/R^2, R = 100 ls

2. If M moves to the left for 10s at a speed of 0.1c, and then stops at the position C, which is 101 ls apart, what is the gravitational force between them at this time?
-----------C ---------------------------101 ls--------------------B
-----------M-----------------------------------------------------m

3. What is the gravitational force between M and m after they stay still for 5s?

4. What is the gravitational force between M and m after they stay still for 1000s?

Last edited: Feb 28, 2021
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3. ### James RJust this guy, you know?Staff Member

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If you're worried about the finite propagation speed of gravitational waves, then using Newtonian gravity is probably not the best approach to use. You should probably be considering the scenario in the context of general relativity, instead.

However, from the point of view of Newtonian gravitation, if we consider the gravitational force on M due to m, for instance, then it just varies continuously according to the formula you gave:
$F=\frac{GMm}{r^2}$.

So, when the separation is $r=101$ ls, for instance, we just plug that into the force formula and that's the force value.

When the separation is $r=100.5$ ls, while M is moving, the force on A just adjusts continuously.

Think of it like this: M is sitting in a "static" gravitational field due to m. When M moves, the field of m doesn't change. The force on M is just $F=Mg$, where $g$ is the field strength at where $M$ is at the time.

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

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What I want to ask is what is the gravitational force on m?

2. If M moves to the left for 10s at a speed of 0.1c, and then stops at the position C, which is 101 ls apart, what is the gravitational force between them at this time? (What is the gravitational force on m?)
-----------C ---------------------------101 ls--------------------B
-----------M-----------------------------------------------------m

If you can’t give a clear formula, then under the premise that the speed of the gravitational field is limited c, for the gravitational force between any two objects, we cannot use F=GMm/R^2 to directly obtain the gravitational force. . Because you have no idea how long the relative position of the two objects has been maintained!

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

Messages:
526
1. Two objects with masses M and m are located at positions A and B. The distance between them is 100 ls. What is the gravitational force on m due to M?
--------------------A ------------------100 ls--------------------B
--------------------M---------------------------------------------m

Answer: F = GMm/R^2, R = 100 ls

2. If M moves to the left for 10s at a speed of 0.1c, and then stops at the position C, which is 101 ls apart, what is the gravitational force on m due to M at this time?
-----------C ---------------------------101 ls--------------------B
-----------M-----------------------------------------------------m

3. What is the gravitational force on m due to M after they keep still for 5s? (After M reaches the C position and stays still for 5 seconds)

4. What is the gravitational force on m due to M after they stay still for 1000s?

If you can’t give a clear formula, then under the premise that the speed of the gravitational field is limited c, for the gravitational force between any two objects, we cannot use F=GMm/R^2 to directly obtain the gravitational force. Because you have no idea how long the relative position of the two objects has been maintained!

Last edited: Feb 28, 2021
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8. ### James RJust this guy, you know?Staff Member

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I just answered you, in post #2.

The gravitational field doesn't have a speed. A field is spread throughout the whole of space. I gave you advice on how to think about it in post #2.

Why does that matter?

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

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You mean the speed of the gravitational field is infinite. The speed of gravitational waves is limited c. Don't you think it is contradictory?
The sun is flying, how does the gravitational field generated by the sun at 100ls follow the sun to fly?

Is it hard to say that the gravitational field is like many sticks inserted on the earth? The earth is like a hedgehog?

If the gravitational field is like a thorn on a hedgehog, then it doesn't matter. But the conclusion that the velocity of the gravitational wave is c is wrong.

Last edited: Feb 28, 2021
10. ### James RJust this guy, you know?Staff Member

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Just use $F=GMm/r^2$. You can calculate the numbers yourself.

Also, in your question #2, you give a range of times and then ask about what happens "at this time", which makes no sense at all.

No. I mean the gravitational field has no speed. i.e. I mean what I wrote.

No, because a gravitational wave is a "disturbance" in the gravitational field.

Think about water waves, as an analogy. You're sort of asking "What is the speed of the ocean?" The ocean doesn't have a speed; the water just sits there (obviously, ignoring currents and so on). But that doesn't mean that waves in the ocean have infinite speed, or no speed.

The field doesn't follow the sun. The field is spread throughout all of space. The sun just creates alters the field somewhat.

I don't know what that means.

The speed of gravitational waves has been measured to be c, so I guess that makes you wrong.

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

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Good.
There is such a scene: a child throws a stone into the sea, and the wave reaches the ship after 100s.
Then, before the waves reach the ship, can the ship feel the waves brought by the stones?

Now I only tell you that there are two objects with masses M and m, and the distance between them is 100 ls. Let you find the gravitational force between M and m. Do you think you can find it?
How do you judge that gravitational waves have reached each other?

Without the presence of waves, how can they feel each other?
As long as the sea exists, can boys on the coast of the United States feel that the boys on the coast of China are swimming?
Without the transfer of energy, how could it be possible to feel energy.

GR tells us that mass has caused the curvature of time and space. Does it take time for this curvature to affect objects at 100ls? What is the speed? Is it a gravitational wave?
The sea has no speed, but energy transfer requires speed. If you recognize this speed, then you need to care about when it starts and when it arrives.
Don't tell me that as long as the ocean exists, American boys and Chinese boys can feel each other.

Last edited: Mar 1, 2021
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12. ### TonyYuanGravitational Fields and Gravitational WavesRegistered Senior Member

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There is no chasing effect between the sun’s gravitational field and the earth, because the sun’s motion direction is approximately perpendicular to the earth’s motion direction, so this chasing effect is almost negligible, but if their speed directions are the same or opposite, then Will feel the obvious chasing effect.

The gravitational field exists statically, but the movement of the gravitational source will cause the generation of gravitational waves, leading to changes in the strength of the gravitational field. Therefore, a chasing effect must be formed between the gravitational field and the object.

my guesses:
If you want to launch satellites more easily on the earth, it will be easier to choose the opposite direction of the earth's revolution. This can reduce the gravity of the earth on the rocket.

Last edited: Mar 1, 2021
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13. ### James RJust this guy, you know?Staff Member

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No.
Yes. $F=GMm/r^2$, if you want the Newtonian force.
Gravitational waves don't come into the picture, if you only want the force between M and m.
In Newtonian physics: it's action at a distance.
In GR: they "feel each other" by sitting in a curved spacetime manifold. Specifically, each mass will follow a geodesic on the manifold, if gravity is the only "force".
In a quantum theory of gravity, virtual particles (gravitons) would presumably carry the relevant "messages".
Probably not. How is that relevant?
It isn't possible to feel energy.
Changes in curvature probably propagate at the speed of light. (I'm not an expert.) I guess you could say that the information about changes is carried by gravitational waves.

Is this "chasing effect" you keep talking about the same as the Doppler effect?

Okay. We seem to be in agreement on that.

I'm still not sure what a chasing effect is. Is it just the Doppler effect?

No. It's easier to launch them in the direction of the Earth's revolution, because that way they benefit from the surface speed of the Earth and require less fuel to reach orbit.

Look it up: rockets carrying satellites are typically launched in an Eastwards direction. They start by going straight upwards, then they turn eastwards.

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

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Yes，it is the Doppler effect.

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15. ### James RJust this guy, you know?Staff Member

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The Doppler effect applies to waves, not fields.

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

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Yes, I know.
I will give the influence of the gravitational waves caused by the movement of the sun on the orbit of the planet.
Now I must be busy with other things.

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17. ### DicartRegistered Senior Member

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The use of the newtonian force is correlated with the assumption that the force is established instantly.
Newton himself was wondering if he should use a finite speed for the interaction or not.

Therefore depending on the context, this approximation can be considered correct or not.
If we want best precision, we know this formula is always wrong and we have to take in account the low speed of light (yes the speed of ligt is very low compared to the distance in space).
We actualy suppose that "the gravitationnal wave" has the same speed as light (why is it so, thats a good question)

Therefore if m is moving, toward or outward of M or even not moving at all, the problem is exacty the same :
The gravitationnal field is dynamicaly evolving and the limit of the reequilibrating zone of the field (which is one field in the entire univers) is a sphere centered on the mass "that moved" (not only moved... because moving speed is a relative point of view, but centered on the mass who was affected by inertia), spreading at the speed of light.

There is no formula to say precisely how the curvature of space-time is at any point of space in real world.
We have to do simulation.
But we can also consider, like for newtonian force, that with some suffiscient precision we can assume that the gravitational field is not evolving any more so much if we got a statical (like the waiting 1000s example) or a repeated dynamical evolution of a system (like the moving of masses on a solar system)

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18. ### phytiRegistered Senior Member

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James R;

If the sun turned dark, when would we know it?

The g-field depends on the distribution of matter, per Relativity.
Since Relativity has eliminated action at a distance, based on a finite light speed, there is a delay in transmitting effects in space. Assuming M the dominate mass, as it moves, the surrounding space/field is altered (via an unknown process). The planets follow the sun through space, per its last direction of radiation.

If Tony has other thoughts, he will correct me.

f=GMm/(x+vt)^2

with x the initial distance at t=0, (x, t) relative to m.
If M stops, then (x+vt) becomes the new initial distance

Graphics usually help.

A constant distance separating m and M could be maintained if m orbited M. That would be a different scenario. No time lag since m would occupy a new position previously charged/conditioned by M, giving an AS IF simultaneous effect.

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

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If we compare the ocean to a gravitational field, compare waves to gravitational waves, and compare ships to the earth. When the ocean is not disturbed, the ocean current maintains a balanced state, and the water surface is calm. When a ship starts to travel, the water surface is disturbed to produce a wave with a certain speed, then can the speed of this wave represent the speed of the ocean current? Obviously it is not possible (it is also inappropriate to use gravitational wave speed to represent the speed of the gravitational field), ocean currents have their own speed, and wave speed will be superimposed on the speed of ocean currents. There will be a chasing effect between ships and waves, and a chasing effect between ships and ocean currents.

So can the gravitational field really be compared to the ocean? What is the "ocean current" velocity of the gravitational field? Can the speed of the gravitational field be calculated backward by the speed of the planet's orbit away? If this speed is limited, although it is very huge, as long as it is a limited speed, a weak chasing effect can be formed, and the universe will continue to accelerate expansion.

F=(GMm/R^2) * [ (x-v)/x ] , x is the Gravitational field velocity (ocean current velocity of gravitational field).

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

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Under this gravitational formula, the mechanical energy of the planet in its orbit will continue to increase, and the orbit will continue to accelerate and expand. If it expands to the entire universe, the chasing effect of the gravitational field will cause the accelerated expansion of the universe.

The essence of GR's space-time model is probably the chasing effect of the gravitational field.
Everything in nature is always very similar. If you understand the ships sailing in the ocean and the waves that are stirred up, you can imagine the planets flying in the gravitational field.

Last edited: Mar 3, 2021
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21. ### DicartRegistered Senior Member

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Not sure, because m and M are very small in comparison of the other mass...
If you want to do this conclusion, you must first formulate some hypothesis about the mechanism of the spreading of the gravitational field.
Per example, if you use the graviton hypothesis, what do this graviton ?
Can a single graviton produce this expansion ? How ? Is it absorbed like a photon ? Do the absorbsion of a graviton produce an other graviton ? Etc.

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

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I just applied the formula to the orbit calculation and found that the orbital energy is continuously increasing and the orbital is constantly expanding. So there is such a thinking.

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23. ### DicartRegistered Senior Member

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Yes but in GR because of the curvature of space-time total energy should decrease (unless some particular cases taken in account). According to Emilie Noether's theorem energy is not constant with time translation if space-time is not flat.
So if you find the reversed result, perhaps it is because you use newtonian formulas.
You can take a look on the demonstration here : https://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html

Also, i think (but i could be wrong) you can not affirm this :
"If it expands to the entire universe, the chasing effect of the gravitational field will cause the accelerated expansion of the universe."
In my opinion, a mathematical result is not enought to conclude that some physical phenomenom will occur, as soon as we go too far from the standard use of the formulas.
You need also to explain the physic that is behind the math.

Per example, and this is why i was telling about the graviton, what if the energy of a single graviton can not be divided by a volume superior to some limit ? Due per example by the plancks limit ?
So, you can not say a single graviton will affect the whole univers... more likely it will affect some limited portion of the univers.
Thats the difference between the math and the physic.

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