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Is force of gravity dependent on speed of one of the body? 
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#1
Mar2912, 10:32 PM

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Is there any such relation as per general theory of relativity?



#2
Mar2912, 11:13 PM

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However, for a moving body, there is no unique definition of a 'stationary' world line. Similarly, if you consider the gravitating mass to be stationary, and the test body moving, there is no unique definition of 'straight as if the body wasn't there'. However, no unique definition doesn't mean no possible definition. For most any plausible definition, the moving gravitating mass exerts more force on the stationary test body ([edit: produces higher peak proper acceleration of the 'stationary' world line]); and the 'straight' moving test body experiences larger proper acceleration (at closest approach) than a stationary test body. Unfortunately, the lack of uniqueness means you can't come up with universally agreed on numbers. Different definitions will lead to different amounts of increase in force [edit: peak proper acceleration of resisting world line] due to relative motion. 


#3
Mar3012, 01:44 AM

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Perhaps one could call this question is force of gravity dependent on speed of one of the body? a Category mistake.
In GR gravity is not a "force". Let's assume we have rather heavy bodies (like stars) and a test body (like a satellite). The energymomentum density T of the stars plus the metric g of spacetime manifold are defining a solution (g,T) of the Einstein field equations. The curvature of the manifold defines geodesics C along which test bodies will move. A different energymomentum density of the stars T' with different positions, different motion comes with a different solution g' for the metric, i.e. a different spacetime with (g',T'). The geodesics C' of this new spacetime are 'different' in a certain sense. But be careful: in principle you are not allowed to compare C and C' b/c they are defined w.r.t. different (g,T) and (g',T'). It's like saying that a Depp is less stupid than a Twompe; that's dangerous b/c both terms are defined w.r.t. to two different languages; Depp is German for Twit, Twompe is (Haitian) Creole. 


#4
Mar3012, 02:08 AM

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P: 337

Is force of gravity dependent on speed of one of the body?
One answer is that if you are moving your mass is greater so the grav force increases, but...(thought experiment) say you have a speed boat at rest, the wieght (due to GRAVITY) equals the force due to water pressure. But what if the boat is moving at speed v with respect to the ocean? The area is reduced by [itex]\gamma[/itex] because of length conrtraction, plus the mass increases and hence the weight of the boat increases...the conclusion is that the boat will sink....



#5
Mar3012, 02:15 AM

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P: 337

The answer is that ALL transvesre forces change under Lorentz tranformations, not just with gravity. You are in motion with respect to a grav body, your mass and hence weight increases, but your transverse force will 'reduce' at the same time your mass and wieght increase.



#6
Mar3012, 02:43 AM

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as I said, gravity is not a force!



#7
Mar3012, 02:49 AM

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Yeah but when the strengh of grav field, speed and masses are so small you got to conceed that such considerations are valid.



#8
Mar3012, 02:50 AM

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it's simple: when the velocity of the test particle changes, its geodesic changes as well



#9
Mar3012, 02:57 AM

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#10
Mar3012, 03:05 AM

P: 939

Also, you need to be more careful here because gravity does not couple to kinetic energy in the same way as it does to mass. It is not clear to me that your analysis is correct at all. Why would the boat driver see his boat sinking? 


#11
Mar3012, 03:25 AM

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P: 337

Length contraction (boat length) reduces and the force due to water pressure decreases and the weight of boat increases becauses the mass increases... factor of gamma^2 involved....this can only explained by transverse force changing...



#12
Mar3012, 04:02 AM

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#13
Mar3012, 04:17 AM

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sorry for the confusion, the formulation is missleading; what I want to say is that test masses at one spacetime point P starting in one direction but with different initial velocities v, v', v'', ... will follow different geodesics C, C', C'', ...



#14
Mar3012, 04:50 AM

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#15
Mar3012, 05:47 AM

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One way to answer this question is to look at things from the perspective of a parameterized postNewtonian (PPN) formalism. Here gravity is still a "force" but it isn't quite Newtonian gravity. In addition to good old F=GMm/r^{2}, there are some first order perturbation terms that result from general relativity in such a formalism. These perturbative effects on Newtonian gravity involve velocity. This approach leads to a very accurate description of the relativistic precession of Mercury. This is also how JPL, the Russian Academy of Sciences, and the Paris Observatory (the organizations that produce the three leading ephemerides) now model the solar system. 


#16
Mar3012, 06:25 AM

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#17
Mar3012, 06:47 AM

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I am not sure whether this is what you are looking for, but here's a paper discussing Einstein’s paper “Explanation of the Perihelion Motion of Mercury from General Relativity Theory”: http://www.wbabin.net/eeuro/vankov.pdf
You will recognize the simplest postNewtonian terms (for this special case). I am not sure if this is what you are looking for; I guess instead of "slow motion + strong gravitational field" you are interested in "fast motion + weak gravitational field"; anyway  the starting point is always the geodesic e.o.m. 


#18
Mar3012, 07:13 AM

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