Gravity question

[bigsplit]

I have a question about Gravity.

If you have three object A, B, C

A and B are stationary at x distance. C is moving away from A towards distance of object B at some fraction of c. When object C reaches the same distance from A as the stationary object B will it experience the same g from object A as object B does?

Why or Why not?

[grant hutchison]

I'm not sure what you're wanting to get at, here.
Your set-up seems to place object C inside object B, since C sets off from A towards B, and arrives at a distance from A that is the same as B's distance from A.
Your phrase "at some fraction of c" just implies that object C has a non-zero velocity, which is implicit in the rest of the question.

But, in general, when two objects are the same distance from a point mass, they'll experience the same gravitational acceleration as a result of its gravity. (But if our massive object is not spherically symmetrical, and therefore can't be treated as a point mass, then the accelerations may differ according to the positions of the two objects, even though they are at the same distance from its centre of mass.)

Grant Hutchison

[tony873004]

I have a question about Gravity.

If you have three object A, B, C

A and B are stationary at x distance. C is moving away from A towards distance of object B at some fraction of c. When object C reaches the same distance from A as the stationary object B will it experience the same g from object A as object B does?

Why or Why not?

No, while object C is in between A and B, itwill never experience the same g from A as it does from B.
g is a vector. In addition to magnitude, it has direction. A and B will always be pulling C in opposite directions when C is between A and B.

If you're allowed to treat g as a scalar, and you assume that A and B have equal mass (and as Grant points out, spherical or point mass), then the gs will be equal at the midpoint of the AB line, regardless of C's velocity.

[grant hutchison]

No, while object C is in between A and B, itwill never experience the same g from A as it does from B.That doesn't seem to be the question, however, unless I'm misreading the OP. What was asked was:When object C reaches the same distance from A as the stationary object B will it experience the same g from object A as object B does?(my bold)
This seems to be asking if objects B and C will experience the same acceleration due to A's gravity, when they are at the same distance from it. (As if B and C were of negligible mass, for instance.)

Grant Hutchison

[Laguna]

That doesn't seem to be the question, however, unless I'm misreading the OP. What was asked was:(my bold)
This seems to be asking if objects B and C will experience the same acceleration due to A's gravity, when they are at the same distance from it. (As if B and C were of negligible mass, for instance.)

Grant Hutchison
Thats how I understand it too...

[tony873004]

I guess I'm outvoted then :)

[bigsplit]

Sorry for any confusion.

The question has nothing to do with any B/C interface and lets assume B and C are the same mass and Spherical. Lets even assume we are comparing AB and AC in different frames.

I am asking, if due to the velocity of object C if the effects of gravity from object A weaken when compare to object A's effect on the stationary object B if they are at equal distances.

Note: I do know that the velocity of object C will increase it relative mass, I am trying to isolate the gravitational effects of the stationary object A only....if this is possible?

I know for a protons, there will be a red shift for C. Is there a gravitational "red shift" as well?

[John Mendenhall]

I do know that the velocity of object C will increase it relative mass, I am trying to isolate the gravitational effects of the stationary object A only....if this is possible?



You are asking then, "If an object C is moving at a high fraction of light speed away from another object A, then does the relativistic increase in the mass of C as seen by an observer at A cause an increase in the gravitational attraction between A and C, as compared to the gravitational attraction between A and B which are comoving?"

If that's the question, I defer to others. KenG, Publius, Grant, etc.

Good question, though.

[Ken G]

It appears the OP is asking, are there relativistic corrections to Newton's gravity based on the motion of the object. I don't know gravity well enough to answer that definitively, but I think the answer is 'it depends how you are treating the problem and what accuracy you are shooting for". You know, like the answer to all physics questions! It sounds like you are treating the problem as if A is magically held in place, and B is a test mass placed in free fall to measure g. But in a relativistic treatment, we don't get proper acceleration that way, we get spacetime curvature, so if you mention g it sounds like you might be talking Newton (hence Grant's answer). Then you introduce C moving relativistically and in free fall in A's gravity, and you are asking if the spacetime curvature it encounters will be the same as for experiments on the test mass B. Since gravity is nonlinear, that would probably only be true if C is also a test mass, so that might put a limit on how close to c it can get before it starts altering the spacetime curvature itself. Also, if A is spinning, I'll bet there is more than curvature going on, so that would also make C respond differently from B. But as all these seem like unnecessary complications, I'll go with "yes".

Maybe the real lesson in all this is-- tailor questions to the theory you want used, not to "reality", because all we can really do are "toy problems", and you might not be playing with the same toy.

[bigsplit]

Thanks for the replies...

I think the most important thing I learned and should have know by now is that there is no such thing as a simple astronomy question.

You are right, I was thinking more newtonian.... I wonder why?

I will ponder on this obvious ambiguity.

[publius]

The only good I can do about "object C moving at relativistic speed" is to say it gets complicated. :lol: Remember in GR we have a full rank-2 tensor source term, the stress-energy tensor, T. T_00, top left corner, is the energy density and that is the source of the Newtonian like part (Newtonian limit is basically all other T_ij terms go to zero. :)

So, in a frame where something is moving at high speed, the 'g' part there is going to increase with the relativistic looking mass. However, the high speed will add other big terms to the other T_ij. There will be a large gravitomagnetic field that will modify things and counter the 'g' when the test particles start falling toward the high speed source.

The gravity field of a relativistic mass would be a strange and complicated thing indeed. A test mass would fall strangely. But go into the rest frame of that mass, and you have a regular Schwarzschild field. There you explain the weird motion of the test masses by the higher order velocity dependent terms in the geodesic equations. That's something I'll stress again. Even in the weak field, relativistic test particles behave very differently from Newton. Newton is the weak field, *low velocity* limit. We sometimes forget about the low velocity part. Get test particles moving near light speed, and things get very non-Newtonian even for weak sources.

Actually, if gravity depending on the rest mass exactly, GEM would be nicely Lorentz invariant in the same manner as EM, and gravity could be a regular vector field. However, that would violate the Equivalence Principle. That was the doosey, and why gravity has to be a tensor field (and why Uncle Al's hair got so wild looking :lol:) . Only at least a rank-2 tensor field can do all the tricks to make that work out in all frames.

-Richard

[Ken G]

Thanks for clearing that up. :)

[mugaliens]

It appears the OP is asking, are there relativistic corrections to Newton's gravity based on the motion of the object. I don't know gravity well enough to answer that definitively, but I think the answer is 'it depends how you are treating the problem and what accuracy you are shooting for".

If I'm not mistaken, I believe the answer to the OP's question is: Yes, as an object increases it's speed, it's relativistic mass also increases, and as that happens, it's gravitational attraction on surrounding objects also increases.

[Ken G]

If I'm not mistaken, I believe the answer to the OP's question is: Yes, as an object increases it's speed, it's relativistic mass also increases, and as that happens, it's gravitational attraction on surrounding objects also increases.But the OP asks for g, and by the equivalence principle, that is independent of the mass of an infinitesmal test mass. But I think publius' remarks show that one simply can't combine language like "the g of gravity" with "relativistic motion", it's an inconsistent language.

[John Mendenhall]

Actually, if gravity depending on the rest mass exactly, GEM would be nicely Lorentz invariant in the same manner as EM, and gravity could be a regular vector field. However, that would violate the Equivalence Principle. That was the doosey, and why gravity has to be a tensor field (and why Uncle Al's hair got so wild looking :lol:) . Only at least a rank-2 tensor field can do all the tricks to make that work out in all frames.

-Richard



Thanks, Richard. Good post.