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How does one interpret the relative motion of an object in orbit as it compares to to the object it is orbiting?

In flat spacetime, it's pretty easy to determine relative motion. If Alice sees Bob as traveling at 100mph, and herself as stationary - Bob sees Alice as traveling at 100mps and himself as stationary. It's perfectly reciprocal.

How does this principle work with gravity introduced?

Assume Alice is at the center of gravity of a large, non-rotating planet, and Bob is in a perfectly circular orbit:

1.) Alice can tell the distance to Bob.

2.) Alice can verify she is not rotating, and at what speed she would need to be rotating to keep facing bob.

Using the above two observations, Alice determines bob is moving at 100mph.


Now, Bob is under the impression he is in an inertial frame. Does Bob view Alice as moving at 100mph relative to him? Or does Bob view Alice as stationary? Intuitively, it seems like Bob would view Alice as stationary.

Which is the correct way to think about this?

EDIT FOR CLARITY:

More precisely, the combined formula for time dilation based on gravity and relative speeds can be written as:

(1−2GM/rc2−v2/c2)^(-1/2)

which reduces to: (1−v2/c2)^(-1/2) when gravity is not present

and reduces to: (1−2GM/rc2)^(-1/2) when relative motion is not present

so when Alice and Bob are calculating the rate at which they observe each other's time as passing, what v (velocity) should Alice record for Bob, and vice versa, in order for the formula to accurately match observations.

Please directly answer this portion of the question before explaining "why", to avoid vague explanations without a clear answer of which velocities should be recorded in order for the formula to match observations.

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  • $\begingroup$ Just make a sketch , with the situation some small time apart, than you see th two relative motions. $\endgroup$
    – trula
    Sep 18, 2023 at 15:47
  • $\begingroup$ Trula, Please answer the question. If I "sketch" this out, Alice sees herself as not moving, and bob as moving 100mph. Bob does not see the reciprocal of this. He sees himself as not moving (because freefall is an inertial frame), and he also sees Alice as stationary (always distance "x" to his left, perhaps rotating in his frame of reference). Is this right, in terms of relativity (for relative time calculations) $\endgroup$
    – Spencer
    Sep 18, 2023 at 16:47

1 Answer 1

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Assuming that Bob is not an idiot then he sees the planet that he is orbiting.

Both Bob and Alice are moving inertially. Both Bob and Alice can choose a reference frame where they are at rest. However, neither Bob's nor Alice's reference frames are inertial. In the presence of spacetime curvature there are no inertial frames (except as a local approximation).

Due to the spacetime curvature Bob and Alice are not equivalent observers. There is no symmetry expected between the two. Both can distinguish the directions towards the center and away from the center, and both know who is in orbit around the planet and who is not. More to the point, both will correctly predict things like the redshift or blueshift of signals sent between them.

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  • $\begingroup$ Please see the edit above and add to your answer, what V should Alice use when formally calculating Bob's rate of time, and what V should Bob use for Alice when formally calculating Alice's rate of time? I cannot extrapolate this from your answer. (Since Alice has inferred Bob's speed to be 100mph based on: 1.) Alice can tell the distance to Bob. 2.) Alice can verify she is not rotating, and at what speed she would need to be rotating to keep facing Bob. ... Then can Bob also calculate Alice's velocity to be 100mph, 0mph, or some other answer in the above formula? ) $\endgroup$
    – Spencer
    Sep 18, 2023 at 22:49
  • $\begingroup$ I want to clarify your answer as well. You say both Bob and Alice are moving inertially however they are not in inertial frames of reference. Is it possible to be moving inertially but not be in an inertial frame of reference, and could you please refer to a source? "The fact that there are inertial frames is essentially an axiom of general relativity. The theory is based on the idea that spacetime has a certain geometric structure, which allows for the existence of geodesics, along which free particles travel." $\endgroup$
    – Spencer
    Sep 18, 2023 at 23:05
  • $\begingroup$ @Spencer the linked answer was clearly talking about local frames. This is what I referenced with “(except as a local approximation)”. Inertial frames are only local in curved spacetime, and the Bob-Alice scale is large enough to be non-local. $\endgroup$
    – Dale
    Sep 19, 2023 at 1:26
  • $\begingroup$ Are you unable to answer which value of "v" should be used as described above? Or does Bob "realize" he is moving and therefore calculates his own time as slower relative to Alice (plus or minus the gravitational effects)? $\endgroup$
    – Spencer
    Sep 20, 2023 at 20:27

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