How does GR predict the Earth's orbit about the sun?

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Summary:

Is Earth's orbit a geodesic function of the Earth's and sun's mass?

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How does GR predict the Earth's orbit around the sun? Newtonian mechanics predict a gravitational force that is a function of the two masses. Is the geodesic path of the Earth a function of both the sun's and Earth's curvature of space?
 

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  • #2
phinds
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Summary: Is Earth's orbit a geodesic function of the Earth's and sun's mass?

How does GR predict the Earth's orbit around the sun? Newtonian mechanics predict a gravitational force that is a function of the two masses. Is the geodesic path of the Earth a function of both the sun's and Earth's curvature of space?
It is a function of the overall geometry(*) of spaceTIME (not just space) that results from these massive bodies.

I believe that the Earth/Sun system is simple enough that GR reduces to Newtonian physics and if it doesn't completely it certainly does out to quite a few decimal places.

* I originally said "curvature" here but as you apparently already know, it's not really curved. It's a straight line in space-time, using the proper math that described space-time. It's only if you insist on applying Euclidean Geometry to space-time that you get "curvature". In either case it's what Euclid would properly call an ellipse in shape it's just that it isn't caused by "curvature".
 
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haushofer
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In theory, yes. In calculational practice often not however; the 2-body problem in GR is, as I understand it, analytically not solvable. So you often treat the earth as a "test mass". Otherwise it wouldn't even follow a geodesic.
 
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Is the geodesic path of the Earth a function of both the sun's and Earth's curvature of space?
Unless we need excruciating accuracy, we only need to consider the mass of the Sun.

Planets of different mass will follow the same orbit as long as the mass of the planet is small compared with the mass of the sun.

Newtonian mechanics explains this by saying that the force needed to keep an object in orbit must be proportional to the mass of the object (easy to see this for circular orbits where the necessary force is ##mv^2/r##) and by a remarkable coincidence the gravitational force has that property.

GR explains this by saying that everything in free fall, regardless of mass, follows a geodesic path. In the vicinity of the sun, the geodesic paths are determined by the curvature produced by the sun’s mass and the mass of the free falling planet doesn’t affect the geodesic path it follows.
(This is the “earth as a test mass” approximation that @haushofer refers to above)
 
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phinds
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In the vicinity of the sun, the geodesic paths are determined by the curvature produced by the sun’s mass and the mass of the free falling planet doesn’t affect the geodesic path it follows.
Would that still be true if the planet were something like 10% to 20% of the mass of the sun or does it only apply if the planet's mass is small? Certainly at some point I would think that the planet's mass would have an effect on the overall geometry of space-time, no?
 
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Would that still be true if the planet were something like 10% to 20% of the mass of the sun or does it only apply if the planet's mass is small? Certainly at some point I would think that the planet's mass would have an effect on the overall geometry of space-time, no?
Correct. As I said in a thread long ago, if Galileo dropped a Jupiter mass BH from the Eiffel tower, it would hit the ground much faster than a rock (because of the earth moving up to the barycenter near the BH).

There is a practical numeric approach to the non-test mass two body problem (beyond post newtonian methods) called "effective one body methods":

https://arxiv.org/abs/gr-qc/9811091
 
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  • #7
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Would that still be true if the planet were something like 10% to 20% of the mass of the sun or does it only apply if the planet's mass is small? Certainly at some point I would think that the planet's mass would have an effect on the overall geometry of space-time, no?
Yes, that is right.

The planet-mass-small approximation shows up in the Newtonian solution when we assume that the central star doesn't move so that a circular orbit will be around the fixed point at the center of the star. We know that in fact the planet and the star are both orbiting their common center of gravity, and that's not quite at the center of the star - but even when the planet is Jupiter we generally don't care. Get up towards 10% or 20% and we start to care.

The analogous approximation in the GR case is to use the Schwarzschild solution, which assumes that all the mass is in a spherically symmetrical and stationary sun.
 
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Correct. As I said in a thread long ago, if Galileo dropped a Jupiter mass BH from the Eiffel tower, it would hit the ground much faster than a rock (because of the earth moving up to the barycenter near the BH).

There is a practical numeric approach to the non-test mass two body problem (beyond post newtonian methods) called "effective one body methods":

https://arxiv.org/abs/gr-qc/9811091
And here is a nice post on Stack Exchange that derives the relevant Newtonian equation. It leaves out the integration details, but I went through it once before, and it looks like I did a "##u##-substitution" with ##u = \sqrt{r}## followed by a trig sub with ##u = \sin \theta \sqrt{r_i} ##.
 

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