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This question already has an answer here:

Alright guys, so I attracted you with the title. Now, to preface the question, I am fully aware of the work of Copernicus, as well as the concepts of Heliocentrism and Barycentric Coordinates. I have a master's degree in Engineering, and have taken many a course in dynamics and kinematics. It is with this experience that this question arises.

When studying bodies in motion, it is customary to pick a stationary reference frame for the basis of your calculations. Since the Sun is extremely massive compared to the rest of the planetary bodies, the barycenter of the Solar System (the stationary origin of choice) is extremely close to (and often inside) the sun. Hence why we commonly say that the planets all revolve around the Sun.

Although this certainly simplifies drawings of the planetary orbits, allowing for mostly non-intersecting ellipsoids in modeling, the base reference frame is, for all intents and purposes, arbitrary. All motion is relative to it's observer, so who's to say we cannot define the origin to be at the center of Earth? Sure, children would no longer be able to make working models out of hangers and Styrofoam balls, but wouldn't the equations of motion remain the same?

I have been searching online for a video, or gif, that illustrated this principle, but was unable to find anyone who took the time to do so. I'd be very interested in seeing what the orbits actually looked like if we re-defined the stationary reference frame from the barycenter to the center of the earth. I'm sure the orbits would look pretty rad! Much like 1:40 into this Solar System Orbit Video.

Is there something I'm missing that would theoretically preclude us from doing so?


Edit:

Ah hah! Finally found a video that shows this simulation.

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marked as duplicate by ACuriousMind, Martin, Kyle Kanos, Qmechanic Jun 11 '15 at 6:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/10933/2451 and links therein. $\endgroup$ – Qmechanic Jun 10 '15 at 16:33
  • $\begingroup$ In mechanics it makes things much simpler to evaluate your equations of motion on the center of mass. $\endgroup$ – ja72 Jun 10 '15 at 16:40
  • $\begingroup$ Isn't this exactly what Ptolemy did? His system gave better predictions than Copernicus did (because Copernicus thought orbits were circular). Just Google for descriptions of the Ptolemaic model. $\endgroup$ – John Rennie Jun 10 '15 at 17:05
  • $\begingroup$ Are we taking about fixing the reference frame to the center of someone's head? Or does the reference frame "follow" the center of earth? (The difference being one you would eliminate a lot of complications by eliminating the rotation of earth, all earths mass would be in orbit around the center) $\endgroup$ – user273872 Jun 10 '15 at 17:09
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    $\begingroup$ I think you miss the point: The whole Universe revolves around ME. $\endgroup$ – Hot Licks Jun 11 '15 at 2:20
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Back before Copernicus (Or rather, before his view was accepted), we used to think the earth was the center of our solar system.

Therefore, if you search for those models, you can find examples such as: Geocentric model

This is, of course, based on observations rather than calculations, but it represents the complication of the solution nonetheless. (Image taken from wikipedia)

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    $\begingroup$ Great image. Good find. I was unable to come up with any images of the sort. All those little loops explain the phenomenon of retrograde, no? $\endgroup$ – dberm22 Jun 10 '15 at 19:17
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    $\begingroup$ @dberm22: yes. You can think of the motion as the normal ellipses of the planets, with Earth's motion "subtracted out", so the result is circles within circles. $\endgroup$ – Jerry Schirmer Jun 10 '15 at 19:29
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    $\begingroup$ Also, I found a simulation of this. youtube.com/… Thanks for the lead. $\endgroup$ – dberm22 Jun 10 '15 at 19:32
  • $\begingroup$ Any reference for the image? I went looking for it and found a great deal of fascinating images, and learned a great deal; but I could not find this particular picture on Wikipedia. But the articles on geocentric model and history of astronomy are, dare I say, stellar. $\endgroup$ – tripleee Jun 11 '15 at 4:37
  • $\begingroup$ en.wikipedia.org/wiki/Deferent_and_epicycle - See the "History" part. (For future reference, you can search by image) $\endgroup$ – Omry Jun 11 '15 at 4:39
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You certainly could define your origin of coordinates to be the center of the Earth. It would be a little tricky, because this would no longer be an inertial frame of reference, so there would be fictitious forces (or Coriolis forces). That is, your equations of motion would no longer look the same.

One reason the standard barycenter frame of reference is useful is because there are no external forces acting on our Solar System. (Well, there are, which is why we orbit the galaxy for example, but we can usually ignore such forces.) And because there's no net force, there's no net acceleration, which means that the barycenter frame is inertial.

But it's not a problem to transform your coordinates. For example, if $\vec{x}_{\text{E-S}}(t)$ and $\vec{x}_{\text{M-S}}(t)$ represent the positions of Earth and Mars relative to the Sun, then $\vec{x}_{\text{M-E}}(t) = \vec{x}_{\text{M-S}}(t) - \vec{x}_{\text{E-S}}(t)$ represents the position of Mars relative to Earth. You're saying that you want to rewrite your equations in terms of just the quantity $\vec{x}_{\text{M-E}}(t)$. To do that, you just replace $\vec{x}_{\text{M-S}}(t)$ with $\vec{x}_{\text{M-E}}(t) + \vec{x}_{\text{E-S}}(t)$, so you'd have these "fictitious" terms of $\vec{x}_{\text{E-S}}(t)$ all over the place. That's acceptable, but not likely to be an easy or particularly useful way to solve the equations.

And if you also want to keep the Earth still in your new reference frame (like in that video), then you would also need to rotate the coordinates, which would involve standard Coriolis effects. You're starting to see why we don't normally do this. But the orbits certainly would look pretty rad. In fact, as observers on Earth, we actually see how they would look, and that's why we see apparent retrograde motion.

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    $\begingroup$ Nice answer...and nice comparison to retrograde motion. Here is a great video about how we see the motion of other planets according to this phenomenon youtube.com/watch?v=72FrZz_zJFU $\endgroup$ – dberm22 Jun 10 '15 at 19:14
  • $\begingroup$ It's not just that we can usually ignore such forces, we can actually completely ignore such forces, to the (very very accurate) extent to which the gravitational potential of the galaxy on the "near" side of the solar system is the same as that on the "far" side of the solar system. The equivalence principle tells us that we can ignore gravitational forces in a freely falling reference frame, and the solar system is exactly that, minus that variation.. $\endgroup$ – Jerry Schirmer Jun 10 '15 at 19:30
  • $\begingroup$ I agree that we can usually treat the barycenter of the Solar System as if it were inertial. But it's precisely that gradient of the potential that keeps us in the Milky Way. So we can't completely ignore such forces. I think it's important not to make blanket statements that might confuse the physics newbies, because they typically make it their job to find objections to such statements -- and rightly so. :) $\endgroup$ – Mike Jun 10 '15 at 19:55
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    $\begingroup$ Note that in astronomy, you do use a geocentric coordinate system every time you aim an Earth-surface telescope at something, or measure the position of something relative to something else on a photograph taken by such a telescope. (All other calculations are of course done in a more convenient coordinate system.) $\endgroup$ – zwol Jun 10 '15 at 21:58

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