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Newton's equation for gravity calculates gravitational force between two bodies $$F= GMm/r^2$$

My question is: How does the force change as these two bodies accelerate together through the universe and approach light speed?

You might argue there should be no change because both bodies are accelerating together and stay the same relative to each other. Consider a planet and a moonbase. They people on each site will experience the same time dilation according to special relativity. Special relativity allows us to adjust the kinematics equation for an object as it approaches the speed of light. The Lorentz factor concisely describes this

$\gamma=1/\sqrt{1 \ - \ v^2/c^2}$.

So, relativistic mass increases with velocity $M=m\gamma$ The force then increases as gamma squared (because the mass of the planet and the moon both increase) $F= G \frac{Mm}{ (1 \ - \ v^2/c^2) \ r^2}$. If I read that right, it says force goes to infinity as a planetary pair reach lightspeed together.

I found this: http://www.einsteins-theory-of-relativity-4engineers.com/support-files/velocity-effects-on-gravityY.pdf But it seems not exactly relevant as it discusses the measurement of gravity between two masses moving at relativistic speeds to each other. I am talking about the gravity within a solar system that is accelerating and reaching a velocity away from ours approaching light speed.

If gravitational force indeed increases then orbital mechanics of the solar system would be different than ours. Solar systems very far away (which happen to be close to light speed due to the Hubble expansion) would have a much greater gravitational milieu. That means that the planets orbit their sun a lot faster than expected at a given distance from their sun.

Please point out the error in my analysis gently. This discussion assumes the gravitational constant,G, does not change as objects accelerate and reach light speed.

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    $\begingroup$ Aquagremlin - can you be a bit more specific about the scenario you consider? It seems you wonder about two massive objects traveling together as seen by a distant observer who moves at relativistic speed compared to the two bodies? However, you also mention "people on each site [on both bodies] will experience the same time dilation according to special relativity". $\endgroup$
    – Johannes
    Commented Jul 28, 2013 at 5:58
  • $\begingroup$ "You might argue there should be no change because both bodies are accelerating together and stay the same relative to each other" you are misunderstanding something here for sure. $\endgroup$
    – TMS
    Commented Jul 28, 2013 at 8:55
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    $\begingroup$ @Johannes, THANK YOU for making my post look readable! Can you tell me how you did it? As for clarification, I think you understood it well. Yes, this is what I meant " It seems you wonder about two massive objects traveling together as seen by a distant observer who moves at relativistic speed compared to the two bodies?". Leave out the bit regarding time dilation, it is not really germane to the question. I just wanted to point out that a distant observer and the set of two massive bodies are separating at relativistic speeds,but that the two massive bodies are staying together. $\endgroup$ Commented Jul 28, 2013 at 12:21
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    $\begingroup$ @Aquagremlin - you're welcome. When new on this site, like you, I had no clue about how to render equations. Others edited the equations in my posts, and I learned simply by clicking 'edit' and looking at the equation codes. It's really simple. Start with a dollar sign and end with a dollar sign (double dollars for non in-line equations) and off you go... $\endgroup$
    – Johannes
    Commented Jul 28, 2013 at 13:50
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    $\begingroup$ When you're talking about relativistic speeds, you have to employ GR to study problems like this, I think. It is not as simple as taking Newton's equation and plug in some gamma factors... $\endgroup$
    – neutrino
    Commented Jul 31, 2013 at 11:42

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Observers in different reference frames may disagree on what they measure, but the consequences of their observations must be consistent. That is, a moving ladder can't be length contracted in one reference frame to fit through a barn, but too long to fit through that barn in another reference frame. With the proper application of Lorentz transforms, you can always find a base truth that all reference frames will agree on.

Else, there's some reference frame where Earth is contracted down into being incredibly dense and we turn into a black hole. Sure, someone in that reference frame will see Earth as being so incredibly dense, but actually turning into a black hole is something that all reference frames (or more particularly Earth's rest frame) must agree on.

Do you see the problem with your proposal now? Your calculation is flawed because you're calculating a force based on your reference frame, but an observer in a different reference frame will disagree.

Not to mention that relativistic mass simply isn't a very good concept and has been generally abandoned by the physics community since it so easily leads to misunderstandings like this.

That's not even getting into how special relativity just fundamentally doesn't play nicely with Newtonian gravity. That's the fundamental reason why Einstein had to develop general relativity in the first place.

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  • $\begingroup$ I have since realized that there can be no tool to report to me the gravity of a system moving away from ours at relativistic speeds. The motivation for this question was trying to understand vera rubin’s findings without inventing dark matter. $\endgroup$ Commented Jun 7, 2023 at 13:32
  • $\begingroup$ I think you have the interpretation of the ladder paradox incorrect. The two observers do disagree about whether the ladder fits in the barn - the ladder-observer does in fact see that the ladder is too large to be fully inside the barn (the barn is shorter than the ladder). What they both agree on is that the barn doors are able to shut without hitting the ladder, but that's because ladder-observer sees them shut at different times, not because they both agree that the ladder is shorter than the barn. $\endgroup$ Commented May 23 at 17:34

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