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The force near a black hole (outside event horizon $r=3r_s/2$) onto a mass $m$ can be calculated by General Relativity:

$$F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}.$$

However, there must be a distance $r$ where the black hole's gravity basically becomes Newtonian only:

$$F=\frac{GMm}{r^2}.$$

At which distance $r$ does this happen?

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    $\begingroup$ Related post by OP: physics.stackexchange.com/q/584522/2451 $\endgroup$
    – Qmechanic
    Oct 7, 2020 at 18:02
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    $\begingroup$ Expand F in inverse powers of r. Show your thoughts. $\endgroup$ Oct 7, 2020 at 18:06
  • $\begingroup$ @Cosmas Zachos: I'm not sure what you mean by this...clearly for a big $r$ the second fraction becomes = 1 and the force becomes Newtonian, but where is the threshold? $\endgroup$
    – Marcus
    Oct 7, 2020 at 18:15
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    $\begingroup$ Ipso facto, this is what I mean by this: how did you get it into your head there is threshold behavior at all? $\endgroup$ Oct 7, 2020 at 18:17
  • $\begingroup$ I mean, when plotting it I get $r = 3r_s$ where the functions tend to overlap. Clearly, $3r_s$ is the minimal radius where circular orbit is possible, but I don't know how to mathematically derive this value. $\endgroup$
    – Marcus
    Oct 7, 2020 at 18:21

4 Answers 4

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From your formula, you can see that $F \approx F_\text{Newtonian}$ when $2GM/c^2r \ll 1 $, or if you rearrange, $2GM/c^2 = r_s \ll r$. In other words, the farther from the Schwarzschild radius, the closer you get to Newtonian gravity.

We can take the ratio $F/F_N$ to get an idea of how far off we are from Newtonian gravity. At $r = 10 r_s$, it is $1/\sqrt{9/10}$ which is about $1.054$, so this is 5.4% off from Newtonian gravity.

At $r = 100 r_s$ it is $1/\sqrt{99/100}$ which is about $1.00504$, so this is 0.504% off from Newtonian gravity.

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  • $\begingroup$ +1, but could you also discuss $3r_s$ as the minimal radius where a circular orbit around a black hole is possible, and why this might be? $\endgroup$
    – Marcus
    Oct 7, 2020 at 19:06
  • $\begingroup$ You're right to ask why F (not F_N) does not suggest unstable orbits for r<3r_s. I haven't questioned where the formula comes from, but I can only assume that this formula is already an approximation only valid within a certain regime, likely already far from r_s. Even if this is not the case, the force you feel says nothing about the shape of geodesics. At this point, we have to turn to GR in full and we can't rely on approximations. $\endgroup$ Oct 7, 2020 at 19:09
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Although UrsaCalli79's answer is very effective at explaining this mathematically, I will attempt to explain it in layman terms.

You do not leave either Newtonian theory nor Relativistic theory, it is only that one serves better to mathematically predict the forces. This can be seen in the similarity of the equations. Newtonian theory can predict the forces with a fair amount accuracy, and Relativistic theory with more accuracy.

What UrsaCalli79 mathematically demonstrated was that as you take distance from the Schwarzschild radius $-$ which also can be said to be the start of black hole's singularity $-$ the Newtonian equation can more accurately describe the forces.

To end, this essentially means there is a difference of how accurately one theory is versus the other, this difference in accuracy decreases as $r$ increases.

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  • $\begingroup$ But how come that these two functions tend to overlap at $r = 3r_s$? Because this would mean that only when General Relativity becomes Newtonian, circular orbits are possible...and this would actually also prove that Newton mechanics includes only circular orbits - celestial bodies if you will... $\endgroup$
    – Marcus
    Oct 7, 2020 at 18:37
  • $\begingroup$ ...and this in the other hand allows us to prove that Newton mechanics does not apply within an $r = 3r_s$. So we have a place in space and time which common physical laws do not apply. This is like quantum mechanics, but large scale ;-) $\endgroup$
    – Marcus
    Oct 7, 2020 at 18:43
  • $\begingroup$ Overlapping between these functions can be looked at similarly as the overlapping of other equations of contrasting theories which attempt to predict forces. $\endgroup$ Oct 7, 2020 at 18:44
  • $\begingroup$ Newtonian theory does not accurately predict within $r < 3r_s$, although it can still be used to predict less-than-accurate force values within $3r_s$. It applies, just not as accurately as Relativistic theories. $\endgroup$ Oct 7, 2020 at 18:47
  • $\begingroup$ Looking at the data, I don't believe that a reliable prediction can be made with Newton mechanics for $r < 3r_s$, it's rather similar to quantum mechanics. Interestingly enough, we are currently struggling to understand atoms and their quantum mechanics, simply because we can't measure anything anymore. But yet there's an huge object in space that is far away, but Newtonian mechanics won't (really) work with that. So why not start to think differently and look at atoms as black holes, or look at atom cores like at black holes, and start from there...TOE, you know. $\endgroup$
    – Marcus
    Oct 7, 2020 at 19:09
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the orbit will be never closed (remember, the system emits gravitational waves), but will spiral towards the center. The smaller $r$ the longer it takes.

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The general relativity equations have the term $r_{s}/r$.

When $r_{s}/r \approx 0$, then the equations reduce to Newton's law of gravitation.

A six solar mass black hole has $r_{s} \approx$ 10.875 miles (about 17,400 meters) At 93 million miles, $r_{s}/r \approx 1.1694*10^{-7}$, which makes Newton's equations an extremely good approximation at this distance.

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