2
$\begingroup$

I know that for compact objects (black holes, neutron stars, etc ) the value of the amount of force acting on any particle can be found by solving Einstein's field equations. So my question is that what is the maximum distance up to which Newtonian mechanics is applicable to find the force near such compact objects. In other words, what is the minimum distance around such objects from where Newtonian mechanics can not longer apply and Einstein's theory of General Relativity comes into action. Please keep the math simple. Thank You.

$\endgroup$
1
  • $\begingroup$ On a circular orbit, the centrifugal force normally offsets gravity and the orbit is stable. As you get closer to a black hole, the centrifugal force first reduces (due to the spacetime curvature) and then changes the direction. At this point both gravity and the centrifugal force pull you down. The orbit becomes unstable at 3 times the radius of the event horizon. For the Earth, it is just over an inch. The last stable orbit perhaps is the very last frontier, beyond which the Newtonian gravity is not applicable in any approximation at all. $\endgroup$
    – safesphere
    Aug 29 '18 at 16:31
3
$\begingroup$

The basic answer is that there is no sharp cut off line between where Newtonian Mechanics will fail and where it won't. At any distance, Newtonian Mechanics will fail, it's just that as you get farther and farther away from the compact object, the failure of Newton becomes smaller and smaller. The equation that describes this is the last equation in this answer. Newton's law is (magnitude only): $$F=\frac{GMm}{r^2},$$ the extra factor of $$\frac{1}{\sqrt{1-\frac{2GM}{c^2 r}}}$$ in that answer is introduced by GR.

If you examine this extra factor given in the last equation of the linked answer, you will see that that factor will approach 1 when r approaches infinity. So in the limit that r is very very large (very far away from the compact object), the force law given by Newton is the same as the one given by GR.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.