Force of gravity in the event horizon

Asked Apr 19 '14 at 18:05

Active Apr 19 '14 at 21:34

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I recently posted a question about black holes and gravity (this) and I discovered the Schwarzschild radius: $r_s=\frac{2GM}{c^2}$.
I thought with a friend of mine about the actual force (in Newtons) that an object with mass $m$ would receive from a Black Hole with Schwarzschild radius if the object was on the Event Horizon, in other words if the object shrinks to the black hole and the distance between the object and the center of the black hole is the Schwarzschild radius (see also the Shell Theorem for other clarifications).
We used the Newton's law: $$F=G \frac{m \cdot M}{r^2}$$ And we proceeded like this: $$F=G\frac{m\cdot M}{{r_s}^2} =G \frac{m\cdot M}{(\frac {2GM}{{c^2}})^2}=\frac{mc^4}{4GM}$$ That would be the force of attraction of a Black Hole with radius equal to Schwarzschild radius, to an object of mass $m$ that is in the event horizon.
Is this right?

Thanks! :)

p.s feel free to edit English is not my mother tongue

edited Apr 13 '17 at 12:39

Community

asked Apr 19 '14 at 18:05

PunkZebra

I changed the question and corrected the equation, could you please remove the hold – PunkZebra Apr 20 '14 at 07:14
Hi Peterix, your question has been answered here. – John Rennie Apr 20 '14 at 08:05
Where, I can't find it? Anyway is $F=\frac{mc^4}{4GM}$ right? – PunkZebra Apr 20 '14 at 09:17
2

No, it's $$F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$ – John Rennie Apr 20 '14 at 09:25
The force goes to infinity at the event horizon. However nota bene that the force calculated here is coordinate dependant and its value is as observed from far outside the horizon. – John Rennie Apr 20 '14 at 09:27

Force of gravity in the event horizon

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