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

  • $\begingroup$ I changed the question and corrected the equation, could you please remove the hold $\endgroup$
    – PunkZebra
    Apr 20, 2014 at 7:14
  • $\begingroup$ Hi Peterix, your question has been answered here. $\endgroup$ Apr 20, 2014 at 8:05
  • $\begingroup$ Where, I can't find it? Anyway is $F=\frac{mc^4}{4GM}$ right? $\endgroup$
    – PunkZebra
    Apr 20, 2014 at 9:17
  • 2
    $\begingroup$ No, it's $$F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$ $\endgroup$ Apr 20, 2014 at 9:25
  • $\begingroup$ 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. $\endgroup$ Apr 20, 2014 at 9:27


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