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To find the gravity at Schwarzschild radius of a black hole, I used the below equations. Assuming escape velocity at event horizon is speed of light c and M is the mass of the black hole

$$ v_e = \sqrt{\frac{2GM}{r}} = \sqrt{2gr\,} $$ or $$ g = {\frac{v_e^2}{2r}} = {\frac{c^2}{2r}} $$

Now substituting the Schwarzschild radius of ${\frac{2GM}{c^2}}$ for r we get

$$ g = {\frac{c^4}{4GM}} $$

Just thinking on a time dilation perspective, I have also read that acceleration can not be distinguished from gravity. So if I can accelerate a body with a rocket at above rate will it cause same time dilation like a black hole? And if at all the answer is yes, shouldn't such a value be independent of the mass of black hole?

Sreekumar R
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  • There is such thing as an acceleration event horizon. http://en.wikipedia.org/wiki/Event_horizon#Apparent_horizon_of_an_accelerated_particle – Aron Nov 25 '14 at 10:12
  • Except the quantity you've defined as "g" is not particularly meaningful. It does not correspond to any type of measurable acceleration. The proper acceleration required to hover near the event horizon asymptotically approaches infinity as $r \to 2GM/c^2$. – Jold Nov 25 '14 at 10:17
  • You've used the wrong equation for the gravitational acceleration. I've linked a question that explains how to derive the correct equation. As jld says, the acceleration goes to infinity at the event horizon. – John Rennie Nov 25 '14 at 10:22
  • Thanks for the comments. @Aron I was exactly thinking of an acceleration event horizon, and from the link JohnRennie gave, I got the explanation why it tends to infinity as it approaches Schwarzschild radius as mentioned by jld – Sreekumar R Nov 25 '14 at 10:44

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