# Is the gravitational acceleration at the event horizon constant?

If the escape velocity at the event horizon of a black hole is equal to the speed of light, does this imply that the gravitational acceleration at the event horizon is also constant?

For example, using Earth $\left(m=5.967 \cdot {10}^{24}\, \mathrm{kg}\right)$ and the equation $c^2 = \frac {GM}{r_{\text{S}}}$, we find a Schwarzchild radius of $r_{\text{S}} =8.86 \cdot {10}^{-3} \, m$. Substituting this value into $a = \frac {GM}{r^2}$ we derived an answer of $a=5.071525376 \cdot {10}^{18} \, m$ at the Schwarzchild radius.

Furthermore, an object's mass will be directly proportional to its Schwarzchild radius, thus the gravitational acceleration would remain constant for all black holes at its event horizon. Is this assumption corrects and does this withhold with all black holes?

• If the gravitational acceleration is proportional to the mass and the inverse of the Schwarzschild radius squared and the mass is proportional to the Schwarzschild radius, then it stands to reason that the acceleration is proportional to the inverse of the radius. What led you to believe it is constant? – Jim Nov 14 '14 at 15:34
• Your formula for acceleration as a function of radius, $\frac{GM}{r^2}$, only works in Newtonian gravity, it doesn't work in general relativity. If an observer uses a rocket or something to hover at a fixed radius r in Schwarzschild coordinates, then as mentioned here, the proper acceleration they experience is given by $a = (1 - 2M/r)^{-1/2}M/r^2$ (in units where G=c=1, so the Schwarzschild radius is just $r_s = 2M$ in these units). – Hypnosifl Nov 14 '14 at 16:05
• Ah, Thank you. However now I see that a = c^4/ 4GM. Does this mean that the gravity decreases at the event horizon as the mass increases? – martin Nov 14 '14 at 16:11
• @martin yes. That's what it means – Jim Nov 14 '14 at 16:16
• Are you saying that $a = c^4 /4GM$ at the horizon? If so, where did you get that? It's definitely not true according to the equation I gave, try plugging in the horizon radius $r=2M$ (note that if you want to use units where G and c are not equal to 1, so the horizon is at $r=2GM/c^2$, that equation would have to become $a = (1 - (2GM/rc^2))^{-1/2} GM/r^2$ in order for the units to work out properly) – Hypnosifl Nov 14 '14 at 16:18