# When will a person falling into a black hole be crashed by tidal forces?

Assume that a person can resist $$30g\approx 300\frac m{s^2}$$ (remark: the record is actually at 46.2). I want to solve for the distance $$r$$ at which a $$1.5m$$ tall slim person falling radially into a blackhole of given Schwarzschild radius $$R_S$$ is destroyed by tidal forces.

From the geodesic equation (as in Sean Carroll's Spacetime and Geometry (5.64), p.208), I have that

\begin{align}\left(\frac{dr}{d\tau}\right)^2 &= E^2 - \left(1-\frac {R_S}r\right)\\ & = \left(1-\frac {R_S}r\right)^2\left(\frac{dt}{d\tau}\right)^2 - \left(1-\frac {R_S}r\right)\end{align}

For $$t=\tau$$, this yields

$$v(r):=\sqrt{\left(1-\frac {R_S}r\right)^2 -\left(1-\frac {R_S}r\right)}$$

As for the average acceleration on the body, I have that

$$a = \frac{v(\text{feet}) - v(\text{head})}{1.5m/v(\text{body CM})}$$

Then, if the feet are at $$R_S+x$$, the CM will be at $$R_S+x+.75m$$ and the head at $$R_S+x+1.5m$$. I can input all this into the equations, but then I don't know how to simplify and solve for $$x$$ in terms of $$a$$.

• $t$ and $\tau$ cannot be set equal, and your expression for $a$ does not have the right dimensions to be acceleration. – G. Smith Mar 12 '19 at 0:05
• Acceleration and tidal force are two different things. You could have 30g of acceleration with no tidal force, for example. – G. Smith Mar 12 '19 at 0:07
• Rodrigo, the answer depends on how much force a normal human being can withstand without being pulled apart. Obviously, such information is very difficult to come by, as it would be extremely unethical and illegal to obtain this information. – David White Mar 12 '19 at 0:45
• To calculate tidal acceleration use geodesic deviation equation. Also, Misner, Thorne,Wheeler §32.6 has a solution. – A.V.S. Mar 12 '19 at 5:45