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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 $a$.

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    $\begingroup$ $t$ and $\tau$ cannot be set equal, and your expression for $a$ does not have the right dimensions to be acceleration. $\endgroup$ – G. Smith Mar 12 '19 at 0:05
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    $\begingroup$ Acceleration and tidal force are two different things. You could have 30g of acceleration with no tidal force, for example. $\endgroup$ – G. Smith Mar 12 '19 at 0:07
  • $\begingroup$ 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. $\endgroup$ – David White Mar 12 '19 at 0:45
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    $\begingroup$ To calculate tidal acceleration use geodesic deviation equation. Also, Misner, Thorne,Wheeler §32.6 has a solution. $\endgroup$ – A.V.S. Mar 12 '19 at 5:45

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