Minimum black hole mass to survive fall to event horizon

[cepheid]

I'm having no end of trouble with this seemingly simple problem:

Homework Statement

What's the minimum mass of a black hole for which you could survive a fall through the event horizon without being ripped to shreds? Why would you be ripped to shreds for smaller black holes?

Homework Equations

\textbf{F} = -G\frac{mM}{r^2}\hat{\textbf{r}}

R_s = \frac{2GM}{c^2}​

(Schwarzschild radius)

The Attempt at a Solution

Assume person of height h (= 2 m) can survive a tidal stretching force of 5mg over the length of his body (g being the acceleration due to gravity on Earth). Assume further that he is to be just barely surviving when his feet touch the event horizon. Then, the difference between the gravitational acceleration on his head and his feet should be:

\Delta a = -GM\left[\frac{1}{R_s^2} - \frac{1}{(R_s + h)^2} \right] = 5g​

When I try and solve this, I get some cubic equation:

\frac{c^2 h^2}{10g} = R_s^3 + 2R_s^2h + R_s\left[h^2 -c^2 \frac{h}{5g}\right]​

I spoke to other people who said they solved this easily and got something on the order of 10⁴ solar masses as the lower limit. Have I done something wrong with my algebra (or worse yet, with the physics)?

 

[Redbelly98]

Without checking your math over, what if you assume h«R_s and drop all the small terms in your equation? (Then double check the h«R_s assumption after you calculate the final answer.)

 

[cepheid]

Without checking your math over, what if you assume h«R_s and drop all the small terms in your equation? (Then double check the h«R_s assumption after you calculate the final answer.)

Yes, if you drop h² terms everywhere, as well as dropping one rh term when it appears added to an r² term, you get the same expression as you do if you approximate the change in force by:

\Delta F = \frac{dF}{dr} \Delta r​

I got a result of M >= 20 496 solar masses, which corresponds to a Schwarzschild radius of 60 536 km (much greater than two metres!).

Thanks for the tip. Also, I noticed that dropping h² terms is equivalent to Taylor expanding (r + h)² to first order.

 

[Redbelly98]

Cool, and I get the same Rs as you. (Didn't check the mass calculation.)

 

[stevebd1]

I used the following equation for tidal forces-

dg=\frac{2Gm}{r^3}dr

set dg=5g where g is Earth's gravity, r=r_s where r_s is the Schwarzschild radius (r_s=2Gm/c^2), dr=2 and solved for m, the answer I got was 20,495 sol.