Tidal Forces inside a black hole

Tidal Forces for a 100 kg Human Falling into Black Holes

We quantify tidal accelerations for a 100 kg person of length $L \approx 2\,\mathrm{m}$ falling feet-first toward two black holes:
a stellar-mass black hole with $M = 10\,M_\odot$ and a supermassive black hole (SMBH) with $M = 4\times 10^{6}\,M_\odot$ .
Outside the horizon of a non-rotating (Schwarzschild) black hole, the dominant components agree with the Newtonian tidal tensor and scale as $r^{-3}$ .

Tidal gradients used
Radial stretching (head vs. feet): $\displaystyle \Delta a_{\rm radial} \approx \frac{2GM}{r^{3}}\,L$
Transverse squeezing (side-to-side): $\displaystyle \Delta a_{\rm trans} \approx \frac{GM}{r^{3}}\,L$
Schwarzschild radius (event horizon): $\displaystyle r_s=\frac{2GM}{c^2}$

A. Tidal Forces at the Event Horizon

Evaluating the radial gradient at $r = r_s$ gives the compact scaling

$\Delta a_{\rm horizon} \;=\; \frac{c^{6}\,L}{4\,G^{2}\,M^{2}} \;\propto\; M^{-2}.$

Black hole	$\Delta a_{\rm horizon}$ (m/s $^2$ )	As multiples of $g$	Force on 100 kg $F = m\,\Delta a$
Stellar, $M=10\,M_\odot$	$\approx 2.06\times 10^{8}$	$\approx 2.1\times 10^{7}\,g$	$\approx 2.06\times 10^{10}\ \mathrm{N}$
SMBH, $M=4\times 10^{6}\,M_\odot$	$\approx 1.29\times 10^{-3}$	$\approx 1.3\times 10^{-4}\,g$	$\approx 0.13\ \mathrm{N}$

Takeaway: Tidal gradients at the horizon fall as $1/M^{2}$ . For stellar-mass BHs they are enormous; for SMBHs they are tiny at horizon crossing.

B. Where Do Tidal Forces Reach Human-Noticeable Levels?

Set a target gradient $a_{\!*}$ and solve for the radius at which the radial component reaches this value:

$r(a_{\!*}) \;=\; \Bigg(\frac{2GM\,L}{a_{\!*}}\Bigg)^{1/3}.$

Compare to $r_s$ to see if this occurs outside or inside the horizon.

Chosen thresholds

Mild discomfort: $a_{\!*}=1\,g=9.81\ \mathrm{m/s^2}$
Severe: $a_{\!*}=10\,g$
Extreme: $a_{\!*}=1000\,g$

(1) Stellar BH: $M=10\,M_\odot$ , $r_s\approx 2.95\times 10^{4}\ \mathrm{m}$

$1\,g$ : $r\approx 8.15\times 10^{6}\ \mathrm{m}\ \approx 276\,r_s$ (outside)
$10\,g$ : $r\approx 3.78\times 10^{6}\ \mathrm{m}\ \approx 128\,r_s$
$1000\,g$ : $r\approx 8.15\times 10^{5}\ \mathrm{m}\ \approx 27.6\,r_s$

Implication: A human is torn apart well before reaching the horizon of a stellar-mass BH.

(2) SMBH: $M=4\times 10^{6}\,M_\odot$ , $r_s\approx 1.18\times 10^{10}\ \mathrm{m}$

$1\,g$ : $r\approx 6.00\times 10^{8}\ \mathrm{m}\ \approx 0.0508\,r_s$ (inside)
$10\,g$ : $r\approx 2.79\times 10^{8}\ \mathrm{m}\ \approx 0.0236\,r_s$ (inside)
$1000\,g$ : $r\approx 6.00\times 10^{7}\ \mathrm{m}\ \approx 0.00508\,r_s$ (deep inside)

Implication: You cross a SMBH horizon with negligible tidal sensation; damaging gradients arise only well inside.

C. Values at Fixed Multiples of the Horizon Radius

Because $\Delta a \propto r^{-3}$ , the gradient grows extremely rapidly as you approach the center.

Case	Radius	$\Delta a$ (m/s $^2$ )	As multiples of $g$
Stellar $10\,M_\odot$	$r=100\,r_s$	$\approx 206$	$\approx 21\,g$
Stellar $10\,M_\odot$	$r=10\,r_s$	$\approx 2.06\times 10^{5}$	$\approx 2.1\times 10^{4}\,g$
SMBH $4\times 10^{6}\,M_\odot$	$r=10\,r_s$	$\approx 1.29\times 10^{-6}$	$\approx 1.3\times 10^{-7}\,g$

D. Geodesic Deviation (Curvature View)

The tidal accelerations are encoded by geodesic deviation:

$\frac{D^2 \xi^\mu}{d\tau^2} \;=\; -\, R^\mu_{\ \nu\rho\sigma}\, u^\nu\, \xi^\rho\, u^\sigma,$

where $\xi^\mu$ is the separation vector between neighboring points on the body, $u^\nu$ the 4-velocity, and $R^\mu_{\ \nu\rho\sigma}$ the Riemann curvature tensor. In Schwarzschild spacetime, the principal components reduce to the gradients used above outside the horizon.

Bottom Line

Stellar BH (~ $10\,M_\odot$ ): lethal tidal gradients occur far outside the horizon.
Supermassive BH (~ $10^{6}$ – $10^{9}\,M_\odot$ ): horizon crossing is uneventful; destructive tides arise only well inside.

Want a quick variant with a different body length (e.g., $L=1.7\,\mathrm{m}$ ) or a different SMBH mass (e.g., $10^{8}\,M_\odot$ )? The same formulas apply; only the numbers change via the $L$ and $M^{-2}$ scalings.