Aren't tidal forces near a black hole so great that long before you reached the event horizon you would already be spaghettified? Consider the example of a fairly typical black hole of 10 solar masses, or 2*10

^{31} kg. The radius of black holes follows a simple formula of R=3M, where M is the multiple of solar masses and R is in kilometers, yielding a radius for this black hole of 30 km. What would the difference in gravity be for an object 31 km distant from the black hole's center versus 31.002 km distant. .002 km is roughly the height of a man, so the question being asked is what is the difference in gravity between a man's head and feet a kilometer distant from the black hole's event horizon?

Just doing the very simple, the ratio in gravity between the two distances only 2 meters apart is 1.00013, so close to 1 as to seem negligible.

But it's not negligible because given the huge gravitational force .00013 of it must be enormous. The gravitational force a kilometer out from the event horizon is 6.67*10

^{14} newtons, so .00013 of that is 8.67*10

^{10} newtons, or translating it into something more familiar, something in the neighborhood of 20 billion pounds.

A difference in gravitational pull of 20 billion pounds between head and toes would be far more than enough to spaghettify a person, so long before you ever reached the event horizon you would already be spaghettified.

I don't have the background to take relativistic effects into account.

--Percy