Question

Calculate the tidal force on a 75 kg person 2 meter height falling into a solar mass black hole just as they cross the event horizon. Compare that to the force of Earth's gravity on that same person (given by their mass multiplied by 9.8).

Answer 1

A "solar-mass" black hole has a radius around 3 km.
That is for a mass of 1.9*10^30 kg


Person is of 75 kg mass and a length of 2 m from head to toe. This places the centre of gravity of the head 1 m from the waist, and the centre of gravity of the feet+ankles 1 m below the waist. Therefore distance between the black hole's centre of mass and centre of mass of the person is 3000 + 1 = 3001 m

Place the person just above the event horizon (as seen from a safe observer's frame of reference)*.

The person's feet are 0 metres above the event horizon and the head is 2 metres above the event horizon (this places the person's centre of gravity exactly 1 meter above the event horizon and exactly 5 km above the centre of the black hole (from the safe observer's frame of reference )

Tidal force is given by
F = G M m / d^2
where
M = 1.9*10^30 kg
m = 75 kg
G = 6.674*10^-11 N m^2 / kg^2

d = 3001 m

Putting the values we get

F = 1.1109 x 10¹⁵ Newtons

whereas the force experienced by the person on earth is mg = 75*9.8 = 735 N

therefore
F_black/F_earth = 1.5114 x 10¹²

therefore force experienced n black hole is 1.5114 x 10¹²times greater than on earth