Calculate the tidal force on a 75-kg person 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).
The tidal acceleration, a is given as:
a = 2GMd/R^3,
where a: tidal acceleration, G: constant of gravity, M: mass of the stellar body (black hole in this case), d: length of object (human being in this case, say 2 m) and R: distance between bodies (radius of earth for someone on earth or radius of event horizon in case of black hole).
radius of event horizon is given as , 2GM/c^2, where c is speed of light.
therefore, for a solar-mass (M=1.98 x10^33 gm) black hole,
R = 2 x 6.67x10^-8 dyne cm^2/gm^2 x 1.98X10^33 gm/(3x10^10 cm/sec)^2 = 294000 cm
Substituting the value of R in previous equation, the tidal acceleration would be:
a = 2 x 6.67x10^-8 dyne cm^2/gm^2 x 1.98X10^33 gm x 200 cm/(294000 cm)^3 =
a = 1.8 x 10^23 cm/s^2
and Tidal Force, F = mass x a = 75000 gm x 1.8x10^23 cm/s^2 = 1.35 x 10^28 dynes.
In comparison, the acceleration due to earth's gravity is just 979 cm/s^2,
thus the tidal acceleration of the black hole is 1.8x10^23 / 979 = 1.84 x 10^20 times than that of earth.
Gravity force on earth = 75000 gm x 979 cm/s^2= 7.34 x 10^7 dynes.