# 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...

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).

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### 1 Answer

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.**

**Sources:**