# Spaghettification of humans near black holes

A few months ago I was discussing the spaghettification phenomenom with my wife, just for the fun of it. This was when the mass of the super massive black hole from M87 hit the news. The black hole was meant to have 6.7e+9 solar masses, so I thought it would be fun to figure out at which distance the tidal forces become so great, that the forces between your head and feet start getting... uncomfortable.

I defined the distance from feet to head to be 2 m, the head having a mass of 10 kg and one foot having a mass of 2.5 kg (I didn't research the numbers, I just did some wild guesses).

So, the force between head and foot should be:

$$G*m_B*m_H \over (d+2)^2$$ $$-$$ $$G*m_B*2*m_F \over d^2$$ where $d$ is the distance from foot to the black hole, $m_B$ is the mass of the black hole, $m_H$ the mass of the head and $2*m_F$ the mass of the feet. Assuming you fall with the feet first.

This is one of the plots I made: (note: the plot only takes the mass of one foot into account.)

I made up this short octave script to plot the forces on this values with the distance in AU as the free variable and thought "ouch, that's gonna hurt pretty soon". I haven't checked which part of the body will rip off first at what forces though...

Anyway, what I wanted to know is: Do my calculations make sense? Is my math correct?

Thanks, Alex.

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your calculation is simplified by using the definition of derivative: f(x+e)-f(x) = f'(x)e. This turns the tidal force equation into a closed form that doesn't involve subtracting two enormous numbers to get a small number. – Ron Maimon Oct 5 '12 at 7:43

I have two feet, therefore I think the force will be double what is calculated.

However, I don't have a head weighing 10kg, therefore I think the force will be half what is calculated. (see http://hypertextbook.com/facts/2006/DmitriyGekhman.shtml and others)

Therefore I think I agree with the conclusion, but for slightly different reasons.

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Good point, I forgot to take into account that there are two feet. So actually I should be giving $2*m_F$ to the formula. – Alexander Janssen Oct 1 '12 at 21:56