Temple of Ranma's Sailor Senshi Seifuku

While performing calculations of a probe approaching a black hole ( http://chebmaster.ru/downloads/bhgravit.exe ) I got some unexpected results:

from https://en.wikipedia.org/wiki/Circular_orbit

orbital period T = 2 * Pi * square_root( r^3 / mu)

where mu = G * M

Practical calculations show: a circular orbits in the danger zone of a black hole may range from one revolution per a few seconds to 10 orbits per second (!), thus a probe has to rotate around its axis once per orbit (emulating being a tidally locked body) otherwise it will experience a brutal vibration with frequency of a few Hertz from the spaghettification force rapidly rotating relative to the probe.

But that adds a centripetal force summing with the spaghettification force trying to pull our probe apart:

From http://en.wikipedia.org/wiki/Centripetal_force

for a material point, Fc = mr * 4 * Pi^2 / (T^2)

for a solid rod, tensile force in the center is Fct = m * (integral of r where r is 0..L/2) * 4 * Pi^2 / (T^2)

where L is length of the rod.

We take only a half of the rod, because if each end pulls with the force of one kilogram, the total tensile force (in the center) is one kilogram.

Integral of r for r from 0 to L/2 = L/4.

So, Fct = m * (L / 4) * 4 * Pi^2 / (2^2 * Pi^2 * (r^3 / mu)) = m * L * mu / (4 * r^3)

From http://en.wikipedia.org/wiki/Spaghettification

for a solid rod, Ft = mu * L * m / (4 * r^3)

Do you see it? THEY ARE FRAKKING EQUAL! Keeping "tidally locked" to avoid vibration precisely *doubles* the force pulling the probe apart. HOW COMES THEY ARE EQUAL!!!

Or did I miss something?

P.S. Some interesting consequences: no, you cannot take a scenic look of a normal stellar-mass black hole. If you approach it closer than where it looks approximately as big as Moon when viewed from Earth, you go splat. And even a sturdiest, most compact probe can't approach the event horizon without going splat. Close to a BH of 10 sun masses a probe 10 centimeters in diameter would make more than a thousand orbits per second while experiencing tensile force of 250 thousand times its mass.

This is totally not my thing and i may be misreading you but:

We take only a half of the rod, because if each end pulls with the force of one kilogram, the total tensile force (in the center) is one kilogram.

Is it?

It is. Intuitive but true. Imagine a rope hanging on a pulley, both ends are weighed with 1kg weights. The rope will be submitted to exactly 1kg of stretching force.

It's the same as with cars colliding head on at 60 kph each. The result for each car will be the same as colliding with a concrete wall at the same 60 kph, proven by MythBusters. Not at 120.

Cheb wrote:It is. Intuitive but true. Imagine a rope hanging on a pulley, both ends are weighed with 1kg weights. The rope will be submitted to exactly 1kg of stretching force.

It's the same as with cars colliding head on at 60 kph each. The result for each car will be the same as colliding with a concrete wall at the same 60 kph, proven by MythBusters. Not at 120.

Oh i´m well aware of the collision thing, i knew that long before mythbusters played around with it.
But that´s because both sides of the crash are essentially doubling as "the wall" due to how the forces interact.

And with the rope thing, it´s the same as if one end was tied to a hook or something, as that essentially exerts enough force to keep the rope hanging still, so effectively 1kg of force yes.

Thing is i vaguely recall something about it sometimes being different, if it involves something rigid, like that rod of yours.

Meh, i don´t remember well enough to say, my dad and eldest brother are the engineers in the family, not me, i´ve never been a fan of maths.

Edit: It might be that i´m thinking of torsion/twisting forces rather than straight pull.

Cheb wrote:It's the same as with cars colliding head on at 60 kph each. The result for each car will be the same as colliding with a concrete wall at the same 60 kph, proven by MythBusters. Not at 120.

As long as their respective mass is equal, IIRC. I rather suspect that if a Lada 1200 doing 60 were to collide head-on with a 2TE70 also doing 60 (let's just say it was a drunk driver), the Lada would lose a lot more spectacularly than if it had hit that wall.