Camera reaching the event horizon This question came to my mind after reading this.

I think it is safe to assume that the supposedly thrown camera would not reach the event horizon in one piece. So my question is : 

If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon, and if so, why?
If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.

For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at
$$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$
where the the second derivatives of the potential, $\Phi_{,jk}$, form the so-called tidal gravitational field. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.

Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.

To be relativistically correct, the tidal forces on an free-falling object are described by the equation of geodesic deviation, in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor:
$$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$
In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called spaghettification.

Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:


$M\sim 10\,\mathrm{M_\odot}$: $\sim 10^6$ Earth gravities;
$M\sim 10^4\,\mathrm{M_\odot}$: $\sim 1$ Earth gravity;
$M\sim 10^6\,\mathrm{M_\odot}$: $\sim 10^{-4}$ Earth gravities.


The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.