How much would the LHC beam be attenuated by the atmosphere? As I understand it, the completed Large Hadron Collider (LHC) will ultimately have a proton beam with $2,808$ bunches of $1.15 \times 10^{11}$ protons each at $7$ TeV, giving a total beam energy of $(2808)(1.15\times 10^{11})(7 \times 10^{12})(1.602\times 10^{-19}$ J$)$ $\approx 362 $ MJ.    
Imagine that instead of using a graphite block for the beam dump, the proton beam is aimed "straight up" at a satellite in geosynchronous orbit 42,160 km above the surface of the earth.  How significantly would be the beam intensity be diminished in the atmosphere, particularly the troposphere?  Could one reasonable expect to be able to blind sensors/cameras on a satellite?
 A: Hmmm...some back-of-the-envelope calculations:
The depth of the air column at sea level is $14\text{ lbs/in}^2 = 2 \times 10^5\text{ g/cm}^2$, so neglecting space-charge effects and assuming minimum ionization the whole way we get about $4 \times 10^5\text{ MeV} = 0.4\text{ TeV}$ energy loss. We are actually above minimum ionization, so we can multiply that by a small integer. Call it 1 TeV, which justifies the approximation.
Another consideration: multiple scattering. The radiation length of air is around $36\text{ g/cm}^2$, so the protons travel through about $5.5 \times 10^3$ radiation length resulting in a RMS position dispersion of about 5 km. Making for a mean areal proton density at geosynchronous orbit of around 1500 per square meter at the top of the atmosphere and expanding in a cone for the next 42,000 kilometers. 
Those protons are rather more energetic than a typical cosmic ray, but not at all unknown in that environment.
Conclusions: no prompt effects, but possible a reduced lifetime.
A: Extremely unlikely.  The beam would diffuse very rapidly.  The LHC beam is condensed into a very small location by magnetic fields and it's energy is maintained through the use of an RF field which replenishes the energy each time the beam circulates while orbiting in a near perfect vacuum. In the absence of such fields the beam would first repel itself through Coluombic repulsion, and then rapidly lose energy through interaction with the atmosphere due to the loss of vacuum.  It would pose no risk to a satellite so far away, especially for a satellite that has frequent interaction with space radiation.  
A: Well, I will give the simple answer of the geometrical intensity fall off with 1/r**2:
r**2=2*10**15meterssquared  approximately for the satelite distance .
Take the 362MJ  for 1 meter squared as an initial area to make life easy, ( at about 1km height?) though in the beginning it is concentrated into microns; only a tiny fraction of a joule will end on the satellite if it has a cross section of 1meter squared: 1.3*10**-7 joules, give or take an order of magnitude because of my sloppiness, will hit the satellite.
I think even if there were no atmosphere there would be absolutely no danger for the satellite.
