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?


3 Answers 3


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.

  • $\begingroup$ Btw, why bother with using a complicated beam dump that draws an 'e' in thick graphite blocks? Why doesn't one simply shoot at a hole in the ground? $\endgroup$ Aug 12, 2011 at 20:46
  • $\begingroup$ What do you mean "draws an e"? $\endgroup$
    – Jen
    Aug 12, 2011 at 20:55
  • 2
    $\begingroup$ The beam dump has to absorb the energy of the beam without melting, contain the inevitable activation in material that can be easily handled (which is why simply pouring the beam into dirt probably isn't a good idea), moderate the secondary neutron flux (and capture as many as possible), control the risk of scary chemistry among the products of the interaction between the beam and the dump, and so on. $\endgroup$ Aug 12, 2011 at 21:00
  • $\begingroup$ @dmckee - In general, how much does the proton beam spread per meter in a vacuum? $\endgroup$ Aug 13, 2011 at 0:20
  • $\begingroup$ That would be the space charge effect that I started trying to compute before I realized that the multiple scattering would dominate. You could check the edit history to see where I got. $\endgroup$ Aug 13, 2011 at 0:27

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.

  • $\begingroup$ Plus, to direct the beam "straight up" you'd need a structure with curvature of the LHC built perpendicular to the plane of the LHC ring, which would mean that you'd need a semi circular structure with a radius 4.3 km, which would would be taller than any building on earth. $\endgroup$
    – Jen
    Aug 12, 2011 at 20:29
  • $\begingroup$ I suppose I should have suggested a diagonal shot where one would tilt the LHC ring by a reasonable angle... $\endgroup$ Aug 12, 2011 at 20:42
  • $\begingroup$ No - I think the best shot would be to fire an abort kicker like the current mechanism and shoot the beam off tangentially to the beam motion. Although to do this you'd have to go through a longer path in the lower atmosphere to get to your satellite. $\endgroup$
    – Jen
    Aug 12, 2011 at 20:53
  • $\begingroup$ @Jen: The whole thing could be buried $\endgroup$
    – endolith
    Aug 12, 2011 at 21:41
  • $\begingroup$ Hmmm. At 4.3 km down with a portal opening, we're talking twice as deep as the deepest known cave. Although, I guess you could build it into a mountain. $\endgroup$
    – Jen
    Aug 12, 2011 at 22:26

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.

  • $\begingroup$ But where do you take as the effective center of the expansion (i.e. how well directed is the initial beam)? The work you show implicitly assume it is only 1 meter from the aperture, but the proper distance is to some degree in the gift of the operation staff and can be put at least as far back as the quadrupole spacing. I couldn't show it in closed form without a little work, but I am pretty sure that the optimal projection of the beam calls for de-tuning the focus: this is similar to what is done for long baseline neutrino work. $\endgroup$ Aug 13, 2011 at 19:24
  • $\begingroup$ @dmckee Yes, the neutrino beams are designed to optimize the focus so that long distances can be covered and flux can be deposited 700km or so away, but still very much diminished cdsweb.cern.ch/record/390779/files/sl-99-034.pdf with respect to protons on target (pot), see table 5. Something like 10**-9 on protons on target. We are talking of 42160 km for a satellite not 700. $\endgroup$
    – anna v
    Aug 14, 2011 at 3:51

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