75. Radius of particle gyration
Hi Dr. Stern,
You have a very useful web site!
I'm trying to estimate parameters of a 50 keV electron's helical motion in
the aurora at about 100 km, assuming no collisions. (I'm just going to
calculate the radius, and distance traveled in one orbit.)
The detail I can't seem to find is, very roughly, the Earth's magnetic field
at that height above the poles. I'm guessing its not much weaker than it is
at the Earth's surface, but I'm not sure.
And is it reasonable to guess the perpendicular and parallel components of
velocity are roughly equal (or can be for some typical particles)?
Reply
To save you some work I looked up "Introduction to the Physics of Space" by Olbert and Rossi, 1970. They give the gyration radius in cm as
R = (144 / B) √ W for protons
R = (3.37 / B) √ W for electrons
where B is the field in Gauss (1 Gauss is the same as 100,000 nanotesla or nT), √ means square root and W is the kinetic energy in ev (electron volts) of the part of the motion perpendicular to the field line. The field above the auroral region may be around 50,000 nT or half a gauss (negligible difference from ground level) so you get about 15 meters.
An energy of 50 keV is rather high for the aurora; 5 keV may be closer, the typical range is 1 to 15 keV. And yes--auroral electrons arrive about evenly distributed in all directions (from above--not from below!) so your assumption of equal parallel and perpendicular components is reasonable.