Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I'm trying to calculate the protection provided by the earth's magnetic field from HZE particles to astronauts in low earth orbit, such as those on the ISS.

HZE particles are often quoted as a danger of travel in deep space. But given their high energies I am skeptical that the Earth's magnetic field provides significant protection to humans outside the atmosphere (the ISS orbits ~370km above the Earth).

I have been able to find data on the energy of the particles expressed in MeV or GeV and the strength of the magnetic field expressed in microteslas.

What formulas would be applicable in calculating whether 1) the particle is deflected, and 2) if it's not deflected the energy reduction by the time it reaches a given altitude?

share|cite|improve this question
And? What is the question here? Are you unsure of the meaning of the units; unsure of the physics or what? I mean, the radius of curvature goes by $p_{\perp}/qB$ (you can look up the full derivation in any E&M text). – dmckee Jan 3 '13 at 22:13
I'm trying to calculate the protection provided by the earth's magnetic field from HZE particles to astronauts in LEO, eg: the ISS. – Patrick Ritchie Jan 4 '13 at 0:00

1 Answer 1

The basic physcs here is the Lorentz force on a moving charge, $q$, with velocity, $\vec{v}$ due to a magnetic field $\vec{B}$. $$ F = q \left( \vec{E} + \vec{v} \times \vec{B} \right) \quad ,$$ where we ignore the electrical field so we get $$ \vec{F}_B = q \vec{v} \times \vec{B}\quad .$$

The cross-product implies that the force acts perpendicularly to both field and velocity which means that the general path is (locally) helix with it's axis along the direction of the magnetic field.

The accelration is $\vec{a}_B = \frac{\vec{F}_B}{m} = \frac{q v_\perp B}{m}$ which implies a radius of curvature of in the plane normal to the magnetic field. Noting that this is a centripital force we use $a = \frac{v^2_{\perp}}{r}$ to find the radius of curvature as $$r_\perp = \frac{v^2_{\perp}}{a} = \frac{m v_\perp}{qB} = \frac{p_\perp}{qB} \quad .$$

Despite the classical derivation the final form involving momentum, charge and field is relativisitcally correct.

Momentum parallel to the field is unaffected.

Now, saying that the path is a helix is only correct for uniform fields which is not true over scales a significant fraction of the Earth's radius, but it is a good approximation of scales of a hundred kilometers or less, which suggests a reasonable step size for a low precision simulation.

Moving beyond the physics you are asking about the the desired result, these particles never lose any energy due to the effects of magnetic fields and are just pointed in different direction. The result is to first order no change in the flux ariving at objects in orbit.

There are some places where that first order result is insuffient (the flux can actually get amplified near the magnetic poles where particles are revesed and could pass the target twice), but it is a place to start.

Note that I put the charge in the wrong place in the comment I dashed off earlier and have used moderator superpower to fix it post facto.

share|cite|improve this answer
Maybe you should emphasize the conclusion that no protection is provided by the earth's magnetic field because the flux does not change locally. Probably the OP is confused by the deviations by the sun's magnetic field, which happen because of the much larger volume that magnetic field affects (c.f.cosmic rays and climate) – anna v Jan 4 '13 at 5:30
that should be "than the one the earth's magnetic field affects" – anna v Jan 4 '13 at 9:42
Thanks, i'll dig into these and see if I can get some results. – Patrick Ritchie Jan 4 '13 at 15:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.