To what extent is the L2 point protected in comparison to the surface, if at all? Since the L2 point might be in the umbra of Mars, it could be shielded against the sun. It would also be helpful if someone could provide the calculation.

  • $\begingroup$ Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun. $\endgroup$ – Magic Octopus Urn Oct 7 '18 at 5:10

The Lagrange points of Mars might be less useful than that of Earth's because Mars' orbit is more elliptical and so there are more perturbing forces.

Nonetheless we can do a simple exercise.

In this answer to How far would the Mars L1 Lagrangian Point be from Mars? I showed how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem (or CR3BP) scenario.

From geometry we can see that if a right circular cone with base radius $r_{sun}$ has a radius $r$ at a distance $D$, then the vertex will be beyond that point by a distance

$$ D\frac{r}{r_{sun}-r}.$$

Using radii for Mars, Earth and the Sun of 3396, 6378, and 696392 km, I get the following, which shows that whether or not the Lagrange points are meaningful and/or useful for an elliptical orbit like Mars.

  • Mars' solar umbra extends about about 30,000 km beyond Sun-Mars L2
  • Earth's the solar umbra ends about 100,000 km before Sun-Earth L2


    MARS      to Sun        to L1      to L2    r    r/(r_sun-r) "end of umbra"
perihelion  206,700,000    981,000    985,000  3,396  0.00490      1,013,000
aphelion    249,200,000  1,183,000  1,187,000    "        "        1,221,000

perihelion  147,100,000  1,466,000  1,476,000  6,370   0.00926     1,360,000
aphelion    152,100,000  1,516,000  1,527,000    "        "        1,406,000

Staying within the umbra near the "L2" point would likely require less delta-v for Mars than for Earth, but either way it's an unstable point and so you need to continually do small station-keeping burns to remain protected. The umbra protects you from line-of sight radiation like photons (light, X-rays, gamma-rays) but may be less protective for ballistic charged particles from a coronal mass ejection or CME because they may be on trajectories that when projected backwards, come from points beyond the disk of the Sun for various reasons. That's an excellent *next question!

For the "normal" solar wind, it may be even more likely to expand into the shadow since wether it's ionized or not, it will have some transverse velocity distribution associated with it.

| improve this answer | |
  • 1
    $\begingroup$ +1 Very good point about the transverse velocity distribution; this is very much the case. "Spiral" might not be the technically correct description but that does help envisage the trajectory of the solar wind. $\endgroup$ – Puffin Apr 7 at 21:55
  • $\begingroup$ @Puffin I left "some transverse velocity distribution" vague because while I had some notion of spiralness in mind I wasn't sure how much of those shapes indicated actual rotation versus just a radial spray from a rotating source. I also was thinking about frozen-out thermal velocity; random motion of the particles left over from when their density was still high enough for collisions at the thermal velocities associated with their temperature. I should probably post a new question about these I'll ping you when I do thanks! $\endgroup$ – uhoh Apr 7 at 23:49
  • 1
    $\begingroup$ sometime in the past I picked up the idea that there is indeed more too it than "just a radial spray from a rotating source", I'm sure I've heard space weather folks talk about the challenge of matching effects measured in GEO with locations on the solar disc in these (i.e. spiral-like) terms $\endgroup$ – Puffin Apr 8 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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