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 I show how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem 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 [CME][3] 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.


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