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Every month the earth-moon lagrangian points must be in the shade of the moon for some time

For L1 and L3 probably on new moons, for L2 probably on full moons, and for L4 and L5 maybe when the moon happens to be in the way of the sunlight

How long do these eclipses at the Lagrangian points last?

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    $\begingroup$ The moon is inclined about 5.14° from the ecliptic, that should make eclipses somewhat rare (except maybe for L1, which might be eclipses by earth when the moon is eclipsed by earth). But this question is surprisingly hard to calculate. $\endgroup$ – Polygnome Mar 3 at 10:19
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    $\begingroup$ Time for someone to build a model with artificial massless objects at the L* points ! $\endgroup$ – Carl Witthoft Mar 3 at 14:08
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    $\begingroup$ @Speedphoenix it's late here and I only talked about eclipses from the Earth. I forgot that the Moon isn't transparent, never mind :-) $\endgroup$ – uhoh Mar 3 at 14:56
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    $\begingroup$ Given that L4 and L5 are as far from the Moon as the Earth is, I imagine, for a specific object, a decent approximation for an umbral encounter would be "About as often as a specific point on the Earth's surface experiences a Total Solar Eclipse, and lasting about as long." $\endgroup$ – notovny Mar 5 at 13:19
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    $\begingroup$ Do you want annular eclipses with full shadow or also the partial eclipses which can last a few seconds ? $\endgroup$ – Cornelisinspace Mar 5 at 18:20
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It's not true that the Lagrangian points are in the shade of the Moon every month, that happens only when the Sun is near one of the orbital nodes of the Moon.

Lunar nodes[1] Author: SuperManu

The nodes are the two points at which the orbit of the Moon intersects the ecliptic. Lunar and solar eclipses can occur when the nodes align with the Sun, roughly 2 times a year, and around those times the Lagrangian points can be shaded.

Earth-Moon

The line of nodes is perpendicular to the image above, and there the Moon has maximum inclination with Earth's orbit around the Sun. In this situation, at the L1 point the Earth will have an angular diameter ($\delta$) of about 2.2$⁰$, the L2 point a $\delta$ of about 1.6 $⁰$ and the L3, L4, and L5 points all will have a $\delta$ of about 1.9$⁰$ (with thanks to @HopDavid for his comment), so all the Langrangian points will be well outside Earth's shade. About half a year later the Moon and the L points will lie below Earth's orbital plane and again well outside its shade.

So during a year the duration of eclipses at the Langrangian points may vary from zero to a certain maximum when the Sun, Earth, Moon and the Langrangian points will all be aligned.

To make calculations of the maximum duration for each Lagrangian point not to complicated, I will assume that the Earth's orbit around the Sun and the Moon's orbit around the Earth are both circular and will not take into account that the lunar nodes precess around the ecliptic. In addition, I assume with eclipses by the Earth and the Moon that their centres will move over the centre of the Sun.

$R$ (distance Earth to Moon) = 384400 km, $L_1$ (distance L1 point to the Moon) = 63000 km.
Diameter of the Earth: $D_e$ = 12742 km, diameter Moon: $D_m$ = 3475 km.
The angular velocity $\omega_s$ of the Sun, seen from each Lagrangian point: $\omega_s = 2\pi/365 days$ = 0.000006156 rad/min.
The angular velocity $\omega_e$ of the Earth or Moon, seen from each Langrangian point: $\omega_e = 2\pi/27.322 days$ = 0.00008225 rad/min.
Angular diameter of the Sun, seen from the Earth-Moon system: $\delta_s = 1.4/150 $= 0.0093333 rad.

L1 point

I. Eclipse by the Earth.

Here the angular diameter of the Earth $\delta_e = D_e/(R - L_1)$ = 0.039645.
Because the angular velocities of the Sun and the Earth are in the same direction, they have to be subtracted from each other to get the angular velocity with which the Earth's disk moves over the Sun's disk: $\omega_0 = \omega_e - \omega_s$ = 0.000076094 rad/min.

Duration of the first penumbra (partial shadow): $\delta_s/\omega_0$ = 122 min.
Duration of the umbra (full shadow): $(\delta_e - 2\delta_s)/\omega_0$ = 275 min.
Duration of the last penumbra also lasts 122 min.

II. Eclipse by the Moon.

Here the angular diameter of the Moon $\delta_m = D_m/L_1$ = 0.05516.
Duration of the first and last penumbra: $\delta_s/\omega_0$ = 122 min.
Duration of the umbra: $(\delta_m - 2\delta_s)/\omega_0$ = 479 min.

L2 point

I. Eclipse by the Moon.

Because the distances of the L1 point and the L2 point to the Moon are about the same, and their distances to the Sun too, the duration of this eclipse will be about the same as the duration of the eclipse for the L1 point.

II. Eclipse by the Earth.

In the ideal case the L2 point, the Moon, the Earth, and the Sun will all be aligned, so the Earth would be "hidden" by the Moon, but in practice the L2 point could also be shaded by the Earth alone.

The angular diameter of the Earth will then be: $\delta_e = D_e/(R + L_1)$ = 0.02848.
Duration of the first and last penumbra: $\delta_s/\omega_0$ = 122 min.
Duration of the umbra: $(\delta_e - 2\delta_s)/\omega_0$ = 129 min.

L3 point

I. Eclipse by the Earth.

Here the angular diameter of the Earth $\delta_e = D_e/R$ = 0.033148.
Duration of the first and last penumbra: $\delta_s/\omega_0$ = 122 min.
Duration of the umbra: $(\delta_e - 2\delta_s)/\omega_0$ = 190 min.

II. Eclipse by the Moon.

In the ideal case the L3 point, the Earth, the Moon, and the Sun will all be aligned, so the Moon would be "hidden" by the Earth, but in practice the L3 point could also be shaded by the Moon alone.

The angular diameter of the Moon will then be: $\delta_m = D_m/2R$ = 0.00452.
This is about half the angular diameter of the Sun of 0.0093333 so in this case there will only be a partial shading.
Duration of the penumbra: $(\delta_m + \delta_s)/\omega_0$ = 182 min.
During this penumbra an antumbra will happen that lasts: $(\delta_s - \delta_m)/\omega_0$ = 63 min.

L4 and L5 point

I. Eclipse by the Earth.

In this case the values of the eclipses are the same as for the L3 point.

II. Eclipse by the Moon.

In theory, the L4 point could be shaded by the Moon when both are moving around the Earth and away from the Sun, and about half a month later the L5 point when it's moving towards the Sun.
Surprisingly the sum of the angular velocities of either the L4 point or L5 point and that of the Moon, seen from the Sun, added with the angular velocity of the Earth-Moon system around the Sun is again $\omega_0$ !
When in these cases eclipses happen, the distance between the Moon and L4 or L5 point will be about the distance from Moon to Earth.

Angular diameter of the Moon: $\delta_m = D_m/R$ = 0.00904.
This is somewhat less than the angular diameter of the Sun of 0.0093333, so there's only partial shading and an annular eclipse.
Duration of the penumbra: $(\delta_m + \delta_s)/\omega_0$ = 241 min.
Duration of the antumbra or annular eclipse: $(\delta_s - \delta_m)/\omega_0$ = about 4 min.

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    $\begingroup$ From L1 earth's angular diameter is about 2.2º. From L3, 4 & 5 it's about 1.9º. From L2 it is about 1.6º $\endgroup$ – HopDavid Mar 7 at 13:44
  • $\begingroup$ @HopDavid Thanks, i forgot to double Earth's radius. $\endgroup$ – Cornelisinspace Mar 7 at 14:00

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