<p>The angle between the inner and outer edges of the penumbra is the same as the Sun's apparent angular diameter.
At 1.0 au from the Sun, that's 32 arcmin or 0.0093 radian.
At the distance from the Moon to EM-L2, this angle spans 0.0093 * 64700 km = 602 km.
The Moon's diameter is 3474 km, so the umbra diameter would be 2872 km and the penumbra diameter would be 4076 km.</p>

CC BY-SA 4.0
2019-04-19T23:53:13.787Z
Maybe you should say "At *a distance equal to the* EM-L2's *distance*"? At EM-L2 it's the Earth that's behind the Moon, not the Sun. EM-L2 is along the Earth-Moon line and has nothing to do with the Sun. A solar eclipse at EM-L2 would be quite rare and transient.
uhoh
CC BY-SA 4.0
2019-04-20T01:30:27.257Z
@uhoh When the Moon phase is full, EM-L2 sees the Sun in conjunction with the Earth and Moon. OP didn't ask about eclipses.
Mike G
CC BY-SA 4.0
2019-04-20T01:55:56.373Z
okay I'm calling EM-L2 being in the Moon's umbra *an eclipse*, I can't see any way that can't be a correct usage of the term. Let's do it this way: [What fraction of the time is EM-L2 in darkness on average?](https://astronomy.stackexchange.com/q/30556/7982)
uhoh
CC BY-SA 4.0
2019-04-19T17:28:16.520Z
2019-04-20T01:29:53.400Z
2019-04-20T01:29:53.400Z
Mike G
<p>Measuring the total width of the Moon's umbra and penumbra at EM-L2, <a href="http://www.oasis-nss.org/articles/2001/08/lagrange.png" rel="nofollow noreferrer">64,700 km</a>, how wide would the shadow be?  </p>

CC BY-SA 4.0
2019-04-19T00:50:56.327Z
Why would you keep your "sources" secret?
uhoh
CC BY-SA 4.0
2019-04-19T00:53:28.283Z
Have a look at [this answer](https://space.stackexchange.com/a/32192/12102) to the question *What is the darkest orbit around Earth?* I think your geometry is wrong. At EM-L2 it's the Earth that's behind the Moon, not the Sun. A solar eclipse at EM-L2 would be quite rare and transient. Also see [Are there any (Lagrange) points in the Solar System in perpetual shade?](https://space.stackexchange.com/q/3565/12102)
uhoh
CC BY-SA 4.0
2019-04-19T03:18:57.980Z
@ uhoh I didn't have much faith in my source that is why I did not cite it. I will look for it again.
Bob516
CC BY-SA 4.0
2019-04-19T03:25:13.367Z
You can quote from [this table](https://en.wikipedia.org/wiki/Lagrangian_point) by subtracting L1: 3.844E05 - 3.264E05 = **58010 km** and L2: 4.489E05 - 3.844E05 = **64500 km**
uhoh
CC BY-SA 4.0
2019-04-19T03:25:26.830Z
@uhoh You are correct, my geometry in wrong.
Bob516
CC BY-SA 4.0
2019-04-19T03:26:58.357Z
It is always okay to answer your own question in Stack Exchange. I'd say go ahead and write it up!
uhoh
CC BY-SA 4.0
2019-04-19T03:29:25.593Z
When I figure it out I will.  Assuming I do not mess up again.
Bob516
<the-moon><lagrange-point><eclipse-non-earth-related><umbra>
How wide is the Moon's umbra and penumbra at EM-L2?
Bob516
# How wide is the Moon's umbra and penumbra at EM-L2?

(Asked by: username_0 on 2019-04-16)

Measuring the total width of the Moon's umbra and penumbra at EM-L2, [64,700 km](http://www.oasis-nss.org/articles/2001/08/lagrange.png), how wide would the shadow be?

**Tags:** the-moon, lagrange-point, eclipse-non-earth-related, umbra

> **Comments:**
> **username_1**: Why would you keep your "sources" secret?  
> **username_1**: Have a look at [this answer](https://space.stackexchange.com/a/32192/12102) to the question *What is the darkest orbit around Earth?* I think your geometry is wrong. At EM-L2 it's the Earth that's behind the Moon, not the Sun. A solar eclipse at EM-L2 would be quite rare and transient. Also see [Are there any (Lagrange) points in the Solar System in perpetual shade?](https://space.stackexchange.com/q/3565/12102)  
> **username_0**: @ username_1 I didn't have much faith in my source that is why I did not cite it. I will look for it again.  
> **username_1**: You can quote from [this table](https://en.wikipedia.org/wiki/Lagrangian_point) by subtracting L1: 3.844E05 - 3.264E05 = **58010 km** and L2: 4.489E05 - 3.844E05 = **64500 km**  
> **username_0**: @username_1 You are correct, my geometry in wrong.  
> **username_1**: It is always okay to answer your own question in Stack Exchange. I'd say go ahead and write it up!  
> **username_0**: When I figure it out I will.  Assuming I do not mess up again.  

---

## 1 answers

### Answer by username_2 on 2019-04-19. Score: 3 | Accepted answer

The angle between the inner and outer edges of the penumbra is the same as the Sun's apparent angular diameter.
At 1.0 au from the Sun, that's 32 arcmin or 0.0093 radian.
At the distance from the Moon to EM-L2, this angle spans 0.0093 \* 64700 km = 602 km.
The Moon's diameter is 3474 km, so the umbra diameter would be 2872 km and the penumbra diameter would be 4076 km.

> **Comments:**
> **username_1**: Maybe you should say "At *a distance equal to the* EM-L2's *distance*"? At EM-L2 it's the Earth that's behind the Moon, not the Sun. EM-L2 is along the Earth-Moon line and has nothing to do with the Sun. A solar eclipse at EM-L2 would be quite rare and transient.  
> **username_2**: @username_1 When the Moon phase is full, EM-L2 sees the Sun in conjunction with the Earth and Moon. OP didn't ask about eclipses.  
> **username_1**: okay I'm calling EM-L2 being in the Moon's umbra *an eclipse*, I can't see any way that can't be a correct usage of the term. Let's do it this way: [What fraction of the time is EM-L2 in darkness on average?](https://astronomy.stackexchange.com/q/30556/7982)  

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