<blockquote>
  <p>But how big is this shadow? How many kilometers is its diameter?</p>
</blockquote>

<p>That's a photo of the umbra and penumbra on the surface of the Earth taken from Space. It's a little distorted because it's not directly under the ISS but far off near the terminator.</p>

<p>It's hard to pin down the size of the penumbra because it's fuzzy and fades near the edges, <em>but if you could see</em> the very edges then it would be twice<sup>†</sup> the diameter of the moon or very roughly 6900 kilometers total.</p>

<p><sup>†</sup>Actually this is only true by coincidence because the <em>angular diameter</em> of the Sun happens to be the same as that of the Moon. <a href="https://astronomy.stackexchange.com/a/35650/7982">@Flaffo's answer</a> does a good job of explaining how to calculate the diameter of the umbra, and that math could probably be extended to calculate the diameter of the penumbra as well.</p>

<p>The umbra has a well defined diameter but the size varies a lot due to variation in the distance from the Moon to the Earth since it's orbit is not circular. Sometimes the Moon is so far away that it can't fill the Sun and there is no umbra at all, that's called an annular eclipse. </p>

<p><a href="https://en.wikipedia.org/wiki/Solar_eclipse#Path" rel="noreferrer">Wikipedia says</a>:</p>

<blockquote>
  <p>Typically, the umbra is 100–160 km wide, while the penumbral diameter is in excess of 6400 km.</p>
</blockquote>

<p><a href="https://i.sstatic.net/5E5fZ.png" rel="noreferrer"><img src="https://i.sstatic.net/5E5fZ.png" alt="enter image description here"></a></p>

<p>Source: <a href="https://en.wikipedia.org/wiki/File:Geometry_of_a_Total_Solar_Eclipse.svg" rel="noreferrer">Geometry of a Total Solar Eclipse</a></p>

<p>You can see an example of a very detailed simulation of just the umbra moving across the Earth's surface in the NASA Goddard video <a href="https://youtu.be/MJgXaqW3md8" rel="noreferrer">Tracing the 2017 Solar Eclipse</a> The precise 3D shape of the Moon generates the shadow and it then moves over the contour of the Earth's topography. If you think the shape is weird I agree! See answers to <a href="https://space.stackexchange.com/q/22373/12102">The Moon's shadow could not possibly look like this — could it?</a> (also see <a href="https://space.stackexchange.com/q/22375/12102">What are the “Moon L, B, C” angles shown in this solar eclipse simulation?</a>)</p>

<p>Here's a screenshot: </p>

<p><a href="https://i.sstatic.net/4Ku7a.jpg" rel="noreferrer"><img src="https://i.sstatic.net/4Ku7a.jpg" alt="screenshot from the NASA Goddard video Tracing the 2017 Solar Eclipse"></a></p>

<p><a href="https://i.sstatic.net/1mDOn.jpg" rel="noreferrer"><img src="https://i.sstatic.net/1mDOn.jpg" alt="screenshot from the NASA Goddard video Tracing the 2017 Solar Eclipse"></a></p>

CC BY-SA 4.0
2020-03-31T01:31:49.040Z
The width of the penumbra is twice the width of the Moon, *minus the width of the umbra*.
Mark
CC BY-SA 4.0
2020-03-31T01:46:42.473Z
@Mark wow thanks! Yes that's right. At zero Moon-Earth distance the umbra and penumbra are coincident and equal to one moon diameter, As the Moon moves away from Earth the umbra gets smaller and the penumbra gets larger by the same amount until the umbra reaches zero and the penumbra is exactly two Moon diameters at which point the Sun and the Moon have exactly equal angular diameters.
uhoh
CC BY-SA 4.0
2020-03-30T11:56:28.477Z
2020-03-31T01:07:51.903Z
2020-03-31T01:07:51.903Z
uhoh
<p>The other answers are great, but if you want an handy tool to explore the next and the past eclipses, check out <a href="https://www.timeanddate.com/eclipse/list.html?starty=2020" rel="nofollow noreferrer">this super cool website</a>!</p>

<p><a href="https://i.sstatic.net/70tub.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/70tub.png" alt="enter image description here"></a></p>

<p>It allows you to search all the eclipses (solar and lunar + Mercury and Venus transits) for the past and the next couple of centuries. Even more interesting are the tracks of the shadows for which you can get both a summary decade-by-decade, and a detailed analyses for each eclipse. And what about simulating the sky view? Yes it does that too!!!</p>

<p>Curiously in the 2040s there will be several weirdly shaped solar eclipses, like the total one taking place on 9 April 2043 over East Russia which is "clipped" because the sun is not yet risen at the beginning of the event:</p>

<p><a href="https://i.sstatic.net/dEN3R.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dEN3R.png" alt="enter image description here"></a></p>

<p>...so I guess that, as most of the time, the full answer is "<a href="https://medium.com/@rauf.rahmancz/dunning-kruger-effect-in-software-development-life-cycle-sdlc-f1cf18181600" rel="nofollow noreferrer">it's complicated...</a>" but have fun!</p>

CC BY-SA 4.0
2020-03-31T20:19:40.880Z
2020-03-31T20:26:19.810Z
2020-03-31T20:26:19.810Z
DarioP
<p>The umbra has a well defined diameter but the size varies due to the eccentricity of the orbits of the Earth and of the Moon. The Moon may be so far away that it can't fill the solar disk at all (for instance Moon at apogee and Earth at perihelion). </p>

<p>We can, however, theoretically determine the diameter of the shadow that the Moon casts on the Earth. The calculation only requires elementary geometry and a nice image. We then obtain the maximum radius:</p>

<p><span class="math-container">$\displaystyle r_{u} = R_{m} - \frac{d_{m}- R_{e}}{d_{e} - d_{m}} (R_{s} - R_{m})$</span></p>

<p>where <span class="math-container">$R_m, R_e, R_s$</span> are the radii of the Moon, Earth and Sun respectively, <span class="math-container">$d_{m}$</span> is the distance Moon-Earth, <span class="math-container">$d_{e}$</span> is the distance Earth-Sun. </p>

<p>We can investigate several cases by varying the distances <span class="math-container">$d_{m}$</span> and <span class="math-container">$d_{e}$</span>, in accordance with the eccentricities of the orbits. </p>

<p>To find the maximum possible radius of the umbra, take <span class="math-container">$d_m = a_m (1- e_m)$</span> (Moon at perigee) and <span class="math-container">$d_e = a_e (1+e_e)$</span> (Earth at aphelion), where <span class="math-container">$e_e, e_m$</span> are the eccentricities of the orbits of the Earth and Moon respectively, and <span class="math-container">$a_e, a_m$</span> their semi-major axis. This makes sense because to get the widest lunar eclipse we want a big (and therefore close) Moon and a small (and far) Sun. 
Substituting some typical values we obtain <span class="math-container">$r_u \approx 120$</span>km (around <span class="math-container">$240$</span> km maximum width). This situation is, however, extremely unlikely (I estimate it happens around once every century).</p>

<p>This equation also tells us that, on average (<span class="math-container">$d_m = a_m, d_e = a_e$</span>), we do not see any eclipse (<span class="math-container">$r_u$</span> would be negative). </p>

<p>We need the Moon to be close to perigee, in which case, assuming  average distance for the Earth (<span class="math-container">$d_e=a_e$</span>), we get <span class="math-container">$r_u \approx 80$</span> km and a width of <span class="math-container">$160$</span>km. A result which might sound familiar by now!</p>

<p>Very similarly, as @uhoh suggested, we can calculate the width of the penumbra. Now, instead of considering the ray <span class="math-container">$T_{s,1} T_{l,1}$</span>, we take the ray <span class="math-container">$T_{s,2} T_{l,1}$</span>. Clearly, this is equivalent to take <span class="math-container">$R_s \rightarrow - R_s$</span>. We then get</p>

<p><span class="math-container">$\displaystyle r_{p} = R_{m} + \frac{d_{m}- R_{e}}{d_{e} - d_{m}} (R_{s} + R_{m})$</span></p>

<p>Now the maximum radius of the penumbra is obtained when <span class="math-container">$d_e = a_e (1-e_e)$</span> (Earth at perihelion) and <span class="math-container">$d_m = a_m (1+ e_m)$</span> (Moon at apogee). In this case we get <span class="math-container">$r_p \approx 3650$</span>km and a width of <span class="math-container">$7300$</span>km.</p>

<p>If instead we take average distance for the Earth we get <span class="math-container">$r_p = 3600$</span>km, so a total of <span class="math-container">$7200$</span>km, not far from @uhoh's answer.</p>

<p>In the case of minimum width, we take <span class="math-container">$d_e = a_e (1+e_e)$</span> (Earth at aphelion) and <span class="math-container">$d_m = a_m (1- e_m)$</span> (Moon at perigee). We then get <span class="math-container">$r_p = 3400$</span>km, so a total of <span class="math-container">$6800$</span>km. </p>

<p>In any case, the width is approximately twice the diameter of the Moon (<span class="math-container">$7000$</span>km), but notice this is only a coincidence and is due to the fact that the angular diameters of the Moon and Sun are very close to each other. Indeed, simplifying the previous equation, we can neglect <span class="math-container">$R_e$</span> at the numerator, <span class="math-container">$d_m$</span> at the denominator and approximate <span class="math-container">$R_s -R_m$</span> with only <span class="math-container">$R_s$</span>. We then see that</p>

<p><span class="math-container">$\displaystyle r_p = R_m + \frac{d_m}{d_e} R_s = R_m \Big(1+ \frac{R_s / d_e}{R_m / d_m} \Big)$</span> </p>

<p>but <span class="math-container">$R_s / d_e$</span> and <span class="math-container">$R_m / d_m$</span> are the angular diameters of the Sun and Moon, therefore the fraction is approximately 1 (actually 1.03 on average) and we recover <span class="math-container">$r_p \approx 2 R_m$</span>.
<a href="https://i.sstatic.net/8qQnX.png" rel="noreferrer"><img src="https://i.sstatic.net/8qQnX.png" alt=""></a>
Image © Flavio Salvati 2020</p>

CC BY-SA 4.0
2020-03-31T01:11:24.850Z
`+1` I wonder if you can also properly calculate the diameter of the penumbra? I've adjusted my answer to note that I used the fact that the Sun's angular diameter is similar to the Moon's to get to the conclusion that the extremes of the penumbra are about twice the diameter of the Moon. I think you are close to calculating the penumbra's diameter correctly here, I wonder if you could add that as well?
uhoh
CC BY-SA 4.0
2020-03-31T08:14:03.263Z
@uhoh yes, just edited my answer!
JustThinking
CC BY-SA 4.0
2020-03-31T08:27:25.580Z
*Beautiful!* :-)
uhoh
CC BY-SA 4.0
2020-03-30T23:23:08.330Z
2020-03-31T08:12:42.207Z
2020-03-31T08:12:42.207Z
JustThinking
<p><a href="https://www.cbsnews.com/pictures/solar-eclipse-august-21-2017/23/" rel="noreferrer">This picture</a> was taken from the ISS during a solar eclipse. </p>

<p><a href="https://i.sstatic.net/z3737.png" rel="noreferrer"><img src="https://i.sstatic.net/z3737.png" alt="enter image description here"></a></p>

<p>You can see the shadow of the Moon on the surface of the Earth. But how big is this shadow? How many kilometers is its diameter?</p>

CC BY-SA 4.0
2020-04-02T19:42:13.587Z
https://xkcd.com/1276/ about as big as the M25 freeway around London
Jacob Krall
CC BY-SA 4.0
2020-04-06T17:00:37.457Z
when the light comes near the shadow comes small
and when the light goes far becomes long
nahom get.
<the-moon><solar-eclipse>
During an eclipse, how big is the shadow of the moon on the earth?
usernumber
# During an eclipse, how big is the shadow of the moon on the earth?

(Asked by: username_0 on 2020-03-30)

[This picture](https://www.cbsnews.com/pictures/solar-eclipse-august-21-2017/23/) was taken from the ISS during a solar eclipse.

[![enter image description here](https://i.sstatic.net/z3737.png)](https://i.sstatic.net/z3737.png)

You can see the shadow of the Moon on the surface of the Earth. But how big is this shadow? How many kilometers is its diameter?

**Tags:** the-moon, solar-eclipse
> **Comments:**
> **username_1**: https://xkcd.com/1276/ about as big as the M25 freeway around London  
> **username_2**: when the light comes near the shadow comes small and when the light goes far becomes long  

---

## 3 answers

### Answer by username_6 on 2020-03-30. Score: 20 | Accepted answer

The umbra has a well defined diameter but the size varies due to the eccentricity of the orbits of the Earth and of the Moon. The Moon may be so far away that it can't fill the solar disk at all (for instance Moon at apogee and Earth at perihelion).

We can, however, theoretically determine the diameter of the shadow that the Moon casts on the Earth. The calculation only requires elementary geometry and a nice image. We then obtain the maximum radius:

$\displaystyle r\_{u} = R\_{m} - \frac{d\_{m}- R\_{e}}{d\_{e} - d\_{m}} (R\_{s} - R\_{m})$

where $R\_m, R\_e, R\_s$ are the radii of the Moon, Earth and Sun respectively, $d\_{m}$ is the distance Moon-Earth, $d\_{e}$ is the distance Earth-Sun.

We can investigate several cases by varying the distances $d\_{m}$ and $d\_{e}$, in accordance with the eccentricities of the orbits.

To find the maximum possible radius of the umbra, take $d\_m = a\_m (1- e\_m)$ (Moon at perigee) and $d\_e = a\_e (1+e\_e)$ (Earth at aphelion), where $e\_e, e\_m$ are the eccentricities of the orbits of the Earth and Moon respectively, and $a\_e, a\_m$ their semi-major axis. This makes sense because to get the widest lunar eclipse we want a big (and therefore close) Moon and a small (and far) Sun.
Substituting some typical values we obtain $r\_u \approx 120$km (around $240$ km maximum width). This situation is, however, extremely unlikely (I estimate it happens around once every century).

This equation also tells us that, on average ($d\_m = a\_m, d\_e = a\_e$), we do not see any eclipse ($r\_u$ would be negative).

We need the Moon to be close to perigee, in which case, assuming average distance for the Earth ($d\_e=a\_e$), we get $r\_u \approx 80$ km and a width of $160$km. A result which might sound familiar by now!

Very similarly, as @uhoh suggested, we can calculate the width of the penumbra. Now, instead of considering the ray $T\_{s,1} T\_{l,1}$, we take the ray $T\_{s,2} T\_{l,1}$. Clearly, this is equivalent to take $R\_s \rightarrow - R\_s$. We then get

$\displaystyle r\_{p} = R\_{m} + \frac{d\_{m}- R\_{e}}{d\_{e} - d\_{m}} (R\_{s} + R\_{m})$

Now the maximum radius of the penumbra is obtained when $d\_e = a\_e (1-e\_e)$ (Earth at perihelion) and $d\_m = a\_m (1+ e\_m)$ (Moon at apogee). In this case we get $r\_p \approx 3650$km and a width of $7300$km.

If instead we take average distance for the Earth we get $r\_p = 3600$km, so a total of $7200$km, not far from @uhoh's answer.

In the case of minimum width, we take $d\_e = a\_e (1+e\_e)$ (Earth at aphelion) and $d\_m = a\_m (1- e\_m)$ (Moon at perigee). We then get $r\_p = 3400$km, so a total of $6800$km.

In any case, the width is approximately twice the diameter of the Moon ($7000$km), but notice this is only a coincidence and is due to the fact that the angular diameters of the Moon and Sun are very close to each other. Indeed, simplifying the previous equation, we can neglect $R\_e$ at the numerator, $d\_m$ at the denominator and approximate $R\_s -R\_m$ with only $R\_s$. We then see that

$\displaystyle r\_p = R\_m + \frac{d\_m}{d\_e} R\_s = R\_m \Big(1+ \frac{R\_s / d\_e}{R\_m / d\_m} \Big)$

but $R\_s / d\_e$ and $R\_m / d\_m$ are the angular diameters of the Sun and Moon, therefore the fraction is approximately 1 (actually 1.03 on average) and we recover $r\_p \approx 2 R\_m$.
[![](https://i.sstatic.net/8qQnX.png)](https://i.sstatic.net/8qQnX.png)
Image © Flavio Salvati 2020

> **Comments:**
> **username_3**: `+1` I wonder if you can also properly calculate the diameter of the penumbra? I've adjusted my answer to note that I used the fact that the Sun's angular diameter is similar to the Moon's to get to the conclusion that the extremes of the penumbra are about twice the diameter of the Moon. I think you are close to calculating the penumbra's diameter correctly here, I wonder if you could add that as well?  
> **username_6**: @uhoh yes, just edited my answer!  
> **username_3**: *Beautiful!* :-)  

---

### Answer by username_3 on 2020-03-30. Score: 24

> But how big is this shadow? How many kilometers is its diameter?

That's a photo of the umbra and penumbra on the surface of the Earth taken from Space. It's a little distorted because it's not directly under the ISS but far off near the terminator.

It's hard to pin down the size of the penumbra because it's fuzzy and fades near the edges, *but if you could see* the very edges then it would be twice† the diameter of the moon or very roughly 6900 kilometers total.

†Actually this is only true by coincidence because the *angular diameter* of the Sun happens to be the same as that of the Moon. [@Flaffo's answer](https://astronomy.stackexchange.com/a/35650/7982) does a good job of explaining how to calculate the diameter of the umbra, and that math could probably be extended to calculate the diameter of the penumbra as well.

The umbra has a well defined diameter but the size varies a lot due to variation in the distance from the Moon to the Earth since it's orbit is not circular. Sometimes the Moon is so far away that it can't fill the Sun and there is no umbra at all, that's called an annular eclipse.

[Wikipedia says](https://en.wikipedia.org/wiki/Solar_eclipse#Path):

> Typically, the umbra is 100–160 km wide, while the penumbral diameter is in excess of 6400 km.

[![enter image description here](https://i.sstatic.net/5E5fZ.png)](https://i.sstatic.net/5E5fZ.png)

Source: [Geometry of a Total Solar Eclipse](https://en.wikipedia.org/wiki/File:Geometry_of_a_Total_Solar_Eclipse.svg)

You can see an example of a very detailed simulation of just the umbra moving across the Earth's surface in the NASA Goddard video [Tracing the 2017 Solar Eclipse](https://youtu.be/MJgXaqW3md8) The precise 3D shape of the Moon generates the shadow and it then moves over the contour of the Earth's topography. If you think the shape is weird I agree! See answers to [The Moon's shadow could not possibly look like this — could it?](https://space.stackexchange.com/q/22373/12102) (also see [What are the “Moon L, B, C” angles shown in this solar eclipse simulation?](https://space.stackexchange.com/q/22375/12102))

Here's a screenshot:

[![screenshot from the NASA Goddard video Tracing the 2017 Solar Eclipse](https://i.sstatic.net/4Ku7a.jpg)](https://i.sstatic.net/4Ku7a.jpg)

[![screenshot from the NASA Goddard video Tracing the 2017 Solar Eclipse](https://i.sstatic.net/1mDOn.jpg)](https://i.sstatic.net/1mDOn.jpg)

> **Comments:**
> **username_4**: The width of the penumbra is twice the width of the Moon, *minus the width of the umbra*.  
> **username_3**: @Mark wow thanks! Yes that's right. At zero Moon-Earth distance the umbra and penumbra are coincident and equal to one moon diameter, As the Moon moves away from Earth the umbra gets smaller and the penumbra gets larger by the same amount until the umbra reaches zero and the penumbra is exactly two Moon diameters at which point the Sun and the Moon have exactly equal angular diameters.  

---

### Answer by username_5 on 2020-03-31. Score: 2

The other answers are great, but if you want an handy tool to explore the next and the past eclipses, check out [this super cool website](https://www.timeanddate.com/eclipse/list.html?starty=2020)!

[![enter image description here](https://i.sstatic.net/70tub.png)](https://i.sstatic.net/70tub.png)

It allows you to search all the eclipses (solar and lunar + Mercury and Venus transits) for the past and the next couple of centuries. Even more interesting are the tracks of the shadows for which you can get both a summary decade-by-decade, and a detailed analyses for each eclipse. And what about simulating the sky view? Yes it does that too!!!

Curiously in the 2040s there will be several weirdly shaped solar eclipses, like the total one taking place on 9 April 2043 over East Russia which is "clipped" because the sun is not yet risen at the beginning of the event:

[![enter image description here](https://i.sstatic.net/dEN3R.png)](https://i.sstatic.net/dEN3R.png)

...so I guess that, as most of the time, the full answer is "[it's complicated...](https://medium.com/@rauf.rahmancz/dunning-kruger-effect-in-software-development-life-cycle-sdlc-f1cf18181600)" but have fun!


---

