During an eclipse, how big is the shadow of the moon on the earth?

Question

This picture was taken from the ISS during a solar eclipse.

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?

Answer 1

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 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:

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

Source: Geometry of a Total Solar Eclipse

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 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? (also see What are the “Moon L, B, C” angles shown in this solar eclipse simulation?)

Here's a screenshot:

Answer 2

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 $r_p \approx 2 R_m$. Image © Flavio Salvati 2020

Answer 3

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!

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:

...so I guess that, as most of the time, the full answer is "it's complicated..." but have fun!