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

• usernumber

2 years ago

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? https://xkcd.com/1276/ about as big as the M25 freeway around London when the light comes near the shadow comes small and when the light goes far becomes long

• 2 years ago

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 $$r_p \approx 2 R_m$$.  +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 yes, just edited my answer! *Beautiful!* :-)