 # II.1 Lunar Eclipse: The Moon's diameter and distance to the Earth

Before you start, beware that most of the following illustrations are wrong! Actually, you can't represent in a readable way at the same scale the Earth, the Moon and the Sun, and at the same time the distances Earth-Sun or Earth-Moon.

For instance, if we represent the Earth by a 1 cm diameter disk, the Moon diameter would be 3 mm and it would be 30 cm away from the Earth. But the Sun would then be 120 m far from the Earth, and its diameter would be more than 1 m!

### The principle of Lunar Eclipses

When the Moon moves around the Earth, it sometimes happens to pass into the Earth's shadow; this is a lunar eclipse. Figure 1

As the Moon's orbital plane is not the same as the Earth's, we have not an eclipse at each Full Moon, the Moon passing either "above", either "below" the Earth's shadow in most cases.

The goal of this exercise is to compute the Moon's diameter and its distance to the Earth from an eclipse picture. You will find several such pictures available within Astronomy On-Line, following the Lunar Eclipse on September 27, 1996. For instance, take a look at the Results from the Lunar Eclipse from where you will have access to some excellent images.

#### a) Diameter of the shadow

From the picture, we want to find the diameter of the Earth's shadow, considered as circular. You may try different methods.

Let's call the diameter of the Earth's shadow DO and the diameter of the Moon DL.

To draw the following pictures, you can use tracing paper, set down on the photo.

* Graphic method: Figure 2 You draw the medians of the chords [CI] and [DI] which cut each other at the centre of the Earth's shadow. You can get by measuring on the picture something like: DO on the picture = 66 mm DL on the picture = 31.5 mm so DO / DL = 2.1

x Computing DO: let's call CH = h and IH = a Figure 3 Trigonometry gives us: The two angles ICH and ILC have the same measure (ICH + CIL = 90 deg; and ILC + CIL = 90 deg;) or ICD and ILD are two inscribed angles that intercept the same arc. tan ICH = IH/CH = a/h tan ILC = CH/HL = h/HL, thus a/h = h/HL, so HL = h2/a and IL = a+h2/a = (a2+h2)/a (We find the same result with the theorem in the triangle ICL with the right angle ICL : CI2 = IH x IL). For instance, we get : h = 14 mm and a = 3 mm, then HL = 68 mm and DO/DL = 2,2

#### b) Diameter of the Moon

We have the diameter of the Earth: DT = 12 700 km

Let's call the Earth-Moon distance r.

First a simple approach. This may be useful if you only want to search for an approximative value. It works by considering the diameters of the Earth's shadow and of the Earth are equal, as if the Earth's shadow was cylindrical.

In that case you have only to know the picture scale to find the diameter of the Moon:

66 mm => 12 700 km

Divide 12 700 with 66 to find:

1 mm => 192 km (so the scale is about 1 / 200 000 000)

Then multiply by 31.5 to find:

31.5 mm => 6000 km

The correct value is 3500 km so this result is far from the real one, but it gives a rough estimate.

In the next example, you may take the correct diameter of the Earth's shadow (8700 km from the following computation) and then get the Moon's diameter from that result.

To be more concise, we now consider the Earth's shadow as a cone, and we compute its diameter at the Moon's distance. There are no big mathematical difficulties, but some of you may have find it more difficult to understand this method.

As it can be measured, the apparent diameter of the Sun is almost the same as the apparent diameter of the Moon (about half a degree). Figure 4

You only need to observe a solar eclipse and see the disk of the Moon covering exactly the disk of the Sun to be convinced of that. Figure 5

(Fu) is pointing to the "lower" edge of the Sun, whereas (Er) and (Fr') point to the "upper" R. We can consider (Er) and (Fr') as parallel, as the distance to the Sun is so big: the Earth-Sun distance is 400 times as big as the Earth-Moon distance and 12 000 times as big as the Earth's diameter! The angle AFG is equal to uFr', i.e. 0.5 degrees, which is the same thing as the Moon's diameter seen at the distance of the Moon. AG = Moon's diameter DL.

We considered (Er) and (Fr') as parallel:

BG = EF = Earth's diameter DT.

BG = BA+AG or DT = DO + DL

We found in question a) the rate DO/DL. With both equations, we can compute the Moon's diameter.

DO/DL = 2.2, so DO = 2.2 DL

DO + DL = DT = 12700

2,2DL + DL = 12700, so DL = 4000 km and DO = 8700 km

(In fact, the Moon's diameter DL is 3500 km)

To get the distance of the Moon now, we use the fact that its apparent diameter is 0.5 degrees. You may try to use trigonometry, but a simple proportion is enough.

With trigonometry: tan(0.5 deg) = DL/r, so r = 4000/tan(0.5 deg) = 460 000 km

With a proportion:

0.5 deg x 4000 km (DL)

Since a full circle is 360 degreees, multiply by 720:

giving 2 880 000 km for the circumference

Then, if 2 x pi x r = 2 880 000 km, r = 460 000 km.

The author of this exercise is Pierre Causeret (Comite de Liaison Enseignants Astronomes - CLEA), 20, rue Renaudot, F-10 160 Aix-en-Othe, France. Please direct any related questions or remarks to: Josee Sert (France) who will pass them on to CLEA.