II.2 How to calculate the duration of a Lunar Eclipse

The Problem: We know the time of the maximum of the eclipse (Moon in position X).We want to compute the remainder of the time-scale, that is, when the Moon is at V, W, Y and Z.

Figure 3

Data

To solve this problem and find the answers, we first need some data:

Let us now try to calculate some other values that are needed to solve this problem:

The radius of the Earth's shadow, at the Moon's distance.

The Moon's speed in its orbit.

2. How to calculate the radius of the Earth's shadow at the Moon's distance

This part of the exercise should not be too difficult (age level 15 years).

The principle

a) Calculate the length of the Earth's shadow cone, TP (see Figure 4).

Figure 4

b) Then calculate the shadow radius at Moon's distance.

We try to find OA or OB. ET and OB are parallel lines.

Figure 5

Data

Some data are needed to solve this part of the exercise.

Constants

Radius of the Sun: 700 000 km
Radius of the Earth: 6 370 km
Radius of the Moon: 1 740 km

Data depending on the eclipse

The distances are measured from centre to centre.

3. How to calculate the Moon's speed (in the Earth's shadow)

From here, it gets harder (age 17 - 18 years).

The Moon orbits the Earth in a period of 29.53 days, relative to the Sun, or to the Earth's shadow. This period is the same as the interval between two successive Full Moons, or two New Moons.

a) Approximate calculation

Let's first assume that the Moon orbits the Earth along a circle with a radius of 384 400 km (this is the average distance between Moon and Earth) with a constant speed. Then we can calculate an average speed of the Moon.

b) More accurate calculation

In fact, the Moon orbits the Earth on an ellipse; its distance to the Earth can vary from 356 000 to 407 000 km. Furthermore, the speed of the Moon is not constant.

Figure 6

Kepler's second law tells us that the radius vector [TL] sweeps equal areas during equal times. For example, if the Moon reaches L2 one hour after L1, and L4 one hour after L3, then the two grey sectors have the same area. Thus, the Moon moves faster when it is in L1 and is closer to the Earth, than when it is in L3. We want to compute the tangential speed, VT, perpendicular to the line of sight (TL), that is to say, following the perpendicular height (LH). VT is proportional to LH. The grey sectors are approximately triangular. The area of the triangle L1L2H equals (TL2 x L1H)/2.

Kepler's second law can then be reformulated: TL * V_T = 2A (constant), where A is the area swept by [TL] during one hour, TL the distance between Earth and Moon (km) and V_T the tangential speed (km/h). We first compute the constant A, then the speed V_T.

Data

Half length of the major axis of the orbital ellipse: 384 400 km
Eccentricity of the ellipse: 0.055

4. Extent of an eclipse

In Figure 7, the Moon is drawn at the time of the maximum of the eclipse, when its centre L is situated at the shortest distance of the centre of the Earth's shadow O.

Figure 7

rO = radius of the Earth's shadow d = distance between the centre of the Moon and the centre of the shadow. rL = Moon's radius DL = Moon's diameter

The extent of a lunar eclipse is equal to: g = (r_O + r_L - d) / 2r_L

As r_O + r_L - d = OA + LC - ( LC + OC) = OA - OC = AC,

we can then write: g = AC / 2r_L or g = AC / D_L

Different types of eclipses

The are different types of eclipses, depending on the geometric conditions, that is the relative size of AC and D_L. Look at Figure 8 and see the cases in the table below.

Figure 8

Astronomical catalogues (or almanacs) usually give the extent of a particular lunar eclipse. Then, we can compute the distance d between the centre of the Moon and the centre of the shadow: d = r_O + r_L - 2 g r_L

The table below lists the extent of the next eclipses:

(*) Partial eclipse (g<1)

5. The solutions

Here, you will find the solutions to the exercises above.

Speed of the Moon

Average speed

(2*pi*384 400)/(29.53*24). We get 3 408 km/h or 56.8 km/min.

Constant A

Let us call the half major axis of the ellipse "a" and the half minor axis of the ellipse "b". Then, the ellipse area equals the area of a disc with radius a, multiplied by b/a, or .

The area swept during one hour is: = 654 010 000 km²/h.

The velocities above are tangential speeds, perpendicular to the line of sight. Consequently, these speeds are situated in the planes of the figures that show the Earth's shadow.

The radius of the Earth's shadow

Length of the shadow cone (Figure 4)

Thalès' theorem gives:

If we set TP = x we get: SP = ST + TP = ST + x:

Shadow radius (Figure 5)

Once more, Thalès' theorem gives us:

We just computed TP, we know OP (TP - TO) and TE (6370 km), so we get OB.

The Result

Time scale

We first compute the distances VX and WX

OV = r_O + r_L (shadow radius + Moon's radius) We know d. Pythagoras' theorem enables us to compute XV.

OW = r_O - r_L (shadow radius - Moon's radius) With Pythagoras' theorem, we find XW.

Calculation of times

Knowing the speed V and the distance, we can compute the different durations.

The Results

The indicated times are in UT (Universal Time). You must add 2 hours in summer to get Central European Summer Time (CEST) and 1 hour in winter to get Central European Time (CET).

(*) Partial eclipse.

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.