Did you observe the lunar eclipse on September 27, 1996? Even if you did not, you may have thought about how astronomers calculate the circumstances of such an eclipse in advance. In this connection, an important aspect is the duration of an eclipse.
This exercise will give you much information about the EarthMoon system and the motion of the Moon around the Earth. It will probably be most suitable for students in secondary schools of ages 14  18 years. It consists of several steps which are not all equally simple; the the degree of difficulty is indicated in each case.
The solutions are given at the end, so that you can check your results.
Below you will find a view of a lunar
eclipse as seen from space. To the right is a view from Earth.

Figure 2 
There are several different phases of a lunar eclipse. They are indicated by letters in Figure 2:
The Problem: We know the time of the maximum of the eclipse (Moon in position X).We want to compute the remainder of the timescale, that is, when the Moon is at V, W, Y and Z. 
Figure 3 
To solve this problem and find the answers, we first need some data:
N° of lunar eclipse 




Date  27/09/1996  24/03/1997  16/09/1997  28/07/1999 
Distance d (km) 




Time of the maximum (UT) 




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




Date  27/09/1996  24/03/1997  16/09/1997  28/07/1999 
Earth  Sun distance (km) 




Earth  Moon distance (km) 




The distances are measured from centre to centre.
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.
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.
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 L_{2} one hour after L_{1}, and L_{4} one hour after L_{3}, then the two grey sectors have the same area. Thus, the Moon moves faster when it is in L_{1} and is closer to the Earth, than when it is in L_{3}. We want to compute the tangential speed, V_{T}, perpendicular to the line of sight (TL), that is to say, following the perpendicular height (LH). V_{T} is proportional to LH. The grey sectors are approximately triangular. The area of the triangle L_{1}L_{2}H equals (TL_{2} x L_{1}H)/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}.
Half length of the major axis of the orbital ellipse: 384 400
km
Eccentricity of the ellipse: 0.055
N° of lunar eclipse 




Date  27/09/1996  24/03/1997  16/09/1997  28/07/1999 
Earth  Moon distance (km) 




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 
r_{O} = radius of the Earth's shadow d = distance between the centre of the Moon and the centre of the shadow. r_{L} = Moon's radius D_{L} = 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}
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.
AC = 0 => g = 0 No eclipse 
0 < AC < D_{L} => 0 < g < 1 Partial eclipse 
AC = D_{L} => g = 1 Total eclipse during a fraction of second 
AC > D_{L} => g > 1 Total eclipse 
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:
N° of lunar eclipse 




Date  27/09/1996  24/03/1997  16/09/1997  28/07/1999 
Extent of the eclipse 




Shadow radius (km) 




Disatance d (km) 




(*) Partial eclipse (g<1)
Here, you will find the solutions to the exercises above.
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 km^{2}/h.
The Moon's speed (in the Earth's shadow)N° of lunar eclipse 




Date 




Speed (km/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.
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
N° of lunar eclipse 




Date 




Length of the cone (km) 


1 381 000 

Radius (km) 




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
N° of lunar eclipse 




Date 




XV (km) 




XW (km) 




Duration from V to X (min) 




Duration from W to X (min) 




Entering into the shadow 




Beginning of the full eclipse 




Maximum of the eclipse 




End of the full eclipse 




Exiting the shadow 




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, F10 160 AixenOthe, France. Please direct any related questions or remarks to: Josee Sert (France) who will pass them on to CLEA.