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

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 Earth-Moon 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.

1. Outline of the problem

Below you will find a view of a lunar eclipse as seen from space. To the right is a view from Earth.



Figure 1



Figure 2

There are several different phases of a lunar eclipse. They are indicated by letters in Figure 2:

V: Entering the shadow (beginning of the eclipse)
W: Beginning of the total eclipse
X: Maximum of the eclipse (middle of the total eclipse)
Y: End of the total eclipse
Z: Exiting the shadow (end of the 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:

  • The Moon's radius: 1740 km
  • Below, in the data base, you will find the minimum distances (d) between the centre of the Moon and the centre of the Earth's shadow at some recent and future lunar eclipses. The time of the maximum of the eclipse is indicated in Universal Time (UT). Remember that you have to adjust this to get your country's local time. For example, if you live in France, you must add one hour during Winter and two hours during Summer to get French time).


  • N° of lunar eclipse
    1
    2
    3
    4
    Date 27/09/1996 24/03/1997 16/09/1997 28/07/1999
    Distance d (km)
    2 081
    3 024
    2 298
    4 914
    Time of the maximum (UT)
    2h54m
    4h40m
    18h47m
    11h34m

    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

    N° of lunar eclipse
      1
    2
    3
    4
    Date 27/09/1996 24/03/1997 16/09/1997 28/07/1999
    Earth - Sun distance (km)
    149 900 000
    149 200 000
    150 400 000
    151 900 000
    Earth - Moon distance (km)
    366 600
    402 200
    357 000
    393 600

    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 * VT = 2A (constant), where A is the area swept by [TL] during one hour, TL the distance between Earth and Moon (km) and VT the tangential speed (km/h). We first compute the constant A, then the speed VT.

    Data

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

    N° of lunar eclipse
    1
    2
    3
    4
    Date 27/09/1996 24/03/1997 16/09/1997 28/07/1999
    Earth - Moon distance (km)
    366 600
    402 200
    357 000
    393 600

    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 = (rO + rL - d) / 2rL

    As rO + rL - d = OA + LC - ( LC + OC) = OA - OC = AC,

    we can then write: g = AC / 2rL or g = AC / DL

    Different types of eclipses

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



    Figure 8

    AC = 0

    => g = 0

    No eclipse

    0 < AC < DL

    => 0 < g < 1

    Partial eclipse

    AC = DL

    => g = 1

    Total eclipse during a fraction of second

    AC > DL

    => 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 = rO + rL - 2 g rL

    The table below lists the extent of the next eclipses:

    N° of lunar eclipse
    1
    2
    3
    4
    Date 27/09/1996 24/03/1997 16/09/1997 28/07/1999
    Extent of the eclipse
    1 245
    0.924
    1.197
    0.402
    Shadow radius (km)
    4 674
    4 500
    4 723
    4 573
    Disatance d (km)
    2 081
    3 024
    2 298
    4 914

    (*) 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 Moon's speed (in the Earth's shadow)
    N° of lunar eclipse
    1
    2
    3
    4
    Date
    27/09/1996
    24/03/1997
    16/09/1997
    28/07/1999
    Speed (km/h)
    3570
    3260
    3670
    3330

    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

    N° of lunar eclipse
    1
    2
    3
    4
    Date
    27/09/1996
    24/03/1997
    16/09/1997
    28/07/1999
    Length of the cone (km)
    1 377 000
    1 370 000
    1 381 000
    1 395 000
    Radius (km)
    4 674
    4 500
    4 723
    4 573

    Time scale

    We first compute the distances VX and WX

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

    OW = rO - rL (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
    1
    2
    3
    4
    Date
    27/09/1996
    24/03/1997
    16/09/1997
    28/07/1999
    XV (km)
    6067
    5458
    6041
    3963
    XW (km)
    2068
    (*)
    1903
    (*)
    Duration from V to X (min)
    101
    100
    98
    71
    Duration from W to X (min)
    34
    (*)
    31
    (*)
    Entering into the shadow
    1h12m
    2h59m
    17h08m
    10h22m
    Beginning of the full eclipse
    2h19m
    (*)
    18h15m
    (*)
    Maximum of the eclipse
    2h54m
    4h40m
    18h47m
    11h34m
    End of the full eclipse
    3h28m
    (*)
    19h18m
    (*)
    Exiting the shadow
    4h35m
    6h20m
    20h25m
    12h45m

    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.


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