These pictures were taken by Antonio Cidadao in Portugal: they show the partial solar eclipse that happened on October 12, 1996.

*Figure 1:
Solar eclipse on October 12, 1996, observed by Antonio
Cidadao. Click to obtain a larger version (JPG, 46k).*

These next photos were also taken by Antonio Cidadao, on September 27, 1996 during the lunar eclipse:

*Figure 2:
Lunar eclipse on September 27, 1996, observed by Antonio
Cidadao. Click to obtain a larger version (JPG, 53k)*

*Try this:* **Make drawings to explain what a solar eclipse
and a Moon eclipse are!**

You may find more information about the **solar** and **lunar** eclipses
within **Astronomy On-Line.**

*Figure 3: One of the solar eclipse photos
obtained by Antonio Cidadao on October 12, 1996.*

Above you will see one of the pictures taken in Portugal during the solar eclipse, now in a larger format.

Print this picture on your printer to be able to work on it. You can also work on the computer screen, but it will be less practical and less accurate.

With a ruler, measure the diameter of the Sun as well as you can. Then, on two sheets of tracing paper, draw with compasses several circles to help you to measure this diameter most accurate: for example, if you have measured 15 cm with the ruler, draw on one sheet concentric circles with radii equal to 7.3 cm, 7.5 cm and 7.7 cm and on the other concentric circles with 7.2 cm, 7.4 cm, 7.6 cm and 7.8 cm. By placing these sheets on top of the printed picture, you'll be able to find easily the radius of the Sun.

With these sheets, you can also find the radius of the Moon on the picture. Compare these two radii.

Can we conclude that the Sun and the Moon have the same
**apparent** diameter as the Ancient Greek did several centuries
B.C.? They did it when noticing during full solar eclipses that
the lunar disk covered almost exactly the solar disk.

* *

*Figure 4: The
apparent diameter of the Sun and the Moon.*

This apparent diameter is measured in degrees as an angle: it is
about 0.5^{o}, both for the Moon and for the Sun.

* *

*Figure 5: A photo of the lunar
eclipse, taken at 02:00 UT in the morning of September 27, 1996 by
Tromsdalen Gymnasium in Northern Norway. Click on it to see a larger
version (JPG, 31k).*

This is one of a series of photos taken at Tromsdalen Gymnasium in Norway on September 27th 1996.

Does it look like a phase of the Moon? What you see on the picture is the Moon, partially immerged in the Earth's shadow. From this view, the Ancient Greek, as they always saw a circular shadow of the Earth on the Moon as a portion of disk, correctly inferred that the Earth has the form of a sphere.

First, you should try to measure the diameter of the Moon on the picture, applying the same method as previously.

Then, try to estimate the diameter of the Earth's shadow. This is more difficult. As you can see it is bigger than the Moon's, and also because the edge of the shadow is not very sharp. So, cut out of paper (or cardboard) different disks to estimate this diameter: for example, cut one with a diameter you think will be somewhat too big, put it on top of the picture. Then cut out a circle with a slightly smaller diameter, try it: if it is still too big, cut out another still smaller disk, if not, cut out of another sheet a disk with an intermediary diameter... and so on! In the end, you will arrive at a rather good estimate of the diameter of the Earth's shadow.

*Hint:* You will find tthat he diameter of the Earth's shadow is
a little more than twice the Moon's diameter.)

Let *D* be the diameter of the Earth's shadow and *d* is
the diameter of the Moon which we measured on the picture (in the same
units, for instance centimetres.

The Sun is very far from the Earth and the Moon, so we may consider that its rays arrive all parallel and so that the shadow of the Earth is a cylinder:

*Figure 6: The Earth's
shadow as a cylinder.*

In this case, we may consider that the Earth's shadow has the same diameter as the Earth: as we know the size of the Earth (radius: 6378 km), we can compute the diameter of the Moon with our previous measurements:

Let's *D _{M}* be the real diameter of the Moon (in
km): then we have the proportionality board:

D | 6378 x 2 km |

(diameter of the Earth's shadow on the picture) | |

d | D_{M} |

(diameter of the Moon on the picture) | km |

From this, you can then get a rough estimate of the diameter of the Moon.

In fact, although it is very far away, the Sun is very big, and its beams are not parallel, so the Earth's shadow is conic and not cylindric:

*Figure 7: The Earth's
shadow as a cone.*

**Warning!** The following illustrations are wrong; they
are not at the correct scale!. You cannot represent in a readable way
on 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's diameter would
be 3 mm and it would be 30 cm away from the Earth. But the Sun would
then be 120 metres away from the Earth, and its diameter would be more
than 1 metre! Nevertheless, these drawings allow us to understand
eclipses and to make computations about.*

*Figure 8*

When you have more mathematical tools, you will be able to compute
that diameter of the Earth's shadow at the Moon's distance. Here, we
will take the value computed by the *"Bureau des
Longitudes"* in France for September 27^{th}: 9348
km. Then, we can make a new proportional computation:

D | 9348 km |

(diameter of the Earth's shadow on the picture) | |

d | D_{M} |

(diameter of the Moon on the picture) | km |

*Figure 9: The Moon's
orbit.*

*AB* is the diameter of the Moon. As the angle *AEB* is
very small, we can say that the length *AB* is equal to the
length of the arc *AB*, and then we can compute the circumference
of the orbit of the Moon with a yet proportionality board:

D_{M} | 0.5° |

(km) | |

Circumference od the orbit of the Moon | 360° |

(km) |

And if we have got the circumference, it's now easy to find the
radius *R* as:,/p>

So, you have now computed the diameter of the Moon
*D _{M}* and the Earth-Moon distance

You may compare them with the real values: *D _{M}* =
3470km and

The author of this exercise is Josée Sert (France). Please
direct any related questions or remarks to: **Josee
Sert**.