The Distance and Orbit of the Moon
We made measurements of the diameter of the Moon at the telescope two weeks ago,
getting an average value for the diameter of the Moon’s image in our telescopes of 11.4
mm. Last week, you measured diameters of photographs of the Moon taken over almost 3
orbits and plotted these on a chart. Tonight, we are going to use these measurements to
estimate a distance for the Moon and something about the nature of the Moon’s orbit.
In order to get a distance for the Moon in physical units (e.g., kilometers), we need to
know its diameter in those units. By about 100 BC, Greek astronomers correctly estimated
the size of the Moon by observing the curvature of the Earth’s shadow on the Moon during
an eclipse. They already had determined the size of the Earth, so they could figure out
from the size of the Earth’s shadow the actual size of the Moon. We will give more details
on how this was done later.
First, take the charts on which you have plotted your measurements of the diameter of the
Moon pictures and try to sketch a smooth curve through the points (do this in pencil until
you think it looks good). Don’t put in wiggles to make it go through the center of each
point, and your curve might even miss going through some of the lines you have drawn
through your points; what you want is a curve that represents the broad distribution of
your points. This curve is your estimate of the actual change in size of the Moon around
the 2+ orbits covered by the pictures.
Of course, the actual numbers you measured are arbitrary, depending on the size the
images were printed. We now want to relate them to the actual physical diameter of the
Moon. This is a three-step process:
1. On the date that you measured the Moon’s diameter at the telescope (10/09/08 HST),
the Moon was at almost exactly the same distance as it was on 10/20/05. Find the point
on the curve you have drawn that corresponds to this date, and go over to the left to get
the Moon diameter from the photograph for the date. Divide this value into our average
telescope measurement (11.4 mm). You should get a number that should be between 0.15
and 0.19; if you get something else, call for help. Note that this is a pure number, since
the units of mm divide out. We will call this number s. If you were to multiply the values
of your graph by s, you would have a graph showing what you would have expected to
measure for the size of the Moon at the telescope on any of the dates shown on the graph.
2. We need now to relate what you measured at the telescope (in mm) to an angular
measurement on the sky (in degrees). Our telescopes have focal lengths of 1200 mm. This
means that the image that our eyepiece is looking at lies on a small part of a sphere with
this radius centered on the center of the telescope mirror. A great circle on this sphere will
have a circumference of 2πr, where r = 1200 mm, or about 7540 mm. Since there are
360° in a circle, each mm is about 0.048°. So now if you were to multiply the values for
your graph by 0.048s, your graph would give the size of the Moon in degrees in the sky
on the various dates. Write down your value for 0.048s, because we will use it in the next
section.
3. The change in the size of the Moon shown by your graph results from changes in the
distance to the Moon. Suppose that the Moon is at a distance such that its diameter is
exactly 0.5°. Now imagine filling up a complete circle around the sky with “moons” this
size, just touching each other. How many would it take? Since there are 360° in a circle,
and each moon is 0.5°, it would take 720 such moons. So, how far away is the Moon in
Moon diameters? The circumference of this circle is 2πr, where r is the distance from the
Earth to the Moon, so r = 720/2π Moon diameters, or about 115 Moon diameters. Note
that, if the Moon appears bigger than this, it is closer: for example, if the Moon were to
appear to be 1° in diameter (which it never does!), it would be at half the distance given
above, or about 57 Moon diameters. So we can come up with a formula (accurate if the
angle is not too large) for the distance of the Moon: r = 57.3/d Moon diameters, where d is
the diameter of the Moon in degrees. But remember that, in §2, above, we showed that
the graph coordinates could be converted to degrees by multiplying by 0.048s. So now
take each of the numbers on the left-hand axis of your graph (62, 63, 64…74), multiply
that number by your value for 0.048s, and divide this result into 57.3. Write this final
result in the margin next to the corresponding line on the right side of your graph. (Note
that you can simplify the calculation a bit by dividing 57.3 by your value for 0.048s first,
then dividing the left-hand graph values into this result directly). Once you have labeled
your graph in this way, you can simply read off the distance to the Moon (in Moon
diameters) directly for any date within the range.
Make sure that the numbers you get are reasonable. Since you know that the Moon is
roughly half a degree in diameter (as in our example above), your distances should be
varying somewhere in the neighborhood of a bit over 100 Moon diameters. If you are
getting values very different from this, go back and check your calculations.
Now we want to see if we can get this distance in actual physical units (kilometers), rather
than simply in Moon diameters. We know that the diameter of the Earth is about 12,800
kilometers (close to the value Eratosthenes found, although he, of course, was using
different units). From photographs of an eclipse of the Moon, we can estimate the
diameter of the Earth’s shadow compared with the diameter of the Moon: divide the
diameter you get for the Moon’s image (in mm) into the diameter you estimate for the
Earth’s image (in mm). Because of the convergence of the Earth’s shadow, you have to add
one more Moon diameter to this number. So, if dE is the diameter of the Earth (in km), and
dM is the diameter of the Moon (in km), and R is the ratio of the diameter of the Earth’s
shadow to the diameter of the Moon (from the picture), then dM = dE / (R + 1). Now that
we know the Moon’s diameter in physical units, we can multiply our distances found
above by dM to give us actual distances in kilometers. Instead of doing this, take the
average value for your distance in Moon diameters and multiply this number by the
diameter of the Moon in km.
Things to include in your report:
1. Show how you calculated the value s, as described in §1, above.
2. Include your plot of measurements from the Moon photographs, including the curve
drawn through your points and the annotation of the corresponding Moon distances in the
right-hand margin.
3. Give your final value for the average distance of the Moon in both Moon diameters
and in km.
4. The points on your graph cover over two orbits. Kepler’s First Law says that the orbit
should be an ellipse, with the Earth at one focus of the ellipse. How would your graph be
qualitatively different if the Earth were at the center of the ellipse instead of at a focus?
5. The Earth’s orbit around the Sun is much closer to being a circle than the Moon’s orbit
around the Earth, so the Sun remains almost the same angular size throughout the year.
Think about what happens if an eclipse of the Sun happens at different points of the
Moon’s orbit. How can this explain differences in the lengths of total eclipses of the Sun,
as well as annular eclipses (look this term up in your textbook if you don’t know what it
means)?
6. When we measure the Moon when it is close to the horizon, it is about 6000 km farther
away than if we were to measure it at the same point in it orbit when it is overhead
(because of the radius of the Earth). Do you think that this a significant error, given the
scatter of points around your curve?