From: Ronald Wong (ronwong@inreach.com)
Date: Fri May 16 2003 - 12:44:26 PDT
Message-Id: <l03102801baeadb4d2b3a@[209.209.18.82]> Date: Fri, 16 May 2003 12:44:26 -0700 From: Ronald Wong <ronwong@inreach.com> Subject: Total lunar eclipse
While watching last night's eclipse, I was reminded of Hipparchus, who is
sometimes referred to as the father of astronomy.
He lived in the second century BC.
Besides inventing trigonometry, he also constructed sighting instruments
and tubes which could measure celestial angles with greater accuracy than
any used before his time.
It was by means of a total lunar eclipse that he was able to determine not
only the moon's diameter but it's distance from the earth.
The method used involved simple mathematical principles and a handful of
facts. I thought some of you might find it of interest - even to the point
of possibly sharing it with your students so I thought I'd tell you about
it.
--------------------------------
1. During a total lunar eclipse, Hipparchus measured the distance
traversed by the moon as it passed through the earth's shadow.
This distance equaled the width of the earth's shadow and
amounted to about 4 moon diameters.
The shadow of the earth can be thought of as a cylinder
projecting from the back of the earth relative to the sun.
In reality, it's a cone but the level of
precision that we are considering here - as
well as the scale - allows us the luxury of
considering it a cylinder.
In this case, the cylinder's diameter is equal to that of the
earth - around 8000 miles.
Erastothenes had calculated this a century
earlier - another activity that involved
simple mathematical principles.
Since this is equal to 4 moon diameters, the diameter of the
moon is about 2000 miles.
2. Knowing how large an object is allows us to determine how far
away it is. All we have to do is measure it's angular size in
our field of vision.
A circle whose radius is 60 miles will have a perimeter of about
360 miles. An angle of 1° measured from the center of this circle
will subtend a distance of about 1 mile along the perimeter. Thus
an object 1 mile long at a distance of 60 miles will subtend an
angle of about 1°.
There is a simple inverse relationship here. At 120 miles, the
same object would only subtend an angle of 0.5° and at 240 miles,
an angle of 0.25°
3. Using Linda Shore's units of measurements, where a "pinky" at arm's
length is 1° of arc (a thumb = 2° of arc, a fist = 10° of arc, and
an open hand = 20° of arc), you will find that the moon is about
one-half of a degree of arc (i.e. half a pinky).
0.5° of arc means that an object one mile in diameter would be
120 miles away.
How far away is an object 2000 miles in diameter? Well, this
diameter is 2000 times the 1 mile diameter so the distance is
2000 times 120 miles.
This gives the earth-moon distance a value of 240,000 miles.
Amazing!
Well, we actually should round this off to 200,000 miles to be consistent
with the level of precision that we have here but the fact is that when
Hipparchus did his measurements, his results were quite accurate, given the
level of precision he had at the time, and so is ours.
Our level amounts to 50% so we did just fine
but then we weren't being very precise - to say
the least. That's typical of back of the envelope
type of work.
Cheers.
ron
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