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Size of the Earth and the Distances to the Moon and the Sun
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Objectives
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Using observations of the:,$L+,+("&-%'&,%()$%*M$,1"0$%
Earth-Moon-Sun system and elementary geometry and trigonometry, we will
duplicate the methods of M/)6'+I+27)6+'6.12.6'96',3'./+'2,14'.,'96)3<'1'=).',-'A1./'.,'4+672)=+',92'93)>+26+:'k2,>)4+4'/+2+')6'1'=2)+-'
the
ancient Greek astronomers in determining the size of the Earth, the size and
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distance to the Moon, andN&,O"-<%C"()%,+("&0>''N)34)3<'./+'21.),',-'.D,'39A=+26'A+136'4)>)4)3<',3+'=@'./+',./+2:'K/+3'D+'G3,D'
the size of and distance to the Sun. In doing so, we will derive the value of
the astronomical unit, the./1.',3+'21.),'D)BB'=+'E2,E,2.),31B'.,'13,./+2'21.),C'134'D+'G3,D'./+'>1B9+6',-'./2++',-'./+'39A=+26C').'A1G+6'
fundamental baseline used for measuring the parallaxes of stars.
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Materials
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• Scientific calculators
• Rulers
Introduction
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As we look into the night\+2+')6'1'6B)</.B@'A,2+'4)--)79B.'+I1AEB+C'D/+2+'./+'
sky, the Moon, planets, and stars appear to be set upon a crystalline sphere.
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We observe the motion of these objects from night to night during the course of a year and discover'
predictable patterns. The Sun, our star, reliably rises and sets each day. From a limited perspective,
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perhaps this is all we need
to know in order to know when to plant our crops, when harvest them,
when
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fishing will be the best, and when we will be able to travel at night beneath a bright full moon. These
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observations, however, will
not give us any information about the size of the Universe nor our place in
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.2)13<B+C')3'./)6'716+'./+'13<B+'2+E2+6+3.+4'=@'./+'F2++G'B+..+2'#<)-#C'/1>)3<'6)4+6',-'B+3<./6'JC'VC'134'W`''
it.
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Even a few centuries agoW,6U1BE/1X'j'14L17+3."/@E,.+396+'U,2'V"WX''
we had no way to know that the Universe was incredibly huge, of order 28
billion light years across M13U1BE/1X'j',EE,6).+"14L17+3.'U,2'J"VX''
our visible limit - the “diameter” of our visible realm. We started with small
steps on what has become'M,'6,B>+'./+'E2,=B+A6'D/+3'./+'13<B+')6'G3,D3'U697/'16'%:e'4+<2++6'-,2'./+'13<9B12'6)*+',-'./+'(93'134'
known as the “cosmic distance ladder.” Our knowledge of cosmic distances
started over 2300 years ago
with Aristarchus of Samos (310-230 BCE) and Eratosthenes (276-196
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./+'2)</.?/134?6)4+:'''
BCE). Aristarchus suggest well before Copernicus that the Earth orbit the Sun, but his ideas were
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rejected. Eratosthenes enjoyed
more success with his measurement of the size of the Earth, employing
A)44B+?67/,,B',2'+12B)+2:'H3'71B79B1.)3<'./)6'+22,2C'D+'7,AE12+'./+'>1B9+'D+',=.1)3+4'.,'./+'177+E.+4'U.29+X'
>1B9+:'M/+'-,2A9B1'.,'96+')6`''
an army to help in his measurements.
Ancient Greeks were able to calculate the size of the Earth,
Moon, and Sun, and to determine the
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distances to the Moon and Sun. Most of these measurements were quite difficult for the ancient
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astronomers to carry out,934+26.134'D/1.')6'/1EE+3)3<'D/+3'@,9'<+.'.,'./+'+34',-'S]93<'$S',-'./+'+I+27)6+:''
as all of this was done without telescope technology and without precise
measuring tools. In this exercise
you will be following the methods of Aristarchus and Eratosthenes.
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Although you will be following their methods, you will be using more precise observations than were
available to the. In fact, you will see that the simple model that these Hellenistic Greek thinkers had for
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the Earth-Moon-Sun system allows quite accurate determination of distances when good observations
are used.
Remember that although today we accept the fact that the Moon orbits the Earth, which is itself orbiting
the Sun, these were all revolutionary ideas in the history of science. Forget all that you currently know
about astronomy and imagine how hard it would be to come up with these ideas by simply looking up
and monitoring the sky.
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cp03a_distance_sun.pages
1
Procedure
Rung 1: The Size of the Earth
This step was first performed by Eratosthenes and involves a very simple model for just the Sun and the
Earth.
1) the Earth is a sphere
2) the Sun is infinitely far away so that its light reaches us in parallel rays
The fact that the Earth is a sphere was known from elementary observations: The shape of the Earth’s
:,&1$.3,$%%
shadow on the Moon during an eclipse and the way a tall ship slowly disappeared as it sailed away. But,
what
is the significance of assuming the Sun to be extremely far away?
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M/)6'6.+E'D16'-)26.'E+2-,2A+4'=@'021.,6./+3+6'134')3>,B>+6'1'>+2@'6)AEB+'A,4+B'-,2'L96.'./+'(93'134'./+'
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-2,A'+B+A+3.12@',=6+2>1.),36'U./+'6/1E+',-'./+'6/14,D',3'./+'8,,3'492)3<'13'+7B)E6+C'./+'D1@'1'.1BB'6/)E'
4)61EE+12+4'16').'61)B+4'1D1@XC'=9.'D/1.')6'./+'6)<3)-)7137+',-'1669A)3<'./+'(93'.,'=+'>+2@'-12'1D1@b''
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Eratosthenes
knew that at noon on the summer
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solstice, the Sun would shine clear to the bottom of a
deep vertical pit in the city of Syene, a city on the '
Tropic of Cancer. This meant that the Sun was at the
021.,6./+3+6'G3+D'./1.'1.'3,,3',3'c93+'$!'U./+'
zenith at that moment. Living in Alexandria, which69AA+2'6,B6.)7+'134'./+'B,3<+6.'41@',-'./+'@+12XC'./+'
(93'D,9B4'6/)3+')3.,'./+'=,..,A',-'1'4++E'>+2.)71B'
was located close to due north of Syene, Eratosthenes
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could measure the length of a shadow cast by an *+3)./')3'./+'6G@'1.'./)6'.)A+:'V+)3<')3'JB+I1342)1C'
D/)7/'D16'B,71.+4'A,2+',2'B+66'49+'3,2./',-'(@+3+C'
obelisk at the same time, noon on the summer solstice.
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This measurement, along with the height of the
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obelisk, gave him the angle that the Sun appeared to6,B6.)7+:'M/)6'A+1692+A+3.C'1B,3<'D)./'./+'/+)</.',-'
the south of the zenith in Alexandria, an angle he ./+',=+B)6GC'<1>+'/)A'./+'13<B+'./1.'./+'(93'
1EE+12+4'6,9./',-'./+'*+3)./')3'JB+I1342)1C'13'13<B+'
determined to be 7.2 degrees.
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(93'D,9B4'6/
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=+.D++3'(@+3
The Sun was not as high in the sky in Alexandria as.)A+',-'41@'134'./+'61A+'.)A+',-'@+12:'M/)6')6')3'-17.'
in
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Syene at the same time of day and the same time of 1'A+1692+A+3.',-'./+'792>1.92+',-'./+'012./d6'6E/+2+'
+3.)2+'012./')
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year. This is in fact a measurement of the curvature=+.D++3'(@+3+'134'JB+I1342)1C'134',37+'4)6.137+',-'
of
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the Earth’s sphere between Syene and Alexandria, and
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4)6.137+'=+.D++3'JB+I1342)1'134'(@+3+C'1'4)6.137+',-'e%%%
once the distance of this small arc is known, the circumference
of the entire Earth could be determined.
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1
To this end, he ordered some soldiers to march off the
distance between Alexandria and Syene, a
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4)6.137+'=+.D++3'JB+I1342)1'134'(@+3+C'1'4)6.137+',-'e%%%'6.14)1'U./,9</.'.,41@'.,'=+'+[9)>1B+3.'.,'1=,9.'
distance
of 5000 stadia, thought today to be equivalent to about 805 km or 500 miles.
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circumference of the Earth could be found by using the following proportionality:
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the answer sheet, calculate what Eratosthenes would have gotten for his answer.
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1
Astronomers
had considerably more political clout in those days!
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What value for the circumference of the Earth would Eratosthenes have determined from his
measurements? What value for the radius of the Earth corresponds to this circumference?
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Your turn: Imagine that you have friends living in Mazatlan who have noticed that each year on the
summer solstice the rays of the Sun fall straight down their water well. You now live in Denver in a
small house on a productive wheat far. In the middle of one of the access roads sits an ancient obelisk.
After hearing from your friends last year, you decide to test the ways of the ancient Greek astronomers
yourself. You know that the longitude of the two places is only slightly different, and that by measuring
the shadow of the obelisk in Denver at the same time the Sun is at the zenith in Mazatlan, you will be
able to calculate the radius of the Earth. You all get on your cell phones as the precise time of the
summer solstice approaches. You click a picture of the obelisk at exactly the right moment. Your image
is shown below.
'
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Using a similar triangles construction like the one shown below,
we can directly determine the radius of
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the Earth. The radius of the Earth is to the height of the obelisk as the distance between Denver and
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Mazatlan is to the length of the shadow.
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5+3>+2'.,'81*1.B13
\+)</.',- 'T=+B)6G
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12/19/11
cp03a_distance_sun.pages
3
;6)3<'1'6)A)B12'.2)13<B+6'7,36.297.),3'B)G+'
Finding the ratio for the right-hand side of this
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equation
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012./:'M/+'B+3<./',-'./+'6/14,D')6'.,'./+'
Using a ruler, measure the length of the shadow and
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the height of the obelisk. Find the ratio of these two
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./+'012./:'K+'713'D2).+'./)6')3'+[91.),3'
measurements, as shown in the equation. Fill in that
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value in the answer sheet.
Finding the ratio for the left-hand side of this
equation
The radius of the Earth is the unknown we are
solving for. To find the distance from Denver to
Mazatlan, find the scaling factor for the map printed
below. At the bottom left on the map is a scale in
kilometers and miles. Using the km scale, find out
haw many km (in the real world) is represented by 1
mm (on the map). Measure the distance from Mazatlan to Denver in mm on the map and convert to km
using the scale factor you calculated.
Putting the two together
Using the above equation and your measurements from the images, calculate the radius of the Earth in
kilometers.
4
Rung 2: The Size of and Distance to the Moon
The distance to the Moon was determined by first finding the size of the Moon relative to the size of the
Earth. This determination of the relative sizes of the Earth and Moon predated the estimate of the
absolute size of the Earth due to Eratosthenes, being first carried out by Aristarchus of Samos. Once
again a model is required to make the determination of the relative sizes of the Earth and Moon. In
particular, we need a good model for what is taking place during a lunar eclipse. It was surmised that
during a lunar eclipse the full Moon is pass through the shadow of the Earth. Aristarchus timed how
long the Moon took to travel through Earth’s shadow and compared this with the time required for the
Moon to move a distance equal to its diameter. This could have been done by timing how long a bright
star was obscured by the Moon as the Moon passed between it and the Earth 2. He found that the shadow
was about 8/3 the diameter of the Moon.
The model we use here must be extended from the one given in Rung 1 because we need to add the
Moon:
1) the Earth is a sphere
2) the Sun is infinitely far away so that its light reaches us with parallel rays
3) the Moon orbits the Earth in such a way that eclipses occur
'''''''''''''''''''''''''''''''
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What is the significance of the 2nd assumption? Compare the left hand image just above -- how the
shadow of the Earth really falls upon the Moon (exaggerated) -- versus that implied by the Sun being
infinitely far away. What is the shape of the umbra of the Earth’s shadow on the left versus that on the
right? Would the assumption of a cylindrical shadow (the one on the right) lead to your calculating the
Moon’s diameter larger or smaller than it actually ? These are not easy questions to answer!
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The above assumption of an infinitely distant Sun will give us a crude estimate of the relative size of the
Moon to the Earth by looking at the curvature of the Earth’s shadow during a lunar eclipse. This
measurement is not without its difficulties and uncertainties, however. The main reason being that the
2
[Watch a total lunar eclipse with time markings: www.youtube.com/watch?v=2dk--IPAi04]
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cp03a_distance_sun.pages
5
Sun is not infinitely far away, and the Moon does not always pass through the middle of the Earth’s
shadow during a lunar eclipse. In addition, you will not be certain as to whether you are measuring the
curvature of the umbra or that of the penumbra.
Here is the Moon just coming out of a total eclipse. You should be able to detect the curvature in the
Earth’s shadow. Use that curvature as a small arc in the total circular shadow being cast, and complete
the circle by using an appropriately sized protractor, paper plate, or whatever fits the arc.
You now have an approximation of the ratio of the radius of the Moon to the radius of the Earth by
finding the ratio of the measured radius of the Moon on the image and the measured radius of the circle
you drew.
Use the radius of the Earth that you determined from Rung 1 of this exercise to find the radius of the
Moon in kilometers, working with the ratio you just found. Then, once we have the radius in kilometers
of the Moon that you just measured, we can easily determine the distance to the Moon by measuring its
angular diameter on the sky. The full angle subtended on the sky by the full Moon is about 0.5 degrees,
making the angle subtended by half of it (to use the radius value) is 0.25 degrees. Calculate the distance
to the Moon using its angular radius (0.25 deg) and the radius you calculated. Here’s the geometry:
6
Rung 3: The Distance to the Sun
Aristarchus also came up with a method for finding the distance to the Sun relative to the distance to the
Moon. This method once again relies on a model for the Earth-Moon-Sun system. In particular, we can
no longer assume the Sun to be very far away as we did for the previous two rungs of our distance
ladder. What we are relying upon here is that the Sun’s rays do not hit both the Moon and the Earth at
the same angle. The rays are NOT parallel over the Earth-Moon distance. This difference in the angle
that the Sun’s rays hit the Moon versus the angle they hit the Earth is extremely small. This is the reason
why this was a particularly difficult measurement for Greek astronomers to perform.
It is important to understand that even though Aristarchus’s estimate of the distance to the Sun relative to
the distance to the Moon was wrong by a factor of 20, his basic method was correct.
Our model now reads:
1) the Earth is a sphere
2) the Sun may be far away, but close enough that its rays hit the Earth and Moon at slightly
different angles
3) the Moon orbits the Earth
Why is the second assumption in our model so important in determining the distance to the Sun? (Hint:
Draw the Earth-Moon-Sun system with the Sun’s rays coming parallel. How many 90 degree angles do
you have?)
This new model and its use in determining the distance to the Sun can be better understood by studying
the following diagram:
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cp03a_distance_sun.pages
7
When the Moon is seen to be exactly in the first-quarter phase, the Sun-Earth-Moon angle is a right or
90-degree angle. If we can measure the angle between the Sun and the Moon when the Moon is
precisely at its first- or third-quarter phase, then we can determine the distance to the Sun. With precise
observations made at first-quarter lunar phase, an Earth-Moon-Sun angle of 89.853 degrees is measured.
How hard to you think it would
be to determine the precise
moment that the Moon is in its
first-quarter or third-quarter
phase?
Since we now know the
distance from the Earth to the
Moon from Rung 2, we can
use that information plus the
measured 89.853 degree angle
to determine the length of the
hypotenuse of the right-angle
triangle and get the distance to
the Sun.
Sin (α) = opposite/hypotenuse or A/C
Cos(α) = adjacent/hypotenuse or B/C
Tan(α) = opposite/adjacent or A/B
SOH CAH TOA
8
Almost there! Now that we have an estimate for the distance to the Sun, it is also possible to determine
the diameter of the Sun in kilometers. We know that the angular diameter of the Sun is almost the same
as the Moon, about 0.5 degrees on the sky. Since obtaining the angular diameter of the Sun directly
would be inaccurate and may make us go blind, we will use the fact that the Moon eclipses the Sun
every so often. Here is the progression of a total eclipse of the Sun as evidence:
Using the distance to the Sun determined above and the angular radius of the Sun, determine the actual
radius of the Sun (your value) in kilometers. You should recall the way in which you calculated the
distance to the Moon from an estimate of its radius in kilometers, and just adjust the equation
accordingly. You know the angle, α, you know D, now find r using a bit of algebraic manipulation.
Summary
As you wrap up your results for this exercise, check the actual quantities listed in the table on your
answer sheet and calculate the percentage error for each value. Fill out the answer sheet completely and
follow the directions given by your instructor in submitting this assignment. We hope you enjoyed
“stepping up” these first few steps. The average distance between the Earth and the Sun is called the
Astronomical Unit. It was the knowledge of this number that led the way to all of the rest of the steps
that eventually took us to an estimate of the distance to the edge of the visible universe!
12/19/11
cp03a_distance_sun.pages
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Preparation for the Exercise - Some Basic Mathematical Review
This exercise
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starts us on the road to using the language of the Universe to describe it. Here is a brief
refresher course.
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Working with ratios: Finding the ratio of two numbers means dividing one by the other. When we
know that one ratio will be proportional to another ratio, and we know the values of three of the
numbers, it makes finding the unknown fourth number easy.
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A simple example:
When working with ratios, some of the units may
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cancel out - this is why ratios are great to work
with.
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Here is a slightly more difficult example, where
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the unknown is in the denominator of one of the
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ratios.
But, after a bit more manipulation.....
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Trigonometry: The hardest part any more about working with trig is knowing which buttons to push on
the calculator. Here’s a refresher for sine, cosine, and tangent of an angle for a right-angle triangle. The
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angle we are working with is designated by the Greek letter alpha. The triangle has side lengths A, B,
and hypotenuse C.
Sin (α) = opposite/hypotenuse or A/C
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Cos(α) =
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adjacent/hypotenuse or B/C
Tan(α) = opposite/adjacent or A/B
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SOH CAH TOA
Percentage Error: Calculating the percentage error is something we all should have learned in middle
school, or at least in theory. We want to compare our value with the accepted or true value.
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