Assignment #2 – Solution
Problems
1) Let’s check the above article and analyze the section on “Spherical Roots” in more detail.
The Size of the Earth according to Eratosthenes
First look at the diagram below. Since the sun is far, far away, the sun’s light rays are almost parallel. At
some locations on Earth these rays hit the in a perpendicular fashion (this is when the sun is in the Zenith),
while at other locations, these rays will arrive at an angle. So if we could “somehow” measure these angles,
we ought to be able to calculate the radius of the Earth…
As mentioned in the article in “Astronomy”, Eratothenes measured the radius of the Earth with the help of a
friend. Rather than measuring two angles, Eratosthenes employed an easier and more elegant method, and
made sure that measurements were performed at a time when the sun would be in the Zenith in Syene. For
this he used a deep well, whose bottom would only be lit if the sun is directly above. He sent his friend to
Alexandria, a city that is about 5000 stadia from Syene (a “stadium” is an old measurement and 1 stadium is
roughly 0.16 km). His friend in Alexandria measured the angle between the Zenith and the sun by using an
upright stick (the angle is the arctan of length of the shadow to the length of the stick – do you remember
this equation from your first lab?). The angle turned out to be roughly 7o.
Your task is to calculate the radius of the earth from these measurements, and then decide if Erastothenes did a
good job. Do this as follows:
a)
b)
c)
d)
First calculate the circumference of the Earth in stadia.
Then calculate the radius of the Earth in Stadia.
And convert this to kilometers.
Compare that value to the more accurate radius of the earth given in the toolkit. Calculate Erastothene’s
percentage error.
e) Is this value “close” to the modern day value? Generally, we use the word “close” if the calculated
value is within 30% of the estimated value.
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Solution to Problem 1 – a step-by-step explanation:
a) As explained in the problem sheet, the ratio of the distance between Alexandira and Syene (5000 Stadia) to
the angle (7o) is the same as the ratio between the circumference of the earth to the angle (360o), i.e.,
Circumference dis tan ce A. toS.
=
360o
Angle A. to S .
dist .
Circ = 360o ⋅
Angle
5000 Stadia
Circ = 360o ⋅
= 257142.8571 = 2.6 ×10 5 Stadia
o
7
The degrees cancel, and you are left with an answer for the circumference in units of Stadia. How many
significant figures do you need in your answer? 360 has a maximum of 3 significant figures, 5000 a maximum
of 5000 (actually it is really only one significant figure), and 7 has one significant figure. So you answer
should have one significant figure, however two are okay. Any more significant figures are too much. So, in
scientific notation your answer is 2.6x105 Stadia.
b) Next look up the relationship between the radius and the circumference (check the toolkit), and then
calculate the radius of the Earth:
Circ = 2 ⋅ π ⋅ radius
1 
 1
aside: 
⋅ Circ = 2 ⋅ π ⋅ radius ⋅

2 ⋅π 
 2 ⋅π
Solving this for the radius gives
radius =
Circ 2.6 × 105 Stadia
=
= 40925.56 = 4.1 ×10 4 Stadia
2 ⋅π
2⋅π
c) Since 1 stadium is roughly 0.16 km, 4.1x104 Stadia would be 4.1x104 times 0.16 km.
radius = 4.1× 104 Stadia ⋅ 0.16
km
= 6.5 × 103 km
Stadia
d) From the toolkit we copy the radius of the Earth, i.e. The percentage error is the difference between the real
and experimental value divided by the real value.
percent . error =
real − exp erimental 6.5 × 103 km − 6378km
=
= 0.027
real
6378km
So the percentage error is only 2.7%.
e)
Since anything within + 30% to –30% is reasonable, a percentage error of 3% is certainly reasonable.
In fact, a percentage error of 3% is amasingly small.
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2) The Size of the Moon and its Distance from us
(Part of this exercise involves going outside and looking at the moon – do not procastinate, you will not be
able to do this measurement ten minutes before you have to hand in the assignment)
Measuring the angular size of the moon is no big problem, however measuring the physical size of the moon
and its distance from us, is a little more complicated. But, if we were patient and waited for the next lunar
eclipse, we could determine the diameter of the moon. The idea is illustrated in the cartoon below. From the
Earth’s shadow cast onto the moon (actually from the duration of the lunar eclipse) we can determine the
size of the moon. Well, you can try to do this experiment during the next lunar eclipe, for now, we’ll just use
the results of others who did the measurements. It turns out that the shadow of the Earth is roughly 8/3 times
larger than the size of the Moon. So the Moon is 3/8th the size of the Earth. (Consult the drawing below.)
a) Go outside and use your body parts (fingers, fists, etc – check the toolkit) to measure the angular size of
the moon. How big is the moon in degrees?
b) Using the fact that the moon is 3/8th the size of the Earth, calculate the diameter of the moon in kilometers. (For the diameter of the Earth use the value you calculated in question 1.)
c) Next calculate the distance to the moon in kilo-meters. (Hint: Use the small angle formula.)
d) Look up the more accurate value of the distance to the moon, calculate the percentage error, and finally
comment on whether your value is “reasonable”.
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Solution to problem 2:
a) The moon is the size of roughly half a finger, or the tip of your pinky (for most people). This means that the
rough angle is 0.5o.
b) We calculated the radius of the Earth to be 6.5x103km. The moon is 3/8th that size, i.e.,
radius of moon = 6.5 × 103 km ⋅
3
= 2.4 × 103 km
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c) Now we need to use the small angle formula form the toolkit. Since we have the angle in units of degrees, it
is easiest to choose the formula:
dis tan ce to moon × angle
size of moon =
57.3o
2.4 × 103 km × 57.3o
dis tan ce to moon =
= 2.8 × 105 km
0.5 o
d) In the toolkit the more accurate value is given to be 3.84x105km. As before the percentage error is:
percent . error =
real − exp erimental 3.8 ×10 5 km − 2.8 × 105 km
=
= 0.26
real
3.8 ×105 km
e) Since anything within + 30% to –30% is reasonable, a percentage error of 26% is still reasonable.
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For Extra Credit
3) The Distance to the Sun
Now you already determined the size of the Earth, the size of the Moon and the distance to the Moon. Let’s
figure out the Size and the Distance of the Sun…
Aristarchus also proposed a method to measure the distance to the sun, which was later carried out by
Hipparchus in ~150 B.C. This is how he did it. Consider the diagrams below. When the quarter moon is
directly in your Zenith, the sun will be barely above the horizon and will soon set. It turns out to be roughly
1.5 of a finger above the horizon. This corresponds to 3o, making the angle A (as in the diagram below) to
be 90o-3o =87 o.
a) Using your body parts measure the angular size of the sun.
b) Calculate the distance between the Earth and the Sun (use the value from question 2 for the distance
between the Earth and the Moon).
c) Calculate the diameter of the Sun using the result from part (b).
d) Compare this value to the modern day value and comment on its accuracy.
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Solution to problem 3:
a) The sun is the size of roughly half a finger, or the tip of your pinky (for most people). This means that the
rough angle is 0.5o. (Actually the sun and the moon are roughly equally big in the sky, which is why we observe
spectacular total eclipses.)
b) To calculate the distance to the sun we use trigonometry and the drawing above.
cos( A) =
dis tan ce earth to moon
dis tan ce earth to sun
dis tan ce earth to sun =
dis tan ce earth to moon 2.8 × 105 km
=
= 5.4 × 106 km
o
cos( A)
cos( 87 )
c) To get the diameter of the sun, we need to use the small angle formula in the same fashion as we did in
question 2c.
dis tan ce to sun × angle
57.3o
5.4 ×10 6 km × 0.5o
size of sun =
= 4.7 × 104 km
o
57.3
size of sun =
d) The radius of the sun 6.96x105 km.
real − exp erimental 4.7 × 10 4 km − 7.0 × 105 km
percent . error =
=
= 0.93
real
7.0 × 105 km
Now, 93% is an unreasonable error.
What went wrong? Well there are several possibilities, either we used the wrong distance to the moon, or
the wrong angle A, or the wrong angular size of the sun. We know that the distance to the moon has a 27%
error, which is big, but apparently still reasonable. Also we know that the angular size of the sun may be off
by a little bit, but certainly it is within the right order of magnitude. So, the most likely error is probably the
measurement of the angle A. In fact, that is true. Lateron, more accurate measurements showed that the
angle should have been 0.3o, rather than 3 o, i.e. the angle A should have been 89.7 o. Let’s check if this
would have given us a better solution. Redoing parts 3b to 3d, and using the more correct value for the
distance to the moon we get:
dis tan ce earth to moon 3.8 × 105 km
dis tan ce earth to sun =
=
= 7.3 × 107 km
o
cos( A)
cos( 89.7 )
7.3 ×10 7 km × 0.5o
size of sun =
= 6.3 × 105 km
o
57.3
real − exp erimental 6.3 ×105 km − 7 ×105 km
percent . error =
=
= 0.095
real
7 × 105 km
and thus the error is 9.5% which is reasonable.
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