Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lesson 20: How Far Away Is the Moon?
175
Student Outcomes

Students understand how the Greeks measured the distance from the earth to the moon and solve related
problems.
Lesson Notes
In Lesson 20, students learn how to approximate the distance of the moon from the earth. Around the same time that
Eratosthenes approximated the circumference of the earth, Greek astronomer Aristarchus found a way to determine the
distance to the moon. Less of the history is provided in the lesson, as a more comprehensive look at eclipses had to be
included alongside the central calculations. The objective of the lesson is to provide an understandable method of how
the distance was calculated; assumptions are folded in to expedite the complete story, but teachers are encouraged to
bring to light details that are left in the notes as they see fit for students.
Classwork
Opening Exercise (4 minutes)
Opening Exercise
What is a solar eclipse? What is a lunar eclipse?
A solar eclipse occurs when the moon passes between the earth and the sun, and a lunar eclipse occurs when the earth
passes between the moon and the sun.

In fact, we should imagine a solar eclipse occurring when the moon passes between the earth and the sun, and
the earth, sun, and moon lie on a straight line, and similarly so for a lunar eclipse.
Discussion (30 minutes)
Lead students through a conversation regarding the details of solar and lunar eclipses.

A total solar eclipse lasts only a few minutes because the sun and moon appear to be the same size.

How would appearances change if the moon were closer to the earth?


The moon would appear larger, and the eclipse would last longer. What if the moon were farther away from
the earth? Would we experience a total solar eclipse?


The moon would appear larger.
The moon would appear smaller, and it would not be possible for a total solar eclipse to occur, because
the moon would appear as a dark dot blocking only part of the sun.
Sketch a diagram of a solar eclipse.
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Students’ knowledge on eclipses varies. This is an opportunity for students to share what they know regarding eclipses.
Allow a minute of discussion, and then guide them through a basic description of the moon’s shadow and how it is
conical (in 3D view), but on paper in a profile view, the shadow appears as an isosceles triangle whose base coincides
with the diameter of the earth. Discuss what makes the shadows of celestial bodies similar. Describe the two parts of
the moon’s shadow, the umbra and penumbra.
Note that the distances are not drawn to scale in the following image.
Discussion
Solar Eclipse
3D view:

The umbra is the portion of the shadow where all sunlight is blocked, while the penumbra is the part of the
shadow where light is only partially blocked. For the purposes of our discussion today, we will be simplifying
the situation and considering only the umbra.

What is remarkable about the full shadow caused by the eclipse? That is, what is remarkable about the umbra
and the portion of the moon that is dark? Consider the relationship between the 3D and 2D image of it.


If a cone represents the portion of the moon that is dark as well as the umbra, then the part that is
entirely dark in the 2D image is an isosceles triangle.
We assume that shadows from the moon and the earth are all similar isosceles triangles.
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
This is, in fact, not the case for all planetary objects, as the shadow formed depends on how far away the light source is
from the celestial body and the size of the planet. For the sake of simplification as well as approximation, this
assumption is made. Then, shadows created by the moon and the earth will have the following relationship:
Earth
Moon
𝑎=𝑏

You can imagine simulating a solar eclipse using a marble that is one inch in diameter. If you hold it one arm
length away, it will easily block (more than block) the sun from one eye. Do not try this; you will damage your
eye!

To make the marble just barely block the sun, it must be about 9 feet (i.e., 108 inches) away from your eye. So
the cone of shadow behind the marble tapers to a point, which is where your eye is, and at this point, the
marble just blocks out the sun. This means that the ratio of the length of the shadow of the marble to the
diameter of the marble is about 108: 1.

In fact, by experimenting with different-size spheres, we find that this ratio
holds true regardless of the size of the sphere (or circular object) as long as the
sphere is at the point where it just blocks the sun from our vantage point. In
other words, whether you are using a marble, a tennis ball, or a basketball to
model the eclipse, the distance the sphere must be held from the eye is 108
times the diameter of the sphere.
Scaffolding:
Emphasize the altitude:base
ratio of lengths by citing other
everyday objects, such as a
tennis ball; for example, a
tennis ball just blocks out the
sun if the tennis ball is viewed
from 108 tennis ball diameters
away.

We conclude that this ratio also holds for the moon and the earth during a solar
eclipse. Since the moon and sun appear to be the same size in the sky (the
moon just blocks the sun in a solar eclipse), we can conclude that the distance
from the earth to the moon must also be roughly 108 times the diameter of the
moon because the earth is at the tip of the moon’s shadow.

Let us now consider what is happening in a lunar eclipse. A lunar eclipse occurs when the moon passes behind
the earth. Once again, the earth, sun, and moon lie on a straight line, but this time the earth is between the
moon and the sun.

Consequently, during a lunar eclipse, the moon is still faintly visible from the earth because of light reflected
off the earth.

Sketch a diagram of a lunar eclipse.
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Lunar Eclipse
3D view:

Based on what we know about the ratio of distances for an object to just block the sun, we can conclude that
the moon must be within 108 earth diameters; if it were not within that distance, it would not pass through
the earth’s shadow, and the earth could not block the sun out completely.

Studying total lunar eclipses was critical to finding the distance to the moon. Other types of eclipses exist, but
they involve the penumbra. The following measurement required a total eclipse, or one involving the umbra.

By carefully examining the shadow of the earth during a total lunar eclipse, it was determined that the width,
or more specifically, the diameter of the cross section of the earth’s conical shadow at the distance of the
1
2
moon, is about 2 moon diameters.
Share the following animation of a lunar eclipse: http://galileoandeinstein.physics.virginia.edu/lectures/eclipse3.htm
Lesson 20:
How Far Away Is the Moon?
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 20
M2
GEOMETRY

With this measurement, a diagram like the following was constructed, where the shadow of the moon was
reversed along the shadow of the earth:

What length and angle relationships can we label in this diagram?
Pose the following questions one at a time, and allow students to discuss and mark their diagrams.

Based on what is known about the shadows of the moon and the earth, what do we know about the measures
of ∠𝐴𝐵𝐹 and ∠𝐵𝐸𝐶?


What is the relationship between 𝐴𝐹 and 𝐴𝐵?


The angle measures are equal since we are assuming the shadows can be modeled by similar isosceles
triangles.
Since the distance an object must be from the eye to block out the sun is 108 times the object’s
diameter, the length of 𝐴𝐵 must be 108 times that of 𝐴𝐹.
What is the relationship between 𝐴𝐹 and 𝐹𝐷?

We know that the diameter of the cross section of the earth’s conical shadow at the distance of the
1
2
moon is about 2 moon diameters, so the length of 𝐹𝐷 is 2.5 times the length of 𝐴𝐹.

Reversing the moon’s shadow completes parallelogram 𝐴𝐵𝐶𝐷. How can we be sure that 𝐴𝐵𝐶𝐷 is a
parallelogram?
Allow students time to discuss why this must be true, using the relationships determined in the diagram. Share out ideas
before explaining further.


̅̅̅̅
𝐴𝐷 must be parallel to ̅̅̅̅
𝐵𝐶 by construction. The diameter of the earth is parallel to the segment formed by the
moon’s diameter and the diameter of the shadow at the distance of the moon.
We know that 𝑚∠𝐴𝐵𝐹 = 𝑚∠𝐵𝐸𝐶 since the shadows are similar triangles. This means that ̅̅̅̅
𝐴𝐵 is parallel to
̅̅̅̅
𝐷𝐶 (alternate interior angles).
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY

Since we know that 𝐴𝐵𝐶𝐷 is a parallelogram, then 𝐴𝐷 = 𝐵𝐶, or 𝐵𝐶 = 3.5 units.

From our work in Lesson 19, we know the Greeks had the circumference of the earth at this time, which means
they also had the earth’s diameter. Then, it was known that 𝐵𝐶 ≈ 8000 miles. With this information, we
conclude:
3.5 units ≈ 8000
1 unit ≈ 2300.

We already know that the distance from the moon to the earth is 108 times the diameter of the moon. If the
moon has a diameter of approximately 2,300 miles, then the distance from the moon to the earth is roughly
(2300 × 108) miles, or 248,000 miles.

With some careful observation and measurement (that of a lunar eclipse) and basic geometry, Aristarchus was
able to determine a fairly accurate measure of the distance from the earth to the moon.
Example (5 minutes)
Example
a.
If the circumference of the earth is about 𝟐𝟓, 𝟎𝟎𝟎 miles, what is the earth’s diameter in miles?
𝟐𝟓𝟎𝟎𝟎
𝝅
b.
≈ 𝟖𝟎𝟎𝟎
Using part (a), what is the moon’s diameter in miles?
𝟐 𝟐𝟓𝟎𝟎𝟎
𝟕
c.
⋅
𝝅
≈ 𝟐𝟑𝟎𝟎
The moon’s diameter is approximately 𝟐, 𝟑𝟎𝟎 miles.
How far away is the moon in miles?
𝟏𝟎𝟖 ⋅

The earth’s diameter is approximately 𝟖, 𝟎𝟎𝟎 miles.
𝟐 𝟐𝟓𝟎𝟎𝟎
⋅
≈ 𝟐𝟒𝟔 𝟎𝟎𝟎
𝟕
𝝅
The moon is approximately 𝟐𝟒𝟔, 𝟎𝟎𝟎 miles from the earth.
Modern-day calculations show that the distance from the earth to the moon varies between 225,622 miles
and 252,088 miles. The variation is because the orbit is actually elliptical versus circular.
Closing (1 minute)

With some methodical observations and the use of geometry, Aristarchus was able to make a remarkable
approximation of the distance from the earth to the moon using similarity.
Exit Ticket (5 minutes)
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Name
Date
Lesson 20: How Far Away Is the Moon?
Exit Ticket
1.
1
On Planet A, a -inch diameter ball must be held at a height of 72 inches to just block the sun. If a moon orbiting
4
Planet A just blocks the sun during an eclipse, approximately how many moon diameters is the moon from the
planet?
2.
1
Planet A has a circumference of 93,480 miles. Its moon has a diameter that is approximated to be that of Planet
8
A. Find the approximate distance of the moon from Planet A.
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
Exit Ticket Sample Solutions
1.
𝟏
On Planet A, a -inch diameter ball must be held at a height of 𝟕𝟐 inches to just block the sun. If a moon orbiting
𝟒
Planet A just blocks the sun during an eclipse, approximately how many moon diameters is the moon from the
planet?
The ratio of the diameter of the ball to the specified height is
𝟏
𝟒
𝟕𝟐
=
𝟏
. The moon’s distance in order to just block
𝟐𝟖𝟖
the sun would be proportional since the shadows formed are similar triangles, so the moon would orbit
approximately 𝟐𝟖𝟖 moon diameters from Planet A.
2.
𝟏
Planet A has a circumference of 𝟗𝟑, 𝟒𝟖𝟎 miles. Its moon has a diameter that is approximately that of Planet A.
𝟖
Find the approximate distance of the moon from Planet A.
To find the diameter of Planet A:
𝟗𝟑𝟒𝟖𝟎
𝝅
= 𝒅𝐩𝐥𝐚𝐧𝐞𝐭 𝐀
The diameter of Planet A is approximately 𝟐𝟗, 𝟕𝟓𝟔 miles.
To find the diameter of the moon:
𝒅𝐦𝐨𝐨𝐧 =
𝟏
𝒅
𝟖 𝐏𝐥𝐚𝐧𝐞𝐭 𝐀
𝒅𝐦𝐨𝐨𝐧 =
𝟏 𝟗𝟑𝟒𝟖𝟎
(
)
𝟖
𝝅
𝒅𝐦𝐨𝐨𝐧 =
𝟗𝟑𝟒𝟖𝟎 𝟏𝟏𝟔𝟖𝟓
=
𝟖𝝅
𝝅
The diameter of the moon is approximately 𝟑, 𝟕𝟏𝟗 miles.
To find the distance of the moon from Planet A:
𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞𝐦𝐨𝐨𝐧 = 𝟐𝟖𝟖(𝒅𝐦𝐨𝐨𝐧 )
𝟏𝟏𝟔𝟖𝟓
𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞𝐦𝐨𝐨𝐧 = 𝟐𝟖𝟖 (
)
𝝅
𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞𝐦𝐨𝐨𝐧 =
𝟑 𝟑𝟔𝟓 𝟐𝟖𝟎
𝝅
The distance from Planet A to its moon is approximately 𝟏, 𝟎𝟕𝟏, 𝟐𝟎𝟐 miles.
Problem Set Sample Solutions
1.
If the sun and the moon do not have the same diameter, explain how the sun’s light can be covered by the moon
during a solar eclipse.
The farther away an object is from the viewer, the smaller that object appears. The moon is closer to the earth than
the sun, and it casts a shadow where it blocks some of the light from the sun. The sun is much farther away from the
earth than the moon, and because of the distance, it appears much smaller in size.
2.
What would a lunar eclipse look like when viewed from the moon?
The sun would be completely blocked out by the earth for a time because the earth casts an umbra that spans a
greater distance than the diameter of the moon, meaning the moon would be passing through the earth’s shadow.
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
3.
Suppose you live on a planet with a moon, where during a solar eclipse, the moon appears to be half the diameter of
the sun.
a.
Draw a diagram of how the moon would look against the sun during a solar eclipse.
Sample response:
b.
A 𝟏-inch diameter marble held 𝟏𝟎𝟎 inches away on the planet barely blocks the sun. How many moon
diameters away is the moon from the planet? Draw and label a diagram to support your answer.
If the diameter of the moon appears to be half the diameter of the sun as viewed from the planet, then the
moon will not cause a total eclipse of the sun. In the diagram, 𝑷𝑸 is the diameter of the moon, and 𝑻𝑼 is the
diameter of the sun as seen from the planet at point 𝑰.
The diameter of the moon is represented by distance 𝑷𝑸. For me to view a total eclipse, where the sun is just
blocked by the moon, the moon would have to be twice as wide, so 𝑻𝑼 = 𝟐(𝑷𝑸). △ 𝑻𝑰𝑼 and △ 𝒀𝑰′𝑿 are
both isosceles triangles, and their vertex angles are the same, so the triangles are similar by the SAS criterion.
If the triangles are similar, then their altitudes are in the same ratio as their bases. This means
substituting values,
𝑰𝑱
𝟐𝑷𝑸
=
𝟏𝟎𝟎
𝟏
𝑰𝑱
𝑻𝑼
=
𝑰′ 𝑨
𝒀𝑿
. By
, so 𝑰𝑱 = 𝟐𝟎𝟎(𝑷𝑸). The moon is approximately 𝟐𝟎𝟎 moon diameters from
the planet.
Lesson 20:
How Far Away Is the Moon?
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
M2
GEOMETRY
c.
If the diameter of the moon is approximately
𝟑
𝟓
of the diameter of the planet and the circumference of the
planet is 𝟏𝟖𝟓, 𝟎𝟎𝟎 miles, approximately how far is the moon from the planet?
The diameter of the planet:
𝒅𝐩𝐥𝐚𝐧𝐞𝐭 =
𝟏𝟖𝟓 𝟎𝟎𝟎
𝝅
𝒅𝐩𝐥𝐚𝐧𝐞𝐭 =
𝟏𝟖𝟓 𝟎𝟎𝟎
𝝅
The diameter of the planet is approximately 𝟓𝟖, 𝟖𝟖𝟕 miles.
The diameter of the moon:
𝟑
𝟓
𝒅𝐦𝐨𝐨𝐧 = 𝒅𝐩𝐥𝐚𝐧𝐞𝐭
𝟑 𝟏𝟖𝟓 𝟎𝟎𝟎
)
𝟓
𝝅
𝒅𝐦𝐨𝐨𝐧 = (
𝒅𝐦𝐨𝐨𝐧 =
𝟏𝟏𝟏 𝟎𝟎𝟎
𝝅
The diameter of the moon is approximately 𝟑𝟓, 𝟑𝟑𝟐 miles.
The distance of the moon from the planet:
𝑰𝑱 = 𝟐𝟎𝟎(𝑷𝑸)
𝑰𝑱 = 𝟐𝟎𝟎 (
𝑰𝑱 =
𝟏𝟏𝟏 𝟎𝟎𝟎
)
𝝅
𝟐𝟐 𝟐𝟎𝟎 𝟎𝟎𝟎
𝝅
The planet’s moon is approximately 𝟕, 𝟎𝟔𝟔, 𝟒𝟕𝟗 miles from the
planet.
Lesson 20:
How Far Away Is the Moon?
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