Measuring the solar system

Introduction

How can we measure the stars if we can’t reach out and touch them?

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In the formative years of astronomy there was a burning desire to measure the size and distance to objects in our solar system. This tutorial will explore the following 5 questions using nothing but naked eye astronomy, geometry and trigonometry:

• How big is the earth?
• How big is the moon?
• How far away is the moon?
• How big is the sun?
• How far away is the sun?

​Let's warm up with a simple yet powerful question: how can you prove the earth is not flat using shadows?

Is the earth flat?

A flat earth was once a commonly held belief, it persisted in some areas until the middle ages. Definitive proof of a spherical earth did not come until  Ferdinand Magellan successfully traveled around the earth (1519-1521). However this belief was challenged much earlier thanks to major advancements in ancient Greek astronomy.

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Plato (427–347 BC) was convinced that the earth was spherical: "My conviction is that the earth is a round body in the centre of the heavens...Also I believe the Earth is very vast". Significantly, he could not prove this was the case. It was his prized student Aristotle who began offering evidence for a spherical earth. One of his more convincing pieces of evidence was the shape of the earth’s shadow visible during a Lunar Eclipse. Notice the subtle curvature of the shadow cast on the moon:

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Another way to prove that the earth is spherical came from Eratosthenes (276–194 BC). His method involves looking at the shadows cast by the sun on the surface of the earth. This is the basis for a sundial:

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But what can shadows tell us about the earth? Eratosthenes did an interesting experiment using two vertical poles located in two distant cities in Egypt. You could do this using telephone poles separated by several hundred kilometers or more. (phone a friend!)

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First wait until the sun is directly overhead point A, which is the moment when the pole casts no shadow. At the same time have a friend observe an identical pole at point B. You will find out that the pole at point B will cast a shadow. Since we can assume the sun's rays reach the earth in parallel lines this is visual proof that the surface must be curved!

Try out the interactive illustration below. You can click and drag the pole to see how it affects the shadow it casts.

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Pause now and think about how these shadows could allow you to determine the radius of the earth. We have a simulation next which will help. You will need to understand how to calculate arc length so let's review that too. Have fun!

Peek-a-boo

How do we know if something in our solar system is further away than something else? For example, the distance to the Moon vs. the distance to Mars. Maybe our Moon is further away from Mars and only appears closer because it’s much larger. How would you prove this theory wrong?

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One simple way to do this is to wait for an evening when the Earth, Moon and Mars form a straight line. This will cause the Moon to block out Mars for a short period of time, known as an occultation. Here is a video showing the occultation of mars by the moon:

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In 365 BC Aristotle observed an occultation of Mars by the Moon. This is undisputable evidence that Mars must be further away from the Moon during the observation.

Occultation time

What else can we learn from an occultation? It turns out there is more information than meets the eye. Consider when the moon is passing over a bright, yet distant star (which we can assume is fixed in place). What could this tell us about how fast the moon is orbiting? Here is an animation of Saturn’s moon Iapetus drifting in front of a background star, captured on Cassini’s narrow-angle camera.

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Consider a star which passes behind the exact middle of the moon. The time it takes for the occultation to occur represents how long it takes the moon to travel its own diameter. If you try this out yourself you will find that it takes approximately one hour. Therefore the moon must be orbiting at approximately 1 moon diameter per hour!

That means if we know the actual diameter of the moon we can figure out its approximate orbital speed!

approximate speed = (diameter in km) km/hour

Now a challenge question for you. Could the occultation time help us determine the size of the moon relative to the earth?  (hint: You will need to time the length of a total lunar eclipse)

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Relative sizes

We previously determined that the diameter of the earth is 12742 km. Now we want to determine the size of the moon relative to the earth.

[[☃ image 1]]Let’s start with an even simpler question: Is the moon bigger or smaller than the earth? How could you prove this? Remember this question wouldn’t seem silly to someone standing on the Moon!

[[☃ image 1]]This question was tackled over 2300 years ago by Aristarchus of Samos (310-230 BC). His measurement begins with the observation of a total lunar eclipse. He correctly assumed that this was the result of the earth casting a shadow on the moon.

[[☃ image 1]]First Aristarchus timed how long the moon took to travel through the earth’s shadow. As Eratosthenes did, he also made a simplifying assumption that the sun’s light reaches us in perfectly parallel lines. Which we know isn’t exactly true as two cone shaped shadows are produced: umbral and penumbral. Though it was close enough for a nice estimation.

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He was actually timing how long between the entry and exit of the darker (umbral) shadow. Which to him appeared to be around 2.6 hours. Next, he compared this with the time it takes the moon to move a distance equal to it’s diameter. In the previous article we determined that this was about 1 hour. So he’s left with two numbers:

• time it takes the moon to travel 1 moon diameter = 1 hour
• time it takes the moon to travel 1 earth diameter  = 2.6 hours

How could we figure out the size of the moon from this?

If the times were equal than it would imply that the moon is the same size as the earth. However the time of a lunar eclipse is much longer, which means the earth must be larger. How much larger? We need to setup a basic proportion:

time #1 / time #2 = moon diameter / earth diameter

1 / 2.6 = moon diameter / earth diameter

This led him to claim that the earth was about 8/3 the diameter of the moon. This is pretty close to the actual difference: the earth is about 3.7 times bigger than the moon. The main reason he underestimated was because the umbral shadow is narrower than the earth.

Actual size of the moon

We now know the relative size of the moon compared to earth. We also know the actual size of the earth from a previous calculation. Finally we can determine the approximate size of the moon!

moon diameter = earth Diameter / 3.7

moon diameter = 12742/3.7 = 3444 km

This is very close to the actual diameter of the moon: 3474.8 km

We determined the size of the earth and moon using nothing but shadows!

A common mistake

A simple yet fundamental concept in astronomy is angular measure. It addresses a common error made by non-astronomers. To illustrate this error, try describing the size of the Sun from your point of view.

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It’s tempting to say it’s about an inch wide, or the size of a quarter. There is a problem with this description however. Do you know why?

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Perhaps my arm is much shorter than yours, and therefore my understanding of “a quarter” is different. The measurement depends on the exact distance to the quarter. In order to use this type of measurement we’d have to say it’s the size of a quarter observed x inches away.

Angular measure

Astronomers use a simpler method based on how many degrees you would tilt your telescope (or head) to scan across an object. This is known as angular measure. Click and drag the circle below to see how the angular measure changes:

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This method leads to the conclusion that both the Sun and Moon are about a half of a degree in size. This means if we put 720 Moons side-by-side in a circle they would complete a ring around the sky! Convince yourself of this, it's very important:

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What about measuring really tiny objects such as planets? Just as we do with microscopic objects, we simply increase the resolution of our measurement. We can divide one degree of arc into 60 arcminutes. We further divide each arcminute into 60 arcseconds:

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Therefore one degree is equal to 60 x 60 = 3,600 arcseconds

Triangulation

When using angular measure we define an isosceles triangle between the observer and the sides of the object we are measuring. As follows:

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Notice we can cut this triangle (and angle) in half to form a right triangle. We love right triangles because it allows us to use trigonometry!

** tan (angular measure/2) = radius / distance**

Quick review (trig in action)

Imagine a pole is 12 meters high and we have to tilt our head 36.8 degrees from the horizon to see the top. How far away are we from the pole?

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tan (ABC) = opposite / adjacent

tan (36.8) =  12 / distance

distance = 12 / tan(36.8)

distance = 16 meters

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Next let's review basic trigonometry & angular measure.

Skinny Triangle Simplification

There is one important way we can speed up our calculations using angular measure. It's an interesting trick all astronomers are familiar with. It applies when the angle of the object we are observing is very small (much less than 1 degree). This is always the case when dealing with celestial objects.

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Skinny triangles have a side length which is almost the same as their heights. Compare this to wider triangles that have much longer sides as compared to the heights. Now here is the trick. If we are dealing with skinny triangles we can assume they are right triangles and use trigonometry to solve for distance.
Let's review.

New shortcut

If some object has a very tiny angular size, $\theta$, then

$tan(\theta) \approx \theta^= \dfrac{d}{D} \
\ $(radians)

where, $d$ = diameter and $D$ = distance.

$1$ radian = $57.3$ degrees = $57.3 \times3600 = 206265$ arc seconds

This gives us the equation that we see most often:

$\theta_{arcseconds}= \left ( \dfrac{d}{D}\right)\times206265 $

Let’s try it out in the next exercise!

Jumpy fingers

Hold up a finger and focus on something in the distant background. While looking at the background alternate opening your left and right eye one at a time. You will find that your finger jumps back and forth (or splits into two). This is the parallax effect in action:

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Parallax is this apparent shift caused by viewing an object from two different vantage points. Next, gradually pull your hand away from your face while still alternately opening and closing your right and left eye. Does the gap between the two images change?

Yes: the greater the distance, the smaller the gap.

Below is an interactive illustration with the red dot representing your finger and a tree representing a distant background. You can click and drag the red dot to bring it towards and away from your eyes to see what happens. Press c to cross/uncross your eyes:

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This simple observation is the key to how we can measure things at a distance. Notice what happens when you direct your eyes to your finger instead of the background (this will cause you to cross your eyes slightly).  Now the background splits into two images which are separated according to how far away your finger is.

The same effect is observed when we look at objects in our solar system from different positions on Earth. In this case the distant stars act as our fixed background and our finger is replaced by a distant object, such as Mars. If you took two pictures of Mars on different sides of the Earth you will find that it seems to jump positions, while the background stars do not. Observing objects from these different positions has the same effect as opening and closing our eyes, except now it’s over a much greater distance.

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This means that the Moon would exhibit a greater parallax (or jump) as compared to Mars. Convince yourself of this using theinteractive illustration **below.**Click and drag the planets and move the time slider to view the parallax effect from different positions on earth:

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Challenge question: Could we use the parallax effect to measure the distance to the Moon?

How far away is the Moon?

In the second century BC Hipparchus derived a very good estimate of the distance to the moon using lunar parallax. It is based on how much the moon shifts relative to the background stars when we observe it from different vantage points on earth.

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To develop our measurement we first need to setup a triangle. Think of the moon as a point in space with two straight lines connecting it to points on Earth:

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In this example the two vantage points are Selsey, UK and Athens, Greece which are separated by 2360 km.

This gives us a triangle. We can simplify things by assuming the moon is exactly between the two points (isosceles triangle).

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Now we need to determine the angle p using the parallax effect.

Finding parallax angle

Here are two photos taken at the same time from Athens and Selsey. We can assume the star (Regulus) near the moon is fixed since it’s 78 light years away.

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The Moon has appeared to shift position. Our goal is to figure out the angular distance of this shift. To do so we combine these images as a stereo pair. To see this, simply cross your eyes (using above images) until the Moons overlap.

If you do this correctly you will see something like this:

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The angular distance between the stars turns out to be approximately 1100 arcseconds, or 0.30 degrees. This looks about right since we know the Moon has an angular diameter of 0.5 degrees. We now have the angle needed. The Moon appears to shift 0.3 degrees when we observe it from two vantage points 2360 km apart.

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Finally, we can split our triangle in half to create a right triangle. This allows us to apply our trigonometric functions to find the distance d directly.

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tan(angle) = opposite/adjacent

tan(0.15) = 1180/distance

1/381.9 = 1180/distance

distance = 1180*381.9

this gives us our estimated distance to the Moon:

Estimated lunar distance = 450 642 km

This estimate is off by only 17% of the actual distance, which is pretty good for a rough estimate! Compare this to the actual average value: 384,000 km

Challenge question: What were the sources of error in our method above? How could we improve?

A burning question

Remember Aristarchus of Samos? He was an early Greek astronomer who suggested that the Earth orbited around the Sun (a heliocentric model). He also answered a fascinating question: how far away is the Sun?

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He realized that we could figure out the distance to the sun relative to the distance to the moon. He noted that the sun, moon, and earth form a right triangle (right angle at the moon) during the first and last quarter moons.

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A quarter moon occurs when the moon appears to be half illuminated from our vantage point. When the moon is seen to be exactly in the first quarter phase the sun-moon-earth angle is exactly 90 degrees. This means a line drawn from Aristarchus’ position to the moon and from the moon to the Sun formed a right angle.

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Aristarchus simply needed to calculate the angle x. That is equal to how far he would have to tilt his telescope between the moon and sun (if telescopes had been invented yet). He calculated this angle as 87 degrees, though the true value is closer to 89.83 degrees (a difficult measurement to get right with the naked eye!).

We already know the distance to the moon is 384,400 km. Finally we have all the information needed! We apply the cosine function to find the hypotenuse of the triangle which is the distance to sun:

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cos(angle) = adjacent / hypotenuse

cos(89.83) = moon distance / sun distance

sun distance = 384,400 / cos(89.83)

sun distance = 129,556,058 km

This is estimate off by 13.3% which is quite good for a rough approximation. Compare this to the actual value:

Average distance to the Sun: 149,600,000 km

This is the basis for the astronomical unit (abbreviated as AU) which is based on the Earth-Sun distance.

1 AU = 149 597 871 km

Challenge question: How could we measure the size of the sun using this new information?

hint: we need to review similar triangles. Let's do that now!

Lucky match

This is an image taken from the Mir space station showing the Moon's shadow during a Solar Eclipse.

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Why do you think a total solar eclipse is important in terms of geometry?

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When the Moon passes perfectly in line with the Sun it blocks it out completely. This means they have the same angular size in the sky. This geometric relationship allows us to apply the properties of similar triangles.

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Thanks to similar triangles we know the ratio of distances from earth to sun and earth to moon must equal the ratio of sun’s diameter to moon’s diameter.

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distance to sun / distance to moon = sun diameter / moon diameter

We’ve already determined these values so we can simply plug them in!

149 600 000 / 384 400 = sun diameter / 3474

sun diameter = 1 352 004 km

This is very close to the actual value! The actual diameter is closer to 1 391 000 km

How cool is that? We measured the size of the sun (about 1.3 million kilometers) using some basic observations and a little geometry! This illustration shows the approximate size of Earth compared to the Sun if they were put directly beside each other.

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Congratulations, you have completed this lesson!

You are ready to tackle some bigger questions...