HOW FAR AWAY IS THAT?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at distances in the universe. So, how DO we know how far things are away from us? How far away is the Moon? How about the Sun? Stars? How can we know the distance to anything we can’t easily visit?

Think about this for a minute. If you want to know how far away your pencil is from you right now, how would you determine that?

How about if you want to know how far your science teacher is from you?

These aren’t super tricky. You’d probably use some type of measuring stick.

How about if you want to know how far you are from home? That might require a different type of “measuring stick” or different method altogether. Think about that one and come up with an idea or two.

I suppose you could drive there and just watch the mileage on your car’s odometer (trip counter). Or maybe you could just map it online and see what the program claims the mileage to be. Or perhaps you could use a GPS.

Here’s a tougher one: Imagine that you want to know how far across a river is, but (and this is the catch) you can’t actually cross it to measure directly. Hmmmmmm – how would you do that? We’ll go one step trickier – let’s say that you’re limited by the technology of 2500 years ago. That means, no GPS, no computers, no modern technology whatsoever.

And while you’re thinking about that one, imagine how exactly one would measure the distance to the Moon without going there. That means that, no, we can’t drive a spaceship there with a long measuring tape behind us. We also can’t drive our spaceship there and check the odometer when we’re done!

OK, both of these problems require the same type of thinking. Take a minute to discuss with your partner how one might be able to attack the river problem.

Have an idea? Here’s one low-tech way to “solve” the problem. (Hey, why did I put solve in quotation marks?)

(See image 1 above.)


So, you’re standing where the diamond is and you notice a tree across the river, indicated by the star. (It doesn’t have to be a tree, but you get the idea.)

Now let’s say that the only equipment you have is a protractor and meter stick. How might you find the distance across the river to the tree? Think about it. Discuss with your partner.

Here’s one approach that will work as long as you’re careful. It involves the use of a simple instrument that you construct, similar to a surveyor’s tool called a transit.

(See image 2 above.)


In this example, you mark a spot close to the shore directly across from the tree. Then you walk a measured distance away from this spot, ending at the diamond spot on the diagram. Here is where the protractor comes to use; also, it may be easier to put it on a notebook so that you can mark off the angle as you measure it. Get on the ground as low as you can. Aim the protractor such that the center (90 degrees) is aimed directly across the river. Try your best to determine the angle (θ) that the tree makes with respect to the shoreline – see the picture above. You may want to use a straw as a “sighting tube”. If so, make sure that the straw crosses the “origin” of your protractor and line it up with the tree. Make sure that you record the angle that the straw makes with respect to the horizontal base of the protractor (which is parallel to the shoreline). You may find it best to ask your teacher for help.

For most students, using a river won’t be very realistic – instead, use this method to calculate the distance across a road (but do very careful – watch for traffic!).

Yes, it’s a little tricky. That’s why I used the word solve in quotes above – this method, like any measuring method, is only as accurate as the tools and the person taking the measurements. In principle, though, it can be quite accurate.

Now, construct a scale diagram on paper. You measured a distance along the shoreline above. Decide upon a realistic scale for the size of your paper. For example, 1 m (of outside distance)= 1 cm (on paper).

Use your protractor to construct the angle that you measured above, in the same position on the diagram that it was outside. Draw additional lines to make a triangle, as shown above.

Now, measure the side of the triangle that represents the distance across the river. Convert it back to meters, using your scale. For example, if your scale was 1 m = 1 cm, and the line is 20 cm – the distance would be 20 m. Get it?

Alternate method – trigonometry!

An alternate method that does not require a scale diagram is to use the mathematics of trigonometry. Everything that you did above is repeated, but a scale diagram is not needed. Since you know one side of a right triangle and an angle adjacent to it, you may use the trigonometric ratios – specifically, the tangent function.

In case you do not know:

In a right triangle, the three sides can be defined as a, b and c. However, it’s sometimes more useful to call them opposite, adjacent and hypotenuse. The hypotenuse is the longest side of the right triangle – the side directly across from the right angle. But what side is opposite and what side is adjacent? This depends on the angle that you’re thinking about – a so-called reference angle.

(See image 3 above.)

Note that opposite means “opposite the reference angle”, and adjacent means “adjacent to the reference angle.”

In a right triangle, several ratios can be defined:

Sine (sin), cosine (cos) and tangent (tan) are the most common. For our purpose, tangent will be most useful. Here are the definitions of the ratios:

sin (θ) = opposite / hypotenuse

cos (θ) = adjacent / hypotenuse

tan (θ) = opposite / adjacent

(This is easy to remember with the pneumonic SOH CAH TOA: Sin equals Opposite over Hypotenuse, Cos equals Adjacent over Hypotenuse, Tan equals Opposite over Adjacent.)

Literally, we read this as (for example):

“Sine of theta (θ) is equal to the opposite side divided by the hypotenuse.” What this means is that no matter how big or small the triangle, the ratio of the sides associated with this angle will always have the same value. For example, the sine of 30 degrees is 0.5 – this means that no matter what the actual size of a right triangle, if it has a 30-degree angle in it, the ratio of side opposite this angle to the hypotenuse of this triangle will always be 0.5. Pretty neat, eh? Sin(θ), cos(θ) and tan(θ) are simply ratios of sides associated with particular angles (θ).

For tangent, our trig ratio of choice, “tangent of theta (θ) is equal to the opposite side divided by the adjacent.”

But how do we apply this to our river problem above?

You have the angle, measured with your protractor. Determine the tangent of this angle by using a graphing calculator (make sure it is in “degrees mode”). Your teacher may help you with this. You know the adjacent side of the triangle and now you can calculate the opposite side:

tan (θ) = opposite / adjacent

Using algebra, we can find that:

opposite = (adjacent) x [tan (θ)]

Try it!

How well did you do? Well, if you were measuring the distance across a river, it may be tricky to get the actual distance. However, if you used a road you could carefully measure the actual distance across the road – use a meter stick, trundle wheel, string or gullible friend to find the actual distance across the road.

Now find the percent that your value is different from the actual value:

[ (Your value) – (actual value) ] / (actual value)

Multiply this by 100 to make it a percent.

Now what is a good percent? That’s hard to say. Different experiments and methods have different amounts of acceptable error. There are no real absolutes here – usually, the scientific community decides on what is acceptable. But here is a rough guide for this particular experiment:

If your value is under 25% (which means that you measured at least 75% of the actual value), that is not bad for a quick experiment of this sort. Higher than that? Try it again.

What are sources of your error? Think about this and write some down. It will be very helpful to talk this over with your partner or teacher.

Astronomical Distances

But what does this have to do with astronomical measurements? The same principle applies, believe it or not. We can measure star angles from the Earth, using instruments that are similar to protractors – sextants, quadrants, etc. This was a classic technique used for centuries. It is usually easier, however, to measure how far stars are from other stars in the sky – this can be viewed as an angle if you think about one line from you to a star, and then another line from you to a different star. Imagine the angle that exists between these two lines.

But there is a little problem.

We still need a measurable distance like the shoreline above, and moving a short distance on the Earth doesn’t give us a significant distance (especially when the stars are pretty far away). So, we change the game a little – we wait for the Earth to move to another part of its orbit, having gone 2 AU. (An Astronomical Unit, or AU, represents the size of the semi-major axis of Earth’s orbit. It is also close to the average distance between Earth and Sun). The 2 AU becomes our known baseline. We take angular measurements at the two points in the orbit and come up with a parallax angle with respect to background stars. The technology is a little different, but the approach is very similar. This technique (called astrometry) works very well for many stars, particularly the nearby ones. Further stars have much smaller parallax angles, making this technique less useful; for the stars farther away, we have other methods for measuring the distances.

By the way, this technique was used to measure the distance to the Moon nearly 2500 years ago by Eratosthenes, among others. Measuring cosmic distances is tricky business, but it’s certainly not new.

That’s all for now – see ya soon, see ya on the Moon!

Further Reading

Kitty Ferguson – Measuring the Universe
Dava Sobel – Longitude
Barbara Ryden - Cosmology


ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009

SO, HOW BIG IS THE EARTH?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at the Earth yet again.

So, now we know that the Earth is a sphere. How big is that sphere?

Quick, how would you determine that? Think about it – ask a friend.

Here are a couple of possibilities:

Get in an amphibious vehicle (!) and drive/sail around the equator, arriving back where you started. Clearly, you should set your odometer (trip counter) to zero and note the reading upon your return to the starting point.

How about this? Get a gigantic piece of string and wrap it around the equator until it reaches the starting point. Alternately, you could wrap the string around adjoining lines of longitude, passing over the north and south poles. This would give you a slightly different value, but the principle is the same.

What’s wrong with these methods? Anything?

Neither of these methods has been utilized, but the size of the Earth has been known with great accuracy for nearly 2500 years! How did someone know this?

As you might expect, it has something to do with mathematics. Let’s go back to our friends, the ancient Greeks – namely, Eratosthenes of Cyrene (276 – 295 BCE). It was known by the locals that Syene (Aswan, in Egypt) was located at or near the Tropic of Cancer – that’s a circle of latitude around the Earth close to 23.5-degrees North. This location is important because the Sun is directly overhead at noon on the day of the Summer Solstice, and no object will cast a shadow then. (The Earth is inclined at an angle of 23.5-degrees, and yes, this has a lot to do with why there will be no shadown here on the Summer Solstice.) With this in mind, Eratosthenes devised a clever experiment. But first, how would you use this information?

How about if you knew of another location roughly along the same line of longitude a certain known distance from Syene? Any clues?

This other location, Alexandria, was north of Syene by some 5000 stades (close to 800 kilometers). At noon on the day of the Summer Solstice, Eratosthenes set up a large stick – it certainly did cast a shadow at noon (when the shadow was shortest), and he determined the size of this shadow and the angle from the top of the stick to the far side of the base. See diagram above, which is clearly NOT TO SCALE.

Think of this – the distance from Syene to Alexandria was known and the shadow and angle were easy to measure very accurately. How would you use this information to determine the size around the Earth? It’s a geometry problem. Think about it for a moment and discuss with your friend.

This part is not so obvious – the two rays of sunlight drawn above are very nearly parallel to each other. If the two sticks were sufficiently long enough, they would intersect at the center of the Earth – creating an angle between them. But this angle is the SAME as the angle between the top of the stick and the ray of sunlight. Again, it looks that way in this diagram, but the proof is a classic one in geometry. If you don’t know the geometry here, you may need to suspend disbelief.

Once you see this, the problem is a simple proportion. How many degrees are in a complete trip around a circle? 360, right? Consider this:

The angle of shadow (measured by Eratosthenes to be 7.2-degrees) is to the angle completely around a circle as the distance between Syene and Alexandria is to the complete distance around the Earth (circumference).

Wow, that’s a mouthful. Symbolically it is so much simpler:

7.2-degrees / 360-degrees = 800-km/ Circumference

I have taken the liberty of writing this as a proportion, though Eratosthenes thought of this just a little differently. Can you solve this for the circumference of the Earth?

So, is this a good estimate? Well, our current value for the approximate circumference of the Earth is 40,030 km (around 25,000 miles). Would you consider this technique to be pretty accurate?

So, what are sources of error in this experiment? And what assumptions did Eratosthenes make. There is a lot to think about here – talk this over with your friend.



Further Reading

Kitty Ferguson – Measuring the Universe

Nicholas Nicastro - Circumference



ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009

How do we know that the Earth is spherical?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at the spherical Earth. So, how DO we know that the Earth is spherical (or close to it)?

So, you live on Earth – at least I presume that you do. That seems pretty reasonable. You’ve known that the Earth is a sphere (or close to it) for as long as you can remember. However, this isn’t really intuitive, is it? Look around you – it’s probably pretty flat where you are. Drive through Ohio or Iowa – yeesh, that’s some flat Earth! How do we know that the Earth actually isn’t flat?

Think about this for a moment – talk with a friend. If you had to prove to someone that the Earth was spherical, someone not from Earth perhaps (?!?), how could you convince him or her? What would be sufficient evidence?

Write down ways that would convince you that the Earth is spherical.

Now look at your list again, but back up 2500 years. Which of these would likely have been plausible then?

Find a partner and explain your choice(s) to him or her. Listen to your partner’s choices. Are these convincing arguments? Is your partner convinced by your reasoning?

Let’s see how your list compares to some of the classic reasoning by the leading thinker of 2350 years ago – Aristotle.

Aristotle, like most ancient Greek philosophers, believed that there were only 4 elements – earth, air, fire and water. Knowing now that there are well over 100 chemical elements, this seems a little kooky. However, the discovery of the chemical elements is a relatively recent thing – chemistry, as we know it, is less than 200 years old. It’s not even as old as the United States! It seemed reasonable to the ancients that all things could be categorized as one of the “big 4”, or some combination of these “elements.” Actually, if you think about if for a moment, you could probably convince yourself that this is pretty smart – or at least not totally ridiculous. Mind you, Aristotle certainly didn’t create the idea of a spherical Earth – his teacher Plato had the notion of spheres within spheres, and others (particularly the Pythagoreans) before both of these gentlemen had spherical theories.

Ancient Greeks (the so-called natural philosophers – predecessors of today’s scientists) believed that these elements would (if given the chance) arrange themselves according to their weights. In an absolute sense, Earth is heavier than water, which is heavier than air, which is heavier than fire. So naturally, things made of earth (which include most things within your reach right now) would tend to collect below the others. Since things can move from all directions, it was natural to imagine that things should collect at a center – that is, arrange themselves in a spherical orientation. (Actually, as we understand gravitation today, that’s not too far removed from how we believe stars and planets are formed.)

Aristotle went even further – he imagined the ideal scenario in which there was a sphere of earth, surrounded by a sphere of water and then spheres of air and fire. Think about – it’s not totally crazy. In fact, it’s somewhat intuitive. After all, if you put something heavy into a swimming pool, it usually sinks, right? (Of course, you can probably come up with objections to Aristotle’s logic, but that’s ok – we don’t pay much attention to Aristotle’s science anymore.)

Are there other reasons to convince someone that the Earth is spherical? Yes, and these are easier to understand and accept.

Imagine that you live near water – like so many early civilizations. Imagine watching boats go away from the shore. What would you see as the boat got further away from you?

Imagine climbing a mountain and staring off at the far away horizon. What do you think you would notice?

Finally, consider the phenomenon of eclipses. Maybe you have not yet seen a lunar or solar eclipse, but when you finally do – wow, they are awesome things to behold. Try to find some pictures of solar eclipses – that is when the Moon gets in the way of the Sun. This happens every year and a half or so, and is usually only visible from a very small portion of the Earth, so it’s likely that you, your family or friends may have never seen one before. Draw what you would imagine this would resemble.

Ask your family or teachers if they have ever seen a solar eclipse. Find some images online. The first thing you notice is the Moon is a dark circle across the Sun (with the Sun’s corona extending around the Moon). This isn’t too surprising – you see the Moon all the time and it certainly looks round, at least during the full Moon phases.

But now look at images of lunar eclipses – that is when the Earth comes between the Sun and the Moon. The shadow of the Earth creeps across the face of the Moon. You may have never seen one directly, but draw what you might expect to see.

Another experiment you could try would be to look at the shadow that a ball casts on a wall when placed in the path of a bright lamp.

Now find some images of lunar eclipses online. Try to find images that show the Earth’s shadow creeping across the surface of the Moon, rather than the time that the shadow completely covers the Moon (though this is interesting, too). What do you notice about the shadow?

What would a lunar eclipse resemble if the Earth were flat? How about if the Earth were a flat disc (as Samuel Rowbotham argued in the 1800s)?

Is this convincing evidence that the Earth is spherical? Can you imagine other arguments that might have worked 2500 years ago?

In any case, Aristotle’s writings convinced nearly everyone (if they could read, of course) and the Earth as a sphere became part of the accepted knowledge base that virtually all people had. Often, Christopher Columbus is given credit for having proven that the Earth is spherical – this is simply not the case. That idea is a myth of more recent origin.

One more thing – I’ve been using the word spherical pretty loosely. The Earth is, in fact, not spherical – it is bulged a bit (oblate) at the equator. The equatorial diameter is about 40 km (27 miles) longer than the polar diameter. Consider how that compares to a billiard ball.

According to the Billiard Congress of America,

16.16. Balls and Ball Rack
All balls must be composed of cast phenolic resin plastic and measure 2 ¼ (+.005) inches [5.715 cm (+ .127 mm)] in diameter.

40 km may seem like a large difference in diameter. However, by percent of diameter, the Earth is closer to a perfect sphere than the average billiard ball.

Wow! Anyway, that’s all for now – see ya soon, see ya on the Moon!

Further Reading

The Beginnings of Western Science
David C. Lindberg

This is about the best text on early science that you will ever find.

Flat Earth?
The wikipedia entry on the Flat Earth Society is certainly worth your time, as is a book by Christine Garwood, titled Flat Earth.

More on the Earth, spherical or flat:

Aristotle - On the Heavens
R. Dicks – Early Greek Astronomy to Aristotle
David Lindberg - Beginnings of Western Science
http://www.bca-pool.com/cgi/site/framegate.cgi?url=http://www.bca-pool.com/play/tournaments/rules/rls_gen.shtml&cat=p

ALL TEXT COPYRIGHT SEAN LALLY 2009