Sounds like tough questions? Nah! YOU are about to learn how to get pretty good estimates of these things!

We are going to only use approximate numbers in these comments. They are not really meant to produce research-quality results, but rather help you to understand the processes of finding them. If you want, you could do the same measurements, REALLY accurately, and you could actually get research-quality results! Wow!

We need to start with the Earth. How big is it? Well, first, we are going to assume that it is spherical, ball-shaped. For the moment, pretend that you live exactly on the Equator, and that you have at least one friend, and you both have cel-phones! You convince the friend to walk exactly East, taking a ruler with him. Once he has walked East exactly 1,000 kilometers (about 640 miles), he is to stop and call you. You each get a perfectly straight piece of pipe, maybe 3 meters (10 feet) long and you each stick it in the ground so it is pointed exactly straight up.

Then you wait for either March 21 or September 21. On those days, the Sun is exactly above the Equator. You two are on your cel-phones, and at one moment (noon, for him) your friend tells you that the Sun is exactly overhead, and that his pipe is not casting any shadow at all. But you see that your pipe is casting a shadow! Actually, when you measure it, (with a protractor you have handy!), you find that the shadow is 9° from being vertical. (If you knew Trigonometry, you could measure the length of the shadow on the (flat) ground, and figure out that exact angle with something called a Tangent table.)

That's all you need! Your friend had traveled east exactly 1000 km, and the two of you now see that the shadows now seem to be 9° different in direction. The light that caused those shadows came from the Sun, which is so very far away that we consider the light to be exactly parallel. What that means, then, is that your friend MUST have traveled far enough to have gone 9° around the Earth! Since a whole circle has 360°, he must have gone 9/360 or 1/40 of the way around the Earth. That means that the total circumference of the Earth must be 1,000 km times 40, or 40,000 km (around 25,000 miles)! You are GOOD!

(Note: If you are NOT on the Equator, this idea still works, but the math can be much more complicated. If your friend went exactly south or north, so you both had the exact same Longitude, the math is easiest, but the answer is not as exact because the Earth is not precisely round in a north-south direction.)

Once you know the circumference of a circle, you can find the diameter. Circumference = PI times the diameter, right? So the Earth's diameter must be 40,000/3.1416 or about 12,800 km (around 8,000 miles). You already read that number in a book somewhere. Now you know where they got the number!

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You now have a second way to calculate the Earth's size! You both already knew that there were 24 hours in a day and each hour has 60 minutes, so there are 1,440 minutes in every day. You two, exactly 1000 kilometers apart, just found that 36 minutes passed between HIS seeing no shadow and YOUR seeing no shadow. That is 36/1440 of a day, or in other words, 1/40 of the time the Earth takes to completely rotate. It rotated 1000 km worth in 36 minutes, and with very simple math, you can see that the day is 40 times that long, so it must rotate 40,000 km worth in a whole day. We now have the same number as we had above, without even needing a protractor this time, only cel-phones and a watch! In fact, if he was REALLY careful in knowing that he had traveled EXACTLY 1000 kilometers, and if you could determine the time of no shadow to within a second each time, you two could determine an amazingly accurate value for the Equatorial circumference and diameter of the Earth!

Let's look at the Moon. It looks like it is about 1/2 degree in size. But what does that mean as to actual size? If you put a penny about 8 feet away from you, it looks like it is about 1/2 degree (also called 30 minutes of arc) in size. Is that how big the Moon is? Probably not!

When the Moon is exactly over your head, you are on the side of the Earth that is nearest the Moon. That is as close as you can be to it. Six hours later, when the Moon is setting, you are on the "edge" of the Earth, looking at the Moon. You are actually farther from the Moon by the radius of the Earth. The Earth's radius is half its diameter, or about 6,400 km or 4,000 miles.

Nearly everyone thinks that the Moon looks bigger when it is near the horizon. they are wrong! It only LOOKS bigger because there are things to compare it to, while when it is high up in the sky, there is nothing near it to compare it to. If you would really carefully measure the actual (apparent) diameter of the Moon at Moonset, you would find that it is only about 29.5 minutes in size rather than the 30 minutes you measure when it is overhead.

Well, you have it! In trigonometry terms, there is a really skinny triangle between the Moon and your two positions as the Earth had turned 1/4 turn. It is such a skinny triangle that it is pretty accurate to say that since you saw that the Moon seemed 1/60 smaller when it was one Earth radius (6,400 km or 4,000 miles) farther away, then the Moon must be about 60 times the Earth's radius away. That means that the Moon is about 400,000 km or 240,000 miles away.

Notice that you are learning these things without having to get a ruler long enough to reach the Moon, and really without even having to leave your home town, and with very simple equipment.

OK! Since the Moon seems to be about 1/2 degree in apparent size, we also can know how big it is. Remember the penny you looked at? If you measure the diameter of the penny and the distance it is away from you, you will find that it is around 115 times as far away as its diameter to appear to be 1/2 degree in diameter. (This is more trigonometry. A RADIAN is 57.3° and so 1/2 degree is about 1/115 radian.)

Well, we already know that the Moon is 400,000 km or 240,000 miles away. So its diameter must be 1/115 of that, or 3,500 km or 2,160 miles.

Let's try the Sun! It also appears to be about 1/2 degree in diameter. We need to know how far away it is.

SAY that we can find out how fast the Earth is moving in its (circular) orbit around the Sun. (It turns out that this is harder than it might seem to measure, and we are not really going to deal with that here.) So you know that the Earth travels at around 30 km/sec or 18 miles/second in its orbit around the Sun. Well, the Earth takes one year to go all the way around that orbit, 365 and 1/4 days. Changing that to seconds is about 31.6 million seconds. So the Earth must go about 30 km/sec times 31.6 million seconds, or about 950 million km in its orbit around the Sun.

That's just the circumference of the orbit, so its diameter is that number divided by PI, which is about 300 million km or 186 million miles. The RADIUS of the Earth's orbit would be half that, 150 million km or 93 million miles, and so that would be the distance between the Earth and Sun.

The distance between the Earth and Sun is REALLY important in astronomy, and astronomers would love to have found some direct way of measuring it. So far, nobody has come up with a way, and this method is the usual one for calculating it.

We learned above that any object that looks like it is 1/2 degree in diameter is around 115 times as far away as it is in diameter. We just figured out that the Sun is around 93 million miles from the Earth. All we have to do then is divide that by 115 to get (93,000,000/115) or 810,000 miles in diameter. We actually fudged a little in saying that the Sun appears to be half a degree in diameter, and it is actually slightly larger than that, and the correct number gives us 865,000 miles for the diameter of the Sun.

Now, if you are still with us, take a deep breath, and we will try to work on something that is a little more complicated. We are going to figure out the mass (weight, sort of) of the Earth and then the Sun! But for this, we need a special number, G, that can be measured in a laboratory, but it involves a fairly difficult experiment. So, in the metric system, G = 6.67 * 10

When something is near the Earth, we can describe the gravitational
force acting on it as g * m, where g is the acceleration due to gravity,
around 9.8 meters/second/second or 32 feet/sec/sec and m is the mass of the
object. We can also describe it with Newton's universal equation for
gravitation, which is G * M * m / (R^{2}). (R is the radius
of the Earth, which we just found up above). The two have to be
the same, and there's little m (mass of the object) on both sides which
cancels out, so we wind up with:

g = G * M / (R^{2}).

We know everything here except M, the mass of the Earth! So we can just
solve that for M:

M = g * (R^{2})/ G.

Using the metric system, we have:

M = 9.8 * (6.4 * 10^{6})^{2} / (6.7 * 10^{-11})

M = 4.05 * 10^{+14} / (6.7 * 10^{-11})

M = 6 * 10^{+24} kilograms.

That's equal to about 6,000,000,000,000,000,000,000 tons!

Are you impressed at how GOOD you are? You are figuring out all this stuff that everyone thinks is impossible, and you are whizzing through it!

We cannot stand on the surface of the Sun to measure g there. So we have to use a different method to find the mass of the Sun. It is similar, but it is based on calculating the "kinetic energy" of the Earth in its orbit. There is an equation that describes the energy in two ways, which have to be the same. It gives:

m * v

Everything is the same as before except that m now means the Earth's mass and M means the Sun's mass, and R is now the radius of the Earth's orbit (which we also calculated above).

This gives:

v^{2} = M * G / R

Putting in the numbers, v = 30,000 meters/second and G and R, and solving
for M, we get:

M = (30,000)^{2} * (1.5 * 10^{+11} (meters) / 6.7 * 10^{-11}

M = 13.5 * 10^{19} / 6.7 * 10^{-11}

M = 2 * 10^{+30} kilograms.

This is about 330,000 times as great as the earth's mass.

YOU just figured out how much a whole star weighs! (actually, its mass) Do you think there is anyone else in your town that could do that?

We already figured the distance of the Sun (150 million kilometers) away, and its diameter (1.3 million km or 865,000 miles). We can go farther. That is about 108 times the diameter of the Earth, so its volume must be 108

If we combine these last two things, a mass of 330,000 times Earth's and a volume of 1,300,000 times Earth's, that tells us that the "density" of the Sun must be 333,000/1,300,000 or about 1/4 that of the Earth.

We will stop here, and let your brain cool down! But think about how many really amazing things you now can calculate, and KNOW, about the Earth , Moon and Sun. You really don't have to just "trust" that the books are right, because YOU can figure these things out yourself!

You didn't need to use any super equipment or impossibly hard mathematics, and yet you confirmed several really amazing things. Congratulations!

This is the case primarily because no one (and no textbook) showed them the incredible usefulness a moderate knowledge of Physics can be.

This series of lessons is meant to correct that situation. Students in High School or College Physics should be able to benefit from and EVEN ENJOY (!!) these Physics lessons. The lessons should help clarify the usages of a lot of those dry subjects and equations the teacher or professor tries to ram down your throat. These lessons are freely made available to teachers and professors for use as they desire, either on the InterNet or in the classroom.

(The preceding paragraphs appears in each lesson, in the event that someone happens to find a single lesson from this series as a result of a search-engine search.)

The High School Physics Lessons - Practical A of this series is:

http://mb-soft.com/public/phys0.html

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E-mail to: cj@mb-soft.com

C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago