Q:

how we measure the distance between moon and earth?

- esraa (age 16)

Alexandria

- esraa (age 16)

Alexandria

A:

There are all sorts of fancy modern ways (time for a laser pulse to bunce of the moon and back, etc.) but there's a way that you can do it with simple ordinary observations. Aristarchus did it some 2200 years ago. I'll go through his argument, accurate to about two decimal places.

Since we can see what angle the moon occupies in the sky, 0.0090 radians, we can easily get the distance D=M/(angle in radians) = 110 M, where M is the diameter of the moon. (Draw the picture to see that.) Now we need to get M measured in some more local units, say E, the diameter of the earth. Then we'll have D/E=110*(M/E).

We'll get M/E by looking during a lunar eclipse to compare the diameter of the earth's shadow (S) on the moon with the diameter of the moon itself. Let's first think of a simplified picture, with the sun a a little dot very far away. Then the light rays from the sun would form parallel lines, so the shadow of the earth on the moon would have diameter S=E. Then we'd just have to look at the moon during an eclipse to find M/S=M/E, just what we want. Our equation D=110 M would become D = 110* E*(M/E) = 110*E*(M/S).

(See for nice pictures of an eclipse. You can measure M/S yourself on it.)

There's a little complication, though. Although the sun is very far away, as Aristarchus also showed, it's not just a dot. It occupies, by remarkable coincidence, the same angular diameter as the moon. So the rays from the sides of it aren't quite parallel, and that makes the size of the full shadow shrink: S=E- 0.0090*D, or E=S+0.0090*D. By our remarkable coincidence (not needed to solve things, just makes the expressions easy), that's E=S+M. So we have D=110*E*M/(S+M)= 110E/(1+S/M). So there's the distance to the moon in terms of the earth's diameter and things you can see. The observed S/M is ~2.7 so D/E is about 30.

Now it might be nice to have the distance in terms of some ordinary distances on earth instead of in terms of E. How many, say, miles is E? Here's where your home town (I'm assuming it's in Egypt) comes in. Eratosthenes used the difference in noon-time shadows between Alexandria and Syene (modern Aswan) to measure the curvature and hence the diameter of the earth, getting close to the modern value of ~8,000 miles. See for a picture of his method.)

So then educated people in your neighborhood knew that the distance to the moon was about 240,000 miles.

Back to Aristarchus, how did he know that the sun was much farther away than the moon? Look at a half-moon. The reason you see just half the moon lit by the sun is that the line from the middle of the moon to the sun forms a right angle with the line from you to the middle of the moon. Now ask, what's the angle you have to turn to go from pointing toward the middle of the moon to pointing toward the middle of the sun. Aristarchus saw that was very close to 90° also. (picture from wikipedia "AristarchusHalfLitMoon2" by andonee)That means that rays from the sun going to the moon and ones going to the earth are almost parallel. So the sun must be much farther away than the moon. Aristarchus seems to have underestimated the distance to the sun by quite a bit, but not enough to make serious problems for his estimate of the distance to the moon. He did figure out that the sun was so far away that it had to be much larger than the earth.

Oddly, for some reason all this was forgotten in the next two centuries, so that when Lucretius wrote about this topic in ~60 BC he had no idea what the size of the moon was.

Mike W.

*(published on 02/13/2015)*