

School kids measure distance to the Moon - RiderOfGiraffes
http://www.technologyreview.com/blog/arxiv/23205/

======
RiderOfGiraffes
An exercise I use as the start of a math masterclass is to calculate the
distance to the moon using a pendulum and a calendar. The answer is remarkably
close to the one obtained by these kids using recordings from the Apollo
missions.

If NASA really did fake the Moon landings, they did a bloody good job!

If you want to know how to calculate the distance, let me know and I'll write
something about it.

~~~
Retric
Shure, I think I could work it out, but I would still like to see your
process.

PS: A _pendulum and a calendar_ and a stopwatch or clock?

~~~
gjm11
Suppose the moon is in a circular orbit with radius R and angular velocity w.
Then its position looks like R exp(iwt), so its acceleration is the second
derivative of this = -Rw^2 exp(iwt), whose magnitude is -Rw^2. (That's a
standard formula but I always have to derive it from scratch.) So GM = R^3w^2.

Now, what do we know? If we have a calendar that marks full moons, we know how
often the occur and therefore we know w. What about the stuff involving GM?
Well, now construct a pendulum. Standard formulae again but let's do it from
scratch again. Length L, angular displacement a: vertical displacement is
negligible to first order, so for small displacements the restoring force just
comes from the tension in the string. To first order again, that equals the
gravitational force on the mass, so the horizontal component of the
acceleration is GM/r^2.La where r is the radius of the earth. So, measure the
period of the pendulum (and of course its length); that tells us a and
therefore GM/r^2. So _provided we know r_ we now have GM and therefore R, and
we're done.

So, one remaining question for RiderOfGiraffes: are you happy to look up r in
a reference book for this exercise, or do you have a classroom-friendly way of
getting that by simple observation and calculation too?

~~~
RiderOfGiraffes
You mean apart from the original definition of the meter as one ten millionth
of the distance from the north pole to the equator passing through Paris? That
means the circumference of the Earth is 40*10^6 meters, so the radis is that
over 2pi.

I guess, then, in addition to the pendulum, stopwatch and calendar we also
need a ruler. We can use these to measure the size of the Earth directly using
either the method of Eratosthenes (who also created the sieve for finding
primes, and invented the system of latitude and longitude) or using the method
used by Abu al-Rayhan al-Biruni.

That last, by the way, is cool, and I intend to use it in my next maths
masterclass field trip.

~~~
gjm11
Yes, I do mean other than apart from the original definition which, after all,
is not the current definition. (You could say "but everyone knows ...", but
then you could equally say that "everyone knows" g ~= 9.8m/s^2.) The
Eratosthenes method is lovely but surely rather difficult in a classroom. (Am
I missing a trick?) All I know about the al-Biruni method is a couple of
sentences from Wikipedia, but it sounds as if it effectively needs some
external (geographical) data that you can't measure in a classroom. Hmm.

~~~
RiderOfGiraffes
OK, I've done some more homework ...

The pendulum method gives the equation

    
    
        l L^2 R^2 = P^2 D^3
    
        l = pendulum length (assume 1m)
        P = pendulum period (roughly 2s)
        L = Lunar orbital time (27.32 * 86400)
        R = Radius of the Earth
        D = Distance to the Moon
    

Now we also know that in a total Lunar eclipse the Moon moves through the
Earth's shadow. Assume it moves directly through the center. Measure the time
for the Moon to move from touching the umbra to completely exiting the umbra.
The distance it moves is then pretty much one Earth diameter, because the
apparent sizes of the Sun and Moon are the same.

So R/D = (T/2) _(2pi/L) where T is the transit time, and turns out from
observations to be around 12500 seconds.

Combining these equations we get:

    
    
        D = l*(pi*T/P)^2
        R = l*pi^3*T^3/(L*P^2)
    

They turn out pretty well.

~~~
RiderOfGiraffes
Bother - forgot that asterisks have a special meaning. Sorry about the
formatting.

------
srveit
For really accurate measurements check out
<http://www.physics.ucsd.edu/~tmurphy/apollo/apollo.html> By bouncing a laser
off several reflectors that were left on the moon, they are able to get
accuracies of one millimeter.

------
mrduncan
I'm reminded of Clifford Stoll's TED talk where he dicusses (among other
things) how he teaches high schoolers to calculate the speed of sound.

[http://www.ted.com/index.php/talks/clifford_stoll_on_everyth...](http://www.ted.com/index.php/talks/clifford_stoll_on_everything.html)

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ZeroGravitas
If you've got two OLPC XO laptops you can measure the distance between them
using a similar techique:

<http://wiki.laptop.org/go/Acoustic_Tape_Measure>

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mhb
Measuring the height of a building using a barometer:

<http://www.snopes.com/college/exam/barometer.asp>

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yread
They should try to measure it the way Ptolemy did! Or the distance to Sun...

------
numbchuckskills
REALLY smart school kids would go to wolfram alpha to get the distance to the
Moon, and let the suckers analyze echo's from old NASA clips.

~~~
RiderOfGiraffes
This deserves a separate discussion.

See <http://news.ycombinator.com/item?id=802856>

