How High The Moon 

How High the MoonA paper for G4G9by Colin WrightIt's surprising, sometimes, how a large collection of apparently unconnected facts can sometimes be woven into whole cloth. Some months ago as I was drifting off to sleep I thought I saw how a bunch of stuff I knew all tied up together into a neat bundle. Then I woke up, but surprisingly, it looked like it all still worked! Even more surprisingly, when I investigated further, it did, and here are the results. As implausible as it may seem, we're going to compute the distance to the Moon using a few wellknown facts, a few simple observations, a pendulum, and a stopwatch. Pretty much everything here was known to Isaac Newton in the late 1600's, and it's even been suggested that he performed pretty much exactly these calculations.
Now, the original definition of the metre was "One tenmillionth of the distance from the North Pole to the Equator through Paris," which means the circumference of the Earth is 40 million metres, so the radius is roughly 6.4 million metres. Substituting this we get
So from a height of 5 metres, the distance to the horizon is about 8000 metres, or 8km.

Now let's turn it around. Suppose we're at sea level and 8000m from the top of a 5m high mountain. If we fire a projectile lineofsight at the peak, ignore air resistance, and it gets there in 1 second (unikely, I know).
It's in orbit. So we've just shown that subject to all our approximations, orbital velocity at grazing altitude is 8 km/s. Quite astonishing how our good friend Pythagoras is, in some sense, "Rocket Science." 

So now let's be a little more general. Instead of being exactly
5m high, let's pick an acceleration a and an amount of time
t and suppose we are $at^2/2$ high.
Our Pythagorean triangle equation is now


or equivalently,
Which is nice. What does this have to do with the Moon? 

If we suppose the Moon to be moving in a circular orbit, and supposing
the radius of that orbit to be M, we can now say that its acceleration
in orbit is $v^2/M.$ So if only we knew how far away it was, and its
velocity, we would know its acceleration.
But if we know its distance then we do know its velocity, because we know it takes 29.53 days from full moon to full moon. Correcting for sidereal time, that means it takes 27.32 days to make a complete circuit of the Earth. Call that time P. Therefore the Moon's velocity in orbit is $(2\pi{}M)/P.$ 

So we have the formula for acceleration in a circle that needs the
distance and velocity, but we now know both of those, so we can
say that the Moon's acceleration is this:
Well, we know that acceleration due to gravity is what holds the Moon in orbit, so if only we knew how hard the Earth is pulling the Moon, then we would know that. But we do.


We know that acceleration at the Earth's surface is g,
and that it falls off as an inverse square. Hence the acceleration
due to gravity at any distance, say M, is given by:

But the Earth's radius is $40\times{}10^6/(2\pi),$ so putting it all together we get:


Therefore
And we know everything on the right hand side, except g.
But we can find g with a pendulum and a stopwatch. (I knew you'd be wondering where they came in.)
We know that the time taken for a complete swing of a pendulum is given by the formula:
Of course we have to go away and construct a pendulum, and then we have to measure how long it takes to swing. Typically we measure 10 swings, both back and forth, and then divide the total time by 10. We should also do that several times to make sure we get error bars on the result, because each one will vary slightly. There's lots to do here.
So what do we get?
Of course, the Moon's orbit isn't circular, the Earth isn't of constant radius, nor is it a sphere, and we've assumed that the metre is one ten millionth of the distance from the North Pole to the Equator. But even so, we're not just in the right ball park, we're smack in the middle of the true range.
Not bad for a few sums.