Edit made on August 13, 2011 by ColinWright at 21:59:12
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WW WM
HEADERS_END
The distance to the moon can be computed as follows.
We know that:
* acceleration due to gravity at the Earth's surface is about EQN:9.8ms^{-2}.
** this can be measured using a pendulum in your living rioom room
* the orbital period of the moon relative to the stars is about 27.3 days.
* acceleration due to gravity decreases as the inverse square.
* the radius of the Earth is 6366km.
** Measured by Eratosthenes
OK, so acceleration in a circle is EQN:v^2/r or EQN:\omega^2r where /r/ is the
radius of the orbit, and EQN:\omega is the rotational velocity. That's
27.3 days divided by EQN:2\pi giving 1/375402 radians/second. Hence acceleration
in the orbit is EQN:(1/375402)^2r where /r/ is the (unknown) distance from
the centre of the Earth to the Moon.
But acceleration due to gravity is EQN:9.8(r/R)^{-2} where /R/ is the radius of the
Earth, so equating these we get:
* EQN:(1/375402)^2r=9.8(R/r)^2
so
* EQN:r^3=9.8(R^2)(375402)^2
This gives an answer of 382517km, which is amazingly close to the figure
quoted on WikiPedia of an average centre-centre distance of 384,403km.
Accurate to 0.1%.
Finally, Orbital Velocity is given by EQN:v=\omega{r} , and that now works ~works
out as 382517/375402 km/s, or almost exactly 1.02km/s.