Editing DistanceToTheMoon
You are currently browsing as guest..
To change this, fill in the following fields:
Username
Password
Click here to reset your password
Who can read this page?
The World
Members
Council
Admin
You have been granted an edit lock on this page
until Fri Apr 26 20:28:33 2024.
Press
to finish editing.
Who can edit this page?
World editing disabled
Members
Council
Admin
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 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 out as 382517/375402 km/s, or almost exactly 1.02km/s.