Pat McBride wrote:
Read Science Fiction, Daniel? You're right on, but the derivation would
probably rattle a few brains, including mine. Haven't done any calculus for
going on nigh 35 years, been better off without it.
No calculus, it's actually just basic algebra. I didn't include it
before because I didn't think anybody would care, but here it is:
Simplifying assumption: Assume the planet is uniform and homogeneous
except possibly in the radial direction.
Acceleration due to gravity is:
g = GM/R^2
Where G is the gravitational constant, M is the mass of the planetary
body, and R is its radius.
M = rho*(4/3)*pi*R^3
Thus: g = G*rho*(4/3)*pi*R
Where rho is the density of the earth.
The period of a pendulum is:
T = 2*pi*sqrt(L/g)
Where L is the length of the pendulum. Let L be the length that
corresponds to a 2-second pendulum. This is one of the two definitions
of a metre that we wish to compare:
L = g * (1s/pi)^2
L = G*rho*(4/3)*pi*R * (1s/pi)^2
L = G*rho*(4/3)* R * 1s^2 / pi
Now consider the alternate definition of a metre (L') which is defined
as 10 millionth of the length of a meridian measured from the equator of
the planet to the pole:
L' = 10^(-7) * (1/4) * (2*pi*R)
2*pi*R being just the circumference of the planet.
We are interested in how L and L' compare. So we will investigate their
ratio:
L/L' = (G*rho*(4/3)* R * 1s^2/pi) / ( 10^(-7)*(1/4)*(2*pi*R) )
L/L' = (10^7*G*rho*(8/3)*1s^2) / (pi^2)
Notice that this ratio does not include an R. So this ratio is
independent of the radius of the planet. The only term that is not a
constant is (rho) which is the density.
Conclusion: The ratio between the two definitions of a metre depend only
on the planet's density.
Cheers,
Daniel.
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