OK, I won't argue any more. Just so long as 1+1=2, payday's on Friday, the

beer is cold (sorry, Ian), and somebody finds out why mosquitoes were

included on the Ark.

We had a late thaw this year, went out this morning to do some work and was

greeted by a swarm of mosquitoes. They were big enough that I could see the

insignia on their wings and bodies, pretty bad when you can count the kill

marks! Had six feeding on one arm, and 20 more lined up making

reservations.

Got so bad, the swallows were hiding from them all; crows were carrying

shotguns in defense, and the eagles thought they'd died and gone to heaven

and thought they were feeding on sparrows.

Have I told you about my cousin, Jasper? Took his bull, Fred, into town the

other day, to get it shod. Trouble was, Farmer Brown was moving his herd of

cows to market. Fred got excited and started chasing the truck, which he

caught about 5 miles down the road. RCMP came along, gave Jasper a ticket

for unsafe operation of a bull, Farmer Brown a ticket for running a truck of

ill repute, and Fred a ticket for speeding. Nothing for the girls, however,

just Fred.

Stay tuned for more Cariboo humor.

Pat

-----Original Message-----

From: Daniel Carrera [mailto:

[hidden email]]

Sent: 2008/05/22 05:29

To:

[hidden email]
Subject: Re: [social] Definition of metre in other planets

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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