Definition of metre in other planets

[Daniel Carrera-2]

It's very interesting that the definition of a metre the length of a
2-second pendulum would be so close to the definition of a metre as 10
millionth of the meridian along a quadrant. So I asked myself whether
this coincidence is unique to the earth or if it would happen in other
planets.

It turns out that what is unique about the earth is its density. Any
planet, whether big or small, will have the same close relationship
between the two definitions of metre, as long as it has the same density
as the Earth. In a big planet, the metre would be longer, but the two
definitions of metre would still agree if the planet's density is right.

I would provide a full formal derivation, but that might be off-topic :-)

Daniel.

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[Paul-16]

> It's very interesting that the definition of a metre the length of a
> 2-second pendulum would be so close to the definition of a metre as 10
> millionth of the meridian along a quadrant. So I asked myself whether this
> coincidence is unique to the earth or if it would happen in other planets.
>
> It turns out that what is unique about the earth is its density. Any
> planet, whether big or small, will have the same close relationship between
> the two definitions of metre, as long as it has the same density as the
> Earth. In a big planet, the metre would be longer, but the two definitions
> of metre would still agree if the planet's density is right.
>
> I would provide a full formal derivation, but that might be off-topic :-)
>
> Daniel.
>

Unsure whether I would follow a formal proof, but it is good to see you back
on the OOo lists...

/paul

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[Pat McBride]

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.  I'm just a simple
Engineering Technologist, the engineers can figure out that stuff, and I'll
tell them that they're wrong by using common sense.

Pat

-----Original Message-----
From: Daniel Carrera [mailto:daniel.carrera@...]
Sent: 2008/05/21 02:15
To: social@...
Subject: [social] Definition of metre in other planets


It's very interesting that the definition of a metre the length of a
2-second pendulum would be so close to the definition of a metre as 10
millionth of the meridian along a quadrant. So I asked myself whether
this coincidence is unique to the earth or if it would happen in other
planets.

It turns out that what is unique about the earth is its density. Any
planet, whether big or small, will have the same close relationship
between the two definitions of metre, as long as it has the same density
as the Earth. In a big planet, the metre would be longer, but the two
definitions of metre would still agree if the planet's density is right.

I would provide a full formal derivation, but that might be off-topic :-)

Daniel.

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[Daniel Carrera-2]

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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[The Janitor]

22 May 2008 - 19:58 (where I live)
Daniel Carrera

Hello Daniel,

Thursday, May 22, 2008, 1:29:07 PM, you wrote:

DC> Acceleration due to gravity is:

etc.

Didn't have all these complications when we used imperial   ;-)

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[Ian Lynch]

On Thu, 2008-05-22 at 20:00 +0100, The Janitor wrote:

> 22 May 2008 - 19:58 (where I live)
> Daniel Carrera
>
> Hello Daniel,
>
> Thursday, May 22, 2008, 1:29:07 PM, you wrote:
>
> DC> Acceleration due to gravity is:
>
> etc.
>
> Didn't have all these complications when we used imperial   ;-)

You mean like an acceleration of 32 ft/sec/sec for gravity and 16 ounces
in a pound and 550 ft poundals per second to make a horse power?

I think approx 10m/s/s for gravity and multiplying that by 5 kg to get
50 newtons and that by say 7 m/s to get 350 watts. Rather less
complicated arithmetic and easier to understand for anyone learning
basic physics. The metric system is only more complicated if you are
entrenched in imperial (and I say that as a UK citizen that has to use
both and inter change between them). Its exactly the same reason that MS
Office users say OOo is not as intuitive or easy to use.

Ian
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[Pat McBride]

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:daniel.carrera@...]
Sent: 2008/05/22 05:29
To: social@...
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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[Pat McBride]

that bit at the bottom sounds like you might be a fellow Celt?  Are you
Welsh, perhaps?

Pat

-----Original Message-----
From: The Janitor [mailto:thejanitor@...]
Sent: 2008/05/22 12:00
To: Daniel Carrera
Subject: Re[2]: [social] Definition of metre in other planets


22 May 2008 - 19:58 (where I live)
Daniel Carrera

Hello Daniel,

Thursday, May 22, 2008, 1:29:07 PM, you wrote:

DC> Acceleration due to gravity is:

etc.

Didn't have all these complications when we used imperial   ;-)

--
Best regards,

The Janitor

mailto:thejanitor@...

Flying with The Bat! Professional
version 4.0.7

Privileged and/or confidential information and/or copyright
material may be contained in this e-mail.If you are not the intended
recipient or the person responsible for delivering it to the intended
recipient you must not copy it,deliver it to anyone else or use it in
anyway whatsoever.To do so is prohibited and may be unlawful.
Instead, kindly destroy this message and notify the sender by reply
e-mail.

Fe all fod gwybodaeth freiniol a/neu gyfrinachol a/neu
ddeunydd dan amodau hawlfraint y neges e-bost yma.
Os nad y chi sydd i fod gael y neges, neu'r un sy'n gyfrifol
am ei throsglwyddo, rhaid i chi beidio a'i chopio na'i
throsglwyddo i neb un arall na'i defnyddio mewn unrhyw fodd o gwbl.
Mae gwaharddiad ar i chi wneud hynny ac efallai y byddech chi'n
torri'r gyfraith trwy wneud hynny. Dinistriwch y neges yma ac
anfonwch neges e-bost at y sawl a'i hanfonodd i roi gwybod iddo fe.

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effective and efficient anti-virus software.


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