# Re: Cooling technique for Jupiter brains

From: Spike Jones (spike66@ibm.net)
Date: Fri Feb 11 2000 - 23:56:43 MST

> From: Spike Jones <spike66@ibm.net> Wed, 09 Feb 2000:
>
> >energy and a big ball of stuff starts to form in the center. Or does
> >it? What is the shape of the gravitational field in that tunnel?

Jeff's right. The G field increases linearly along the axis of the tunnel
and and is unchanging in any direction away from the axis. So then,
the junk tossed inside the tunnel, upon losing kinetic energy, would
form a flattened disk, right? The junk pile would attract itself, and
would have a G field axial along the tunnel only. So the disk would
be flattened along the direction of the tunnel until it touched the sides.

> Amara Graps wrote:
> Hey you know this problem! Fun isn't it?

Ja!

> First of all, regarding the time it takes to fall from one
> end to the other (it takes 42 minutes each way) it doesn't
> matter _where_ the ends of the cylinder are located, i.e. where
> "yahoo_A" and "yahoo_B" are pitching their trash at each other.
> The time to fall from one end to the other (by gravity alone)
> is independent of the two places.

Ja, assuming of course the tunnel passes thru the center
of the nonrotating sphere.

> Second of all (and this is even more cool :-), the
> object when it falls has an oscillatory motion (single
> harmonic motion about the center of the earth).

and the time it takes to oscillate one full cycle is equal to the time
to orbit around the sphere, assuming the lowest possible orbit
around a smooth sphere with no atmosphere. {8-]

> ==> T = 85 minutes for the full period (from AtoB and return)

If you are in a boring meeting and need to derive the mass of the
earth, one need only remember the orbit time of 85 minutes, and it
can be backed out with only that information. The lowest satellites
take a little longer, since they need to be a couple hundred km above
the deck so they dont get dragged down immediately by the atmosphere.

There is another way, if you need the earth's mass to only one
digit of precision: you know the volume of a sphere is 4/3pi*r^3,
but for single digit accuracy, pi~3, so V~4*r^3, and you know
that the km was initially proposed as 1/10,000 the distance from
the equator to the pole, so the radius of the earth can be calculated
to be 10,000/pi*2~6370 km. Cube that and quadruple it, and the
volume of the earth comes out to close enough to 1E12 km^3, and
since the density of iron and nickel are in the 8000 kg/m^3 range
and granite and other rocky stuff is in the 4000 range, then assume
an average density of about 6000 kg/m^3 and you get an earth
mass of 6E24 kg, which is quite close enough.

>From that, knowing that a 1 kg mass weighs a force of 9.8 newtons,
one can calculate the universal gravitation constant without having
any reference material handy, assuming you remember that
F=GMm/R^2, since you know F, M, m and R.

But I digress. {8-]

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