Re: Cooling technique for Jupiter brains

From: Amara Graps (amara@amara.com)
Date: Sat Feb 12 2000 - 11:34:21 MST


From: Spike Jones <spike66@ibm.net> Fri, 11 Feb 2000 :

>From me:
>> 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.

No, no, that's one of the beautiful things about this problem.
It _doesn't matter_ where the ends of the cylinder are located.
It doesn't have to go through the center of the sphere. You
simply have to assume that the density of the sphere is constant
and that there's no rotation.

So the period for dropping through this Earth tunnel and bouncing back,
B_to_A and return, is 85 mins. Dropping through a tunnel any shallower
or deeper is _also_ 85 mins.

                         * B*
                     * / *
                    * / *
                   * /
                   * / * *
                   * /
                   * / *
                    A *
                      * *
                          * *

Try making it into a geometry problem, and dividing the sphere into
conic sections, like what you do in 2D with the Kepler problem.
The gravitational force on the mass is directly proportional
to the mass and inversely proportional to the square of the distance, so
the regions of conic sections provide equal and opposite forces on the
mass due to the trade-off between mass and distance.

[professors love to give this problem in qualifying exams]

>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. [...]
>>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.

What a nerd!

:-)

Amara

*****************************************************************
Amara Graps | Max-Planck-Institut fuer Kernphysik
Interplanetary Dust Group | Saupfercheckweg 1
+49-6221-516-543 | 69117 Heidelberg, GERMANY
Amara.Graps@mpi-hd.mpg.de * http://galileo.mpi-hd.mpg.de/~graps
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        "Never fight an inanimate object." - P. J. O'Rourke



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