Re: Cooling technique for Jupiter brains

From: Amara Graps (
Date: Thu Feb 10 2000 - 15:36:42 MST

From: Spike Jones <> Wed, 09 Feb 2000:

>Here's one I thought about: suppose you had a solid uniform sphere,
>Earth-sized, with a 600 km diameter hole drilled thru the middle.
>Suppose the construction workers toss stuff in the hole just for the
>fun of seeing it pop back up 88 minutes later. At first it does so, but
>some yahoo on the other side has the same idea and pretty soon
>a lot of people are throwing garbage in the enormous bottomless pit,
>and consequently the stuff starts to hit other junk and lose
>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?
>We already know if one were inside a huge hollow sphere there
>is no gravitational field, easily proven by one integral, but what
>about a sphere with a cylindrical tunnel? spike

Hey you know this problem! Fun isn't it? I draw some notes
here from one of my old homework. Maybe the answer to your question
will pop out here somewhere along the way.

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.

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). So yahoo_A
will get his/her/it's trash thrown *right back* at him.
Think of the games!

It looks like this.

          B (y should be perpendicular, but ascii drawing is hard)
tunnel->/x=0 x theta=angle from x=0
       / y -- / to where mass m slides through
      / / tunnel
    (x=A) F = -G M' m /r^2 rhat

                       M'=mass enclosed by the sphere
                        =(4/3) pi rho r^3
                       m = mass falling through tunnel
                        rhat = r direction
                   F = (-Gm/r^2)(4/3 pi rho r^3) rhat
                   F=-Gm(4/3) pi rho r rhat

  The x component of F is the only one that matters.

  F = Fx = Fsin(theta)
  but sin(theta) = x/r
  Fsin(theta) = F(x/r)
              =-G(4/3 pi rho r m) x/r

     F = -G(4/3 pi rho m) x see! indep of theta
         this is like a spring constant k

  i.e. F = -kx, shows the oscillatory behavior of the mass
      thrown down the tube (so in 42 min, it will come right
      back at you like a rubber band!)

      Period of oscillator of mass m:
      T = (2pi/2) SQRT(m/k) substitute m
      T = 2pi(SQRT(3/(4 pi G rho))

      At A or B, the body is at the surface of the Earth
      where gravitational force on m is
      F = GMm/R^2

      And we know M=(4/3)pi rho R^3

      Substitute M: F = (4/3) G m pi rho R = mg
      Note g/R = (4/3) G pi rho

      Therefore the period T = 2pi SQRT(R/g)
      R = 6.37E6m, g = 9.8 m/s^2

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


Amara Graps email:
Computational Physics vita: finger
Multiplex Answers
"If you gaze for long into the abyss, the abyss also gazes into
you." - -Nietzsche

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