# MATH: Gravity in the tunnel (was Re: cooling JB's)

From: Jeff Davis (jdavis@socketscience.com)
Date: Fri Feb 11 2000 - 14:09:07 MST

Futurist folk,

Spike wrote:

>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?

You now the equation for gravitation:

F= GmM/(r^2). Eq #1

Inside the tunnel, r becomes a variable, with a range of values from r=
R(surface) to r=0, with M also varying as a function of r, thusly:

M(as a function of r)=M(r)=(4/3)(pi)(r^3) (rho) Eq#2

where rho is the average density of the tunneled sphere.

Sustituting #2 into #1 gives:

F=Gm(4/3)(pi)(rho)(r^3)/(r^2), which reduces to

F=Gm(4/3)(pi)(rho)r Eq #3 gravitational force in the tunnel.

Ergo, the force of gravity in the tunnel varies in a simple linear
fashion, going from zero at the center to its maximum at the surface.

This, of course, also means that any object in the tunnel experiences a
linearly varying rate of acceleration.

F=ma combined with Eq #3 above, F=Gm(4/3)(pi)(rho)r , gives

ma = Gm(4/3)(pi)(rho)r which reduces to

a(as a function of r)= G(4/3)(pi)(rho)r Eq #4: rate of accel. of
object in tunnel

And not an integral or differential in sight.

By the way, spike, I'm often out of sync with the discussion, due to my
delay in getting to the more current posts. I wasn't trying to drag the
discussion back to my (by then) older version of the idea. A common
problem. Even now am I in the same situation.

Many folks took this thread and ran with it, which is precisely as it
should be. In the spirit of free ideaware, I salute you!

Best, Jeff Davis

"Everything's hard till you know how to do it."
Ray Charles

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