Re: ballistic subterranean trains

From: Spike Jones (spike66@attglobal.net)
Date: Tue Sep 25 2001 - 21:51:06 MDT

> Damien Broderick wrote:
> >A year or so back, there was a brief discussion on the list of the
> >sub-orbital dynamics of a vehicle connecting distant places via an
> >evacuated tube under the ground
>
> Andrew Clough wrote:

> Anyway, assuming a
> straight path, you won't feel any free fall. In fact, you'll always feel
> at least some gravity, always perpendicular to the tunnel floor!

Disagree, partially. Imagine the tunnel thru a non-rotating uniform
sphere (a close enough model of the earth for the sake of
argument). The car reaches maximum velocity when its
floor is perpendicular with the radius vector of the sphere.
If it is going sufficiently fast, the centripetal acceleration
equals the acceleration due to gravity at that point, so the
occupants of the car are temporarily wieghtless. If the
tunnel is any steeper, the occupants experience negative Gs.

Ill estimate the angle required for weightlessness, without
looking up anything. If I recall correctly, the earth is about
6E24 kg, with a radius of 6370 km. I know it gets denser near
the core, but lets ignore this. The potential energy of particle
at the surface is MG/R. If an object is in a deep hole, the
potential energy is MhG/Rh, where Rh is the distance from
the bottom of the hole to the center of the earth and Mh is
the mass of the sphere with radius equal to Rh. So the
train car converts energy from potential to kinetic, and
assuming no other energy conversion and if I calculated
right with the above, then G is about 6.7E-11 and the
velocity of a round orbit is (MhG/Rh)^.5, so set

KE = .5 M V^2 = .5*(MhG/Rh) = MG/R-MhG/Rh

1.5Mh/Rh = M/R

The average density of the earth is calculated at about
5500 kg/m^3, so

Mh = 4/3*pi*Rh^3, so

Rh = (2*pi*rho*M/R)^.5 = 5200 km

so if the angle of a straight tunnel with the ground is equal to