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*> Damien Broderick wrote:
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*> >A year or so back, there was a brief discussion on the list of the
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*> >sub-orbital dynamics of a vehicle connecting distant places via an
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*> >evacuated tube under the ground
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*>
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*> Andrew Clough wrote:
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*> Anyway, assuming a
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*> straight path, you won't feel any free fall. In fact, you'll always feel
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*> 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

arccos(5220/6370) = .61 radians or about 35 degrees,

the occupants get a momentary weightlessness. Greater

than that angle, they get negative Gs.

Welcome to the list Andrew! spike

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