From: hal@finney.org
Date: Tue Sep 25 2001 - 23:07:22 MDT
Spike writes:
> 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.
I don't believe this is correct. Centripetal acceleration only applies
to circular motion, and the car is going in a straight line. Imagine a
test particle within the car going along a straight tunnel; no matter
how fast the car goes, the test particle will only fall towards the
bottom side of the car. No force will deflect it upwards.
Actually, assuming zero friction, the felt weight is constant throughout
the trip. If the maximum depth of the tunnel is D and the radius of
the earth is R, the felt weight is (R-D)/R times the normal weight.
You can see this from my earlier diagram:
A *
| * R
| *.
| . *
| . * M
| . /*
| . / *
| . / *
|. / *
|/ *
--------------------*
C B
(In the diagram above, if you have a constant size font similar to mine,
C is the center of the earth and the tunnel "*" goes from A to B which
are on the surface of the earth. The "/" line is the perpendicular from
the center of the earth to the midpoint M of the tunnel. The "." is a
line to a typical point R on the tunnel.)
At point R the gravitational force is decomposed into a component along
the direction of the tunnel AB, which is not felt and causes acceleration,
and a component perpendicular to the tunnel, parallel to the line CM,
which is the felt weight.
The strength of the gravitational force is proportional to the distance
CR, for the reasons described earlier. The magnitude of the felt weight
can then be found by projecting vector CR onto the line through C and M,
which projection is exactly the line segment CM. Hence the magnitude
of the felt weight is constant no matter where you are on the the trip.
The felt weight is easiest to calculate at the midpoint where it is
proportional to distance CM, which is where the formula (R-D)/R comes
from above, as that is the ratio of the distance CM to the radius of
the earth.
Hal
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