The 42 minutes appears to be correct. We can show this in two parts.
The first is to show that the time is independent of the distance as
long as the tunnel is a straight line.
A *
| * R
| *.
| . *
| . * M
| . /*
| . / *
| . / *
|. / *
|/ *
--------------------*
C B
In the diagram above, if you have a constant size font similar to time,
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.
The gravitational force at point R is proportional to the distance from
the center of the earth. This is because the portion of the earth
outside point R cancels itself out gravitationally. Only the part
at radius less than CR counts. Mass goes as the cube of the radius,
gravitational attraction as inverse square, and the net result is that
gravitational force goes as the radius. (This neglects the increase in
density towards the core.)
You then have to project this force along the direction of the tunnel
to determine the acceleration the train will feel. You can see in the
diagram that projecting vector CR onto line AB will be the segment
RM. Therefore the acceleration along the tunnel at point R will be
proportional to the distance RM.
Since RM is proportional to the size of the whole tunnel, it follows
that the time it takes will be independent of the tunnel length.
A tunnel which is X% longer will have exactly X% larger accelerations
and therefore take exactly the same amount of time.
This means that we can get the answer to any tunnel problem by considering
the longest possible tunnel, one which passes straight through the center
of the earth and comes out the other side. This is simpler to analyze.
Let x be the distance from the center of the earth, starting at about
6.4E6 meters. Acceleration is initially -9.8 m/s^2, the acceleration
of gravity. We have that force is negatively proportional to x, which
can be met by the formula x = k cos (wt). k is the initial radius value
of 6.4E6. Differentiating this twice gives accel = -w^2 x. Setting
accel at -9.8 and x at 6.4E6 gives w = .00124. So the formula is
x = 6.4E6 cos (.00124 t), with x in meters and t in seconds.
To fall through the earth and come out the other side, the argument to
cosine must go from 0 to pi, so we have pi = .00124 t for the time to
get from one end to the other. This leads to about 42 minutes.
Hal
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