On Sat, 31 Dec 2011, STJensen wrote:

Let us say that you had a 62,000-mile-long Earth-anchored space
elevator and let us say it has an electro-magnetic repulsion
accelerator along its entire length. If you were to accelerate a
human so that he experienced only 2g (twice the force of gravity)
during the entire length of the space elevator, what velocity would
that human be expelled from the space elevator and how long would it
take the human to travel the entire length?

If you disregard angular momentum by assuming the Earth doesn't rotate,
air resistance and weakening of gravity by distance from center of Earth,
the problem becomes:

If a person accelerates at one g for 62,000 miles, what is the
terminal speed and how long does it take to reach terminal speed?

d = distance, v = velocity, a = acceleration

Set d(0) = 0 = v(0), a = g.

v = integral(0,t) a dt = at
d = integral(0,t) v dt = integral gt dt = gt^2 / 2

2d = gt^2; t = sqr 2d/g (= 76 min)

v = at (= 28 miles/sec, if I've done the calculations correctly.)

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If you disregard angular momentum by assuming the Earth doesn't rotate,
and air resistance, then the time will be shorter, the terminal speed
larger and the calculation more complex.

r = radius of Earth; s = distance above earth

Set s(0) = 0 = v(0).

Force of gravity at s is gr^2 /(r + d)^2.
(Is there a physicist in the house?)

Net acceleration at s is 2g - gr^2 /(r + d)^2.

2g - gr^2 /(r + s)^2 = dv/dt = d^2 s/dt^2 = s"

2g(r + s)^2 - gr^2 = (r + s)^2 s"

gr^2 + 4grs + 2gs^2 = (r + s)^2 s"

If I set it up right, that's a hard to solve differential equation.

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