On 2012-01-01, STJensen wrote:
So 64 feet per second squared = 2 gravities?

Yes, though in a coordinate system based at the Earth's surface, a
stationary object feels one gravity already.


But that's for the first second. For the second second, the human
would have to be going at 128 feet per second squared to continue to
feel the force of two gravities, right?

No. Feet per second squared is an acceleration, and 128 ft/s^2 would
be approximately 4 gees.


So without the impact of Earth's gravity and then centrifugal force
past 22,000 miles, how fast would a human be traveling at the end of
62,000 miles and how long would he take to travel that 62,000 miles?
Now if someone knows how Earth's gravity and post-22,000-mile-point
centrifugal force would change the previous answer, I would love to
know that too.

You can calculate it from a work-done point of view:

delta-E = Integral F . dl

This is easiest if you treat the person as being subject to thrust
(real) force, and gravitational and centrifugal pseudo-forces. We may
assume that the Coriolis forces have negligible net effect, due to the
fact that the person's path stays along the beanstalk with its vastly
greater mass preventing significant sideways motion.

Then the forces at a given radius from the center of earth r a
thrust = m a, with a = 2 gees ~20 m/s^2;
gravity = -m g (R/r)^2, with R = Earth's radius ~6.4 Mm;
centrifugal = m w r^2, with w = rate of Earth's rotation ~73 urad/s.

The range of r is from R to R + h, with h = 60,000 miles ~100 Mm).

Then
delta-E / m = Int (a - g (R/r)^2 + w^2 r) dr
= a h + g R (1/(1+h/R) - 1) + (1/2) w^2 ((R+h)^2 - R^2).

E = (1/2) m v^2 and starts at zero, so you can just substitute the
above formula into v = sqrt(2 E/m).


--
Tim