In a hypothetical case if an object was dropped in a pure vacuum in
say, a 5 or 10 mile high tube how fast would it be going when it hit God?s
green earth?  Also does gravity have a speed where it ?tops out?

The scientific term for an object that is in free fall that has
reached a spead where it stops accelerating is called terminal
velocity.  This is the spead at which the friction(also called drag)
with other matter equals out the the force of gravity.  Under
newtonian physics, an object in a vacum would continue to acclerate at
a velocity by 9.8 meters/s^2 since there is nothing to interact with
to cause friction.  I don't have the formulas but after one second,
the object has covered 4.9 meters and is traveling at a speed of 9.8
meters a second. (remeber it started at a spead of 0 so the distance
covered is not 9.8 meters). After two seconds, it is traveling at 19.6
meters/second. etc.

Eistein came up with a new set of rules called special relativity,
where objects accelerate, the kinetic energy increases its mass. 
Under special relativity, an object would not be able to reach the
spead of light because its mass would become infinitly large as it
approached the spead of light.

Under Special relativity, anything with mass should not be able to
reach the spead of light, but there are hypothosiis that there are
particles that can travel faster than the spead of light such as the
tachyon but that is way beyond the scope of the question.

Hi, mcisencraft-ga:

In one sense the "speed of gravity" would mean the speed at which the
force of gravity propagates, which theoretically is the speed of
light.  But from reading your Question, it seems that is not what
you're asking.

The Earth's radius is a bit less than 4,000 miles, so at an altitude
of 5 or 10 miles the gravitational force exerted by the Earth on an
object (which varies inversely with the square of distance) is
slightly weaker than at the Earth's surface.  Over longer distances
the weakening of gravity would be more significant.

To look at this phenomenon from another perspective, consider the
"escape velocity", i.e. the speed necessary for an object rising away
from Earth to leave orbit.  At the Earth's surface the escape velocity
is roughly 7 miles per second or 25,000 miles per hour:

[What is escape velocity?]
http://www.physlink.com/Education/AskExperts/ae158.cfm

"Escape velocity is defined to be the minimum velocity an object must
have in order to escape the gravitational field of the earth, that is,
escape the earth without ever falling back."

Naturally at higher altitudes the escape velocity is smaller, but one
answer to the last part your Question could be this escape velocity. 
An object "at rest" falling to Earth in a vacuum from arbitrarily far
away would never reach the escape velocity (this time directed toward
Earth, rather than away from it), so in this sense that's the speed at
which Earth's gravity "tops out".

regards, mathtalk-ga

Since it is in a vacumme it would be d=.5 a t^2  where 
d= distance
a = Accelleration
t = time

On Earth a = 32 feet per second squared.  So in one second we get
d = .5 X (32) X 1^2 = 16 feet. and so forth.  To fall 10 miles in a
vacume would take 58 seconds or almost a minute.  Ouch!  It would be
falling at a rate of 10 miles a minute(about), which is not fast
enough to be noticible with regards to special relativity.

This is assuming that the object that is falling never bumps into the
sides of the tube, which would slow it's descent somewhat.

On a personal aside, it is sad to see that, even in the year 2004, the
United States is still saddled with the archaic imperial measurement
system.  I hope that the next president finally gets the ball rolling
on educating people on the merits of the metric system.