If a satellite is in a circular orbit above Earth at an altitude of
4000 miles, how fast is it moving and how long would it take to make
one revolution around the Earth?

Interesting question, halejrb-ga!

First, let's tackle the speed, since it's the hard part of your
question.

The diameter of the earth is about 12.7 million meters, or 6.35
million meters diameter, plus about 6.5 million meters above the
surface gives not quite 13 million meters from the earth's center of
mass to the satellite.  To achieve a stable orbit, the gravitational
force pulling the satellite towards the earth must equal the
centripetal force pushing it away.  If I remember my physics right:

M1*M2*G/r^2 = gravitational attraction
M1v^2/r) = centripetal force

We factor out M1 (the mass of the satellite), and r, leaving and are
left with M2 * G / r = v^2, or M2 kg * 6.673 * 10^-11 m^3 s^-2 kg^-1 /
13EE6 m = v^2 m^2 s^-2.  (meters, kilograms, and seconds).  The mass
of the Earth is about 6EE24 kg.  Checking the units, they cancel (the
left hand side gives m^2/s^2, as does the right).  Performing the
square root, we find v = ~5550 m/s (about 12.5 thousand miles per
hour).

Now, as for one revolution around the earth.  The circumference of a
circle is pi * diameter.  The diameter of the circle the satellite
sweeps out is about 26 million meters, so the circumference is about
81.5 million meters.  81.5 million meters / 5550 meters / second =
15000 seconds = about 4 hours.  This makes intuitive sense since 4
hours is one sixth of 24 hours, and a geosynchronous orbit, which
takes 24 hours, is 6 times as high.  After all those units, we get "go
twice as high, and it takes twice as long".

I hope whatever you need this for is as interesting as finding the
answer was!

-Haversian

A great answer and a great service for those of us who never took physics.

Please note that the above answer assumed that the body is orbiting
Earth in a circular path ON THE SAME PLANE AS THE EQUATOR.

Currently an Earth orbit with radius 4000 miles is only useful for
earth sciences and astronomy. Its period of about 4 hours make it
virtually useless for communication purposes and its height makes it
useless for Earth observation.

For example, there were two satellites called Lageos-1 and Lageos-2
which were deployed in 1976 and 1992 respectively. These tiny
satellites orbited our planet at an altitude of nearly 4000 miles. The
lageos satellites are passive vehicles covered with retroreflectors
designed to reflect laser beams transmitted from ground stations. By
measuring the time between transmission of the beam and reception of
the reflected signal from the satellite, stations could precisely
measure the distance between themselves and the satellite.

Regards.

"Go twice as high and it takes twice as long" is incorrect.  The
period of a satellite is proportional to the radius of the orbit to
the power 1.5.  So "go 4 times as high, takes 8 times as long" is
correct (if you measure height as the distance from the center of the
earth), though not very useful.  If you measure height from the
surface of the earth, the period of the satellite is not proportional
to any power of the height.  It is just happenstance that the period
of a satellite at 1/6 the altitude above the earth's surface as a
geosynchronous satellite is about 1/6 of a day.  Actually, such a
satellite would orbit 6.32 times each day.