well i just did the math 
. (should have warned ya... its a bit long )
Ok here is the intial tangental velocity an vehicle would have at the equator
Vlaunch site = R * angular velocityof earth
Where R is the radius from the spin axis (radius to the center of the earth)
Vlaunch site = 6378.137 (km)* .0000729212 (rad/s)
Vlaunch site = .4651 (km/s)
ok now for the tangental velocity at the geo sync orbit
Vcircular orbit = Sqrt(mu/R)
Where mu is earths gravitational constant, 3.986*10^5 and R is the radius of the orbit to the center of the earth (altitude (km) + 6378.137(km)
Vcircular orbit = Sqrt(3.986*105(km3/sec2)/42158(km))
Vcircular orbit = Sqrt(9.4549(km2/sec2)
Vcircular orbit = 3.074 km/sec
so as you can see you would not be able to maintain a circular orbit.
ok now to prove that it will burn up
first we must find the specfic mechanical energy which is always negative for elliptacl orbits, 0 for parabolic, and positive for hyperbolic (which makes sense as it takes more energy than what it take sto just take a parabolic orbit)
SME=V2/2 - mu/R
where R is the radius of the orbit, and v is the velocity at that orbit
SME=.4651(km/s)2/2-3.986*105(km3/sec2/42158(km)
SME = -9.23855 (km2/s2)
from here we can calculate the semi major axis
a = - mu/(2*SME)
a = - 3.986*105(km3/(2*-9.23855)
a = 21572.76 km
and now to determine the radius at perogee (closet to earth)
a=(Ra + Rp)/2
we know the Ra (radius at apogee) is 42158 as this is the furthest distance from earth
21572.76(km)=(42158 (km)+ Rp)/2
Rp = 987 (km) < 6378.137(km) and thus will make contact w/ earth.
to prove it has a highly elliptical orbit we must check it eccentricity (e). when e=0 the orbit is circular, 0<e<1, the orbit is elliptical with values closer to one becoming more elliptical, e=1 the orbit is prabolic (leaving earth), e>1 the orbit is hyperbolic (leaving earth, used for interplanatary travel)
e = (Ra - Rp) / (Ra + Rp)
e = (42158(km) - 987(km)) / (42158(km) + 987(km))
e = .95 (unitless)
i would tell you the time it would take but that takes shyt load of time
hope that helps (that took alot of time to type arg lol)
I'll need to go through this later on when I'm more awake, but I can see what Ken Mattingly meant when he said "I don't know how to do 90% of this mission."
I would hate to be the poor sod who has to translate this into roll and pitch angles and burn durations...
Anyway, thanks for taking the time out for that, I appreciate it and will go through it step by step when I'm fully awake in a couple of hours...
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