How would North Korea launch a geostationary satellite?
A speculative example - Jonathan McDowell
Let's assume we launch due east from Sohae.
Let the second stage reach 7.342 km/s at cutoff,
which puts it in a mildly suborbital -465 x 300 km x 39.7 deg orbit.
Next, use spin motors (rockets mounted tangentially) to
spin up the vehicle so it is spin-stablized and
maintains the same inertial pointing direction without need
for attitude control. Separate the third stage and coast.
The third stage (and discarded second stage) cross the equator
23 minutes after launch at an altitude of 224 km and
velocity vector pointing slightly down (-3 deg).
Fire the third stage which we assume was separated in an attitude
which makes it now pointing local horizontal.
Size the third stage to give you a delta-V of 2.64 km/s, which
puts you in a 219 x 35793 km x 39.7 deg geotransfer orbit.
Separate the payload, still spinning.
Suppose the satellite mass is 1000 kg 'wet' and the propellant
has an effective Isp of 280s (like 1960s UDMH engines),
and guess a third stage dry mass of 300 kg. Then 2100 kg
of propellant is needed, for a total mass of 3400 kg inserted
into the near-orbital second stage trajectory.
First apogee is at T+5.5 hr over Brazil. First perigee is at T+11H over
the Horn of Africa Second apogee is at T+16.5h over 152E 0N, which
should be just about visible from Korea but it's a bit low on the
horizon.
Instead, let's burn stage 3 slightly more, 2.69 km/s, to get a 39300 km
apogee, 2nd apogee is over 130E at T+18h and you have a great view from
Korea. Inertial attitude of payload, since it's still spinning, is the
same as at second stage separation. You have to yaw the payload to the
right direction, this is the tricky part. At 2nd apogee you are
travelling northeast at 1.5 km/s inertial (1.0 km/s north, 1.1 km/s
east). You want to be going due east at 2 km/s, so you need to burn 1.0
km/s south and 0.9 km/s east. With a more detailed analysis I get a
burn of about 2.0 km/s needed to get to an orbit of 34000 x 39000 km x
1.0 deg, drifting 10 deg W a day.
There'll be a big uncertainty on the resulting orbit. Say they have an
orbit determination a day later. Then the sequence of events will depend
on the actual orbit. In my example, they'd burn half a day later, about
T+50 hours, as it passed through the GEO altitude at 115E. A burn 50 m/s
up and 20 m/s retrograde would reduce the drift rate to 0.1 degree per
day. Now you can take your time, because the longitude won't change much.
Another day later, a very small burn can reduce the drift rate to less
than 0.01 deg per day and we're done.