Can an Artificial Satellite Move Opposite in a Geostationary Orbit?

[luckis11]

Helpful notes:
The speed v=√(GM/(R+h)) which is required for an artificial satellite to be set at orbit, is one that its Δx is drawn on the reference frame (call it FR2) that does not move together with the self-rotating movement of the earth. Because whereas it has the speed v=√(GM/(R+h)), the Δx of its speed which is drawn on the reference frame (call it FR1) that moves together with the self-rotating movement of the earth, is zero.
The geostationary orbit can only happen at the height of ~35,000km above the ground (see http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970408d.html).

My question is:
Is it possible to set an artificial satellite at an orbit at the geostαtionary height (35,000), and the direction of its motion (its motion Δx that is drawn on the FR2) to be the opposite of the geostationary satellites? (It IS possible as it seems at the moving drawing at http://en.wikipedia.org/wiki/Satellite). If-since it is possible, then the speed that it must have is again v=√(GM/(R+h))?

 

[Vanadium 50]

Is it possible to set an artificial satellite at an orbit at the geostαtionary height (35,000), and the direction of its motion (its motion Δx that is drawn on the FR2) to be the opposite of the geostationary satellites?

Of course. But it won't be geostationary, since it's going in the wrong direction.

 

[luckis11]

The answer I ended up with so far, is:
The speed required should or might be v=√(GM/(R+h) for both cases of a satellite which moves at the same direction as the self-rotation of the earth, and of a sattellite moving at the opposite direction. But providing initial speed v=√(GM/(R+h) alone, cannot result in an orbit for both cases, because the air resistance those two meet, is not the same. And that air resistance difference is not small. The motion of the air drawn on FR2 at sea level is a wind of 465metres/sec=1,674km/hour. Same speed drawn on FR2=>different speed drawn on FR1=> different speed in relation to the air. Now, at the height of the geostationaries 35,000km, the air or eather resistance might be almost zero, but the v=√(GM/(R+h) refers to all heights.

 

[Chronos]

Huh? Rocket scientists do not overlook air resistance in satellite launches. They also generally include booster rockets on the payload to nudge satellites into the desired orbit.