# How to calculate the trajectory needed to enter geostationary orbit?

Im building a rocket simulation engine and I would like to try and calculate the components of acceleration required to get the rocket into geostationary orbit. For example, the yellow and green lines in the photo below.

Given that geostationary orbit is around 39000km, how do I calculate the trajectory that the rocket would need to travel through become tangential to the geostationary orbit.

Take a look at the Hohmann transfer. This technique can be used to construct an orbit which will be tangent to your initial circular orbit, and final circular orbit - similar elliptical orbits in your diagram assuming they started from a circular orbit at the perigee side. It should be fine for what it looks like you're doing.

• @Ronikos and good luck! By the way 42164.5km is used more commonly. – Puffin Dec 22 '16 at 19:08

The Hohmann orbit tangent to both LEO and GEO would have an perigee of 6678 km and an apogee of 42164 km.

I hope it's evident that the semi-major axis, a, of this transfer orbit is (6678 km + 42,216)/2.

The vis viva equation can be used to calculate the speed of a spacecraft on this orbit provided you know the distance from earth's center, r.

$v=\sqrt{GM(2/r - 1/a)}$

G=Gravitational Constant
M=Mass of Earth (see right column under Physical characteristics)
a=24,421 km for this transfer orbit.
r=6,678 km when ship's at perigee and 42,164 km at apogee

So there's the tools to get transfer orbit speed at perigee and apogee.

But what's the speed of the LEO and GEO orbits?

Well, with a circular orbit, r = a.

$v=\sqrt{GM(2/r - 1/a)}$
Substituting a = r,
$v=\sqrt{GM(2/r - 1/r)}$
reduces to
$v=\sqrt{GM/r}$

You can use the above to get speeds of both LEO and GEO orbits. At LEO r=6678, at GEO r=42,164.

The burn from LEO would be transfer orbit's perigee speed minus LEO speed. The circularization burn at apogee would be GEO orbit speed minus the transfer orbit's speed at apogee.

I hope I've given you enough to work this out on your own.