# Why is the SES launch going to an apogee of 80,000 km?

Why is the apogee of the Falcon 9 SES launch going clear out to 80,000 km? GEO is only 36,000 km or so, it seems like a waste to go so far beyond it.

The launch vehicle upper stage only delivers the satellite to a transfer orbit, typically with a low perigee and apogee near GEO - in this case, the target orbit is 295x80000 km at 20.75$^\circ$ inclination. After that, it's up to the satellite itself to use its onboard propulsion system to get itself into geostationary orbit, which is circular at 35,786 km and as near 0$^\circ$ inclination as possible.
The $\Delta V$ for an impulsive combined maneuver to change altitude and inclination is given by the following equation: $$\Delta V=\sqrt{V_i^2+V_f^2-2 V_i V_f \cos \Delta i}$$ Where $V_i$ is the initial orbit velocity, $V_f$ is the final, and $\Delta i$ is the inclination change. Since the inclination change is what's interesting in this problem, let's just look at a pure inclination change with no change to the orbit's semi-major axis. Note that in the above equation, if $V_i = V_f$, it simplifies to $$\Delta V = 2V_i\sin\left(\frac{\Delta i}{2}\right)$$ So for a given $\Delta i$, the required $\Delta V$ changes linearly with $V_i$. For a circular orbit, $V=\sqrt{\frac{\mu}{r}}$, where $\mu$ is the standard gravitational parameter for the central body (Earth in our case), and $r$ is the circular orbit radius. From this, we see that $V$ decreases as $r$ increases (for non-circular orbits, we get the same effect).
Putting these together: increasing $r$ decreases your velocity $V$, which in turn decreases the amount of $\Delta V$ to reorient your orbital plane. So this means if you do your maneuver(s) to take out your inclination, you stand to save a lot of propellant (thus adding potentially months or years to your life) by doing it at a higher inclination. In this specific case, it's a no-brainer for SES, since the Falcon 9 easily has the performance to inject to the higher orbit.