Escape velocity of satellites I know that the equation for it is $$v^2 = \frac{2GM}{r},$$ and with that, the rocket should be launched at that speed. But could it go much slower spending much more fuel to escape from gravity right? 
Wouldn't it be easier to calculate it with energy? 
 A: You've made the common mistake of thinking that the velocity needed to launch a satellite is the (initial) velocity needed to raise it to its orbital radius.
If you raise a satellite to e.g. 300km then let go the satellite will immediately fall straight back to Earth. You need to do two things:


*

*raise the satellite to 300km

*increase its tangential velocity to $\sqrt{GM/r}$
The rocket used to launch a satellite doesn't travel straight up. It travels in a curve that looks something like:

(Image from nasaspaceflight.com)
A: Any launch profile will suffice (as long as you do not try to go through the Earth of course) as long as the velocity at the end meets the following criteria,
$$
\|\vec{v}\| \geq \sqrt{\frac{2GM}{\|\vec{r}\|}}
$$
where $\vec{r}$ is the radius (position relative to the center of mass of the Earth) at that moment.
For this I also assume that its trajectory will not go through the Earth as well and is sufficiently out of its atmosphere.
You could few escape velocity from an energy point of view, since when escape velocity is reached the specific orbital energy becomes zero:
$$
\epsilon = \frac{v^2}{2} - \frac{GM}{r},
$$
because the gravitational potential is defined such that it goes to zero when $r$ approaches infinity. So at escape velocity, if all kinetic energy would be converted into potential energy, then you would have to go infinitely far away.
A: Of course it could go much slower spending much more fuel to escape from gravity ! 
If you look at Ariane 5 speed after 2minutes in the air, it is "only" 2km/s, far from the 11km/s required to leave the Earth's attraction (at ground altitude).
It could indeed go much slower, it would just to go much further (if you decrease v, you increase r)
