# Why aren't orbits transferred at the perigee?

In example 10.6 titled "Satellite Orbit Transfer 1" of Kleppner and Kolenkow the author says

The most energy-efficient way to put a satellite into circular orbit is to launch it into an elliptical transfer orbit whose apogee is at the desired final radius. When the satellite is at apogee, it is accelerated tangentially into the circular orbit.

At the apogee, because the satellite is farthest from Earth, it's velocity is lowest. However since the aim of the boost is to increase the kinetic energy and since $$\Delta K = \frac{1}{2}m[(\mathbf{v} + \Delta \mathbf{v})^2-v^2]$$ $$= m\mathbf{v} \cdot \Delta\mathbf{v} + \frac{1}{2}m (\Delta v)^2$$ the easiest way to increase kinetic energy with a small $$\Delta v$$ is by applying it when the velocity is maximum which is at the perigee, not the apogee. So why aren't orbits changed at the perigee?

Your aim is the circular orbit of radius $$R$$. If you do as you suggest, then this would mean that you first launch your satellite into an elliptic orbit with $$R_p=R$$ being its perigee and $$R_a>R$$ its apogee. But this also means that this transfer orbit's major axis will be $$R_p+R_a>2R$$, so you need to spend so much energy to get there, and then lose all the unneeded energy to circularize the orbit.
On the other hand, how K&K explain in your quotation, requires one to launch the satellite to a transfer orbit with apogee $$R_a=R$$ and perigee $$R_p, which will have major axis of $$R_a+R_p<2R$$, and then to gain some more energy to get to the circular orbit.
I didn't calculate, but I suppose you won't be able to compensate for the extra energy you spent in your accelerate-at-perigee method by using perigee instead of apogee. Notice that in both cases the distance, at which you accelerate to circularize the orbit, is the same, $$R$$.