Could we say that since it fired briefly, the potential energy doesn't change in that brief time-span but the kinetic energy does at that brief moment?
Ah right I got lost. That means you would need a smaller amount of Delta V (the minimum) in the perigee case than you need in the apogee case to achieve the same value 1/2m(v1+ΔV)^2 = 1/2m(v2+ΔV)^2.
I guess I am lost on how to express this mathematically.
This is how I went about it:
At apogee:(1/2)mV2^2-GM/r2 = 1/2m(V2+ΔV)^2
At perigee:(1/2)mV1^2-GM/r1=1/2m(V1+ΔV)^2
Where r2>r1, V1>V2
At the perigee, the rocket's energy changed the most since it was most negative and changed to a positive...
I am really lost on how to deal with this. Since this is an elliptic orbit, the mechanical energy is negative. For the rocket to escape orbit, we have to get the mechanical energy to be equal to or greater than zero. I thought at first that it would escape in the perigee, since that's where the...