Energy is always conserved, but different oberservers will disagree about how much energy there is, and what forms it takes. Also you have to be sure to include the whole system. Let's return to the Phobos example from the linked question, but be a bit more careful.

Suppose what we actually do is use our teraton nukes to split Phobos into two equal halves of mass roughly $5\times 10^{15} kg$ each, and push them apart at a mutual velocity of 1.8 km/s. Now we can analyse these events from the viewpoint of a couple of different observers. 

 1. An observer in spaceship which starts out close to, and at rest relative to, Phobos. They see zero initial kinetic energy and $0.5\times 10^{16}\times 900^2 \sim 4\times 10^{21}J$ of final kinetic energy, which nicely matches the energy released by a 1 teraton explosion. 

 2. An observer at rest relative to the centre of mass of Mars.They see an initial KE of $0.5\times 10^{16}\times 2100^2 \sim 2.2\times 10^{22} J$ from the orbital velocity of Phobos.  Afterwards, they see one part escaping at 3 km/s carrying $2.5\times 10^{15}\times 3000^2 \sim 2.25\times 10^{22} J$ and the other about to crash into Mars at 1.2km/s carrying $2.5\times 10^{15}\times 1200^2 \sim 3.2\times 10^{21} J$.  Thus they can see about $2.6\times 10^{22}J$ of KE a gain over the starting conditions of about $4\times 10^{21}$ again matching the nukes. 

So you can use energy calculations, but you need to keep track of the reaction mass and both the initial and final KE of all the various elements.  Incidentally, if you wanted to launch Phobos relatively intact using a much smaller amount of reaction mass and a higher exhaust velocity you'd need much **more** energy.