Death Star Physics Revisited (part II)

Saturn's moon Mimas looks eerily like the Death Star.

Given the very in depth analysis some of my readers sent me after the last Death Star post, I hesitate to follow up with this one since there's clearly some Star Wars physics that should to be debated before we continue.  However, I'm posting this any way because (1) I already typed it up and I'm pretty lazy about these kind of things and (2) if you allow my initial assumptions, there's still at least one more question that begs to be asked: how is all that laser energy stored?  Is there some giant car battery hidden somewhere?  After all, it would take 40 million years to collect enough solar power to blow up a planet.  Wouldn't one shot sap all the energy out of the Death Star?  According to relativity, one could very efficiently store energy as mass.  If that were the case, what percentage of the Death Star's mass would be lost with each shot?   

Luke mistakes the Death Star for a small moon, which suggests the mass of the Death Star is about that of a moon.  I'll assume the Death Star's mass is equivalent to that of Saturn's moon Mimas (M=3.7×10¹⁹ kg) since it kinda looks like the Death Star.  As we calculated before, the total energy required to blow up a planet is E = 2.1×10³² J.  According to relativity, the amount of energy stored in a piece of matter with mass m is given by the equation

E = mc²,

where c = 3×10⁸ m/s is the speed of light.  If we solve for the percentage of mass lost with each laser shot, we find

fraction of mass = m / M = E / (M c²)

= ( 2.1×10³² J ) / [(3.7×10¹⁹ kg) (3×10⁸ m/s)²]

= 0.0063%.

It's a very small percentage.  At this rate, the Death Star could blow up over 15,000 planets before it ran out of mass.  This of course assumes that the energy is directly transferred to the planet without loss.