Weird Harold wrote:You apparently missed the words "theoretical value" -- the theoretical minimum velocity of a dropped object through Standard Temperature and pressure of a planetary atmosphere.
That is the determinator for the theoretical minimum yield for a 650Kg RMN KEW.
ETA: Of course since there is a minimal drive and guidance package, the drive could be used to decelerate the KEW to very near the target and simply drop a 650Kg weight on an individual standing in the open so they target only suffers damage equivalent to a piano pushed off a four or five story building -- no "Boom," just "Splat!"
If we must...
Ideally, gravitational potential energy is simply mass times gravity times height. This ignores the change in local gravity over orbital height distances, uses Earth gravity which may differ greatly from local gravity, and ignores air resistance entirely (technically, air resistance doesn't affect GPE, but it would affect how much of that energy bleeds off on the way to the ground).
Assuming a convenient 2000 km (low earth orbit) stationary drop, 650 kg would give you 0.00304 kt upon impact on the ground - remarkably similar to the low yield SLN strike at the beginning of SoF. Air resistance would reduce that a bit, but probably only a couple percent if the dart is nicely aerodynamic.
The much bigger issue is that gravity reduces with distance squared. In an Earth-standard gravitational field, a KEW at 2000 km is going to start falling at 7.46 m/s^2, not 9.8 m/s^2. This changes continuously during the fall and only reaches 9.8 m/s^2 at ground level. That is going to significantly reduce the energy, but my calculus is way too rusty to even attempt that calculation (and I couldn't find one online).
Obviously higher drop points and larger masses give bigger booms.