[Spaceships] Conventional and EM Gun ranges in atmosphere?

[Anthony]

Originally Posted by vicky_molokh

I'm not sure, that would imply that an object will drop its kinetic energy to zero at the same distance regardless of the kinetic energy it had. How would that happen?

Kinetic energy is proportional to v^2. Drag is proportional to v^2. Loss of energy from drag is equal to (drag force) * (distance). Thus, dE/dd = kE. This is a well known differential equation, and means E ~= e^(kd) (where E is energy, d is distance, and e is the base of the natural logarithm) and half-energy range is ln(2) / k (note: doing this quick by memory, so some signs may be off).