Yeah, I was ignoring sideways velocity. I was taking "just look up what happens when you drop a 10ton tungsten rod from outer-space" perhaps a bit too literall

Yeah, I was ignoring sideways velocity. I was taking "just look up what happens when you drop a 10ton tungsten rod from outer-space" perhaps a bit too literally. If we add in 1/2 * 10,000 kg * (7500 m/s)^2, to account for sideways LEO-equivalent momentum, I have it coming out as about 0.0906 kilotons: about 4 times more energy. Which isn't too surprising, since the hardest part about getting into orbit isn't the going up part, it's the going sideways fast enough enough part. My whole formula is this: (((10000 * 9.8 * 1000000) + (0.5 * 10000 * (7500 * 7500))) / 3600000) / 1162200 Basically, (mgh + 1/2 mv^2) converted from joules to kilowatt hours to get it into a larger unit I can sort of relate to, and then converting that in turn to kilotons equivalent TNT in order to compare with nuclear yields. That's assuming I didn't leave off some zeroes somewhere or apply a formula incorrectly.