Ball-park calculations

Home Forums Moon Why are most of the Moon’s craters circular? Ball-park calculations

Paul Leyland

According to the energy yield of that impact, with an assumed relative density of 3 (about that of a stony asteroid) is 75 megatons TNT which will dig a crater a kilometre in diameter and a quarter of that in depth. The calculator assumes a terrestrial impact. The lower gravity on the moon ensures that somewhat more (but not a lot more, because mass is proportional to the cube of the size of the excavated material) excavation can be performed at the same energy cost.

I do not know the depth to which it would penetrate before exploding, partly because it depends on the structural strength of both the impacting body and the lunar regolith, but note that until the asteroid hits something it travels its own diameter in 5 milliseconds, so perhaps 500m might be a reasonable guess, a distance which requires a travel time of well under a tenth of a second.