To estimate the kinetic energy of a GBU-57 Massive Ordnance Penetrator (MOP) bomb, we need its mass and velocity at impact. The GBU-57 weighs approximately 30,000 pounds (13,600 kg). The exact impact velocity is classified, but sources estimate it exceeds Mach 1 (approximately 767 mph or 343 m/s at sea level) when dropped from a B-2 Spirit bomber at 50,000 feet.
For calculation, let’s assume a conservative impact velocity of 343 m/s (Mach 1). Kinetic energy is given by:
$ KE = \frac{1}{2} m v^2 $
Where:
$ m = 13,600 \, \text{kg} $
$ v = 343 \, \text{m/s} $
$ KE = \frac{1}{2} \times 13,600 \times (343)^2 $
$ KE = \frac{1}{2} \times 13,600 \times 117,649 $
$ KE = 6,800 \times 117,649 $
$ KE \approx 800,013,200 \, \text{Joules} $
So, the kinetic energy is approximately 800 megajoules (MJ). Some sources suggest it could be 800–900 MJ depending on higher velocity estimates (e.g., slightly above Mach 1).
This is a rough estimate, as actual velocity varies with drop conditions, drag, and terminal velocity dynamics. For context, this energy is equivalent to the kinetic impact of a 285-ton Boeing 747 landing at 170 mph, concentrated into a small area for penetration.