This is not what I was claiming. I was talking about soil. Concrete, I believe they claimed 18 meters. Obviously, different ground types will be more complex/easier to penetrate.
The GBU-57 is dropped from a high-altitude B-2 Spirit bomber, which can fly at altitudes of up to 50,000 feet. This high drop altitude is crucial for the bomb to reach a very high terminal velocity. Some sources suggest it reaches supersonic speeds, potentially around Mach 1.29 (approximately 440 m/s).
Let's conservatively assume a terminal velocity (v) of 400 m/s (approximately 895 mph).
Calculating Kinetic Energy (KE):
The formula for kinetic energy is:
KE = 0.5 * m * v²
Plugging in our values:
KE = 0.5 * 13,600 kg * (400 m/s)²
KE = 0.5 * 13,600 kg * 160,000 m²/s²
KE ≈ 1.088 billion Joules
This is an enormous amount of energy that must be absorbed by the ground to stop the bomb.
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Resistive Force of Soil (60m penetration estimation):
To simplify, we can use empirical formulas developed from extensive testing. One of the most well-known is Young's empirical formula, which provides a way to estimate penetration depth based on the projectile's characteristics and the soil's properties.
resistive force is as a pressure (force per unit area) acting on the front of the MOP. Let's call this the dynamic soil resistance. The total resistive force (F) would be this pressure multiplied by the cross-sectional area (A) of the bomb.
The cross-sectional area of the MOP (with a diameter of 0.8 m) is
A = π * (radius)² = π * (0.4 m)² ≈ 0.5 m²
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Calculating Penetration Depth (d):
The work done (W) by the soil to stop the bomb is:
W = F * d
Setting the initial kinetic energy equal to the work done:
KE = F * d
Therefore, the penetration depth is:
d = KE / F
To achieve a 60-meter penetration, the average resistive force would have to be:
F = 1,088,000,000 J / 60 m
F ≈ 18,133,333 Newtons
This is equivalent to a force of over 4 million pounds. While this seems immense, it's plausible given the energies involved.
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Now, we can calculate the resistive force:
Convert PSI to Pascals (Newtons per square meter):
15,000 psi(assuming) × 6,895 Pa/psi ≈ 103.4 Million Pascals (MPa)
Calculate the MOP's Cross-Sectional Area:
Diameter = 31.5 inches (0.8 meters)
Radius = 0.4 meters
Area (A) = π × (radius)² = π × (0.4 m)² ≈ 0.503 m²
Calculate the Total Resistive Force (F):
Force = Pressure × Area
F = 103,400,000 N/m² × 0.503 m²
F ≈ 52 Million Newtons
So we see that 18 Million Newtons is not enough and the bomb would have to be significantly supersonic, or my calculations are too conservative, or they are overestimating the 60m soil penetration, but we ARE in the same ballpark.
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now, you might ask how can an object achieve over 1 Mach terminal velocity?
At high altitudes (like 30,000-50,000 feet): The air is much colder and less dense. For instance, at 35,000 feet, the temperature can be around -54°C, and the speed of sound drops to about 295 m/s (about 660 mph).
In this high-altitude, low-density environment, the MOP's terminal velocity is incredibly high. It can easily accelerate past the local speed of sound (which is already lower due to the cold) and go supersonic, then slowing down when near ground.
The bomb is also likely designed like a super aerodynamic dart to achieve maximum terminal velocity.