One of the ideas for a space weapon back in the Cold War was dropping a tungsten rod from orbit onto an enemy target. The Wikipedia article suggests a 20 foot long, 1 foot diameter tungsten rod—about 8300 kg of tungsten, if it were cylindrical. What would that sort of impact do to a target? Since it's a small diameter ‘bullet’, would the damage be highly localized, or would the damage be mostly the result of a shockwave?

  • $\begingroup$ "Since it's a small diameter ‘bullet’, would the damage be highly localized" An explosion is very localized too: the energy is all released where the bomb is, but the effects can be felt a long way off if the bomb is energetic enough. Work out how much energy a 8 ton object has at, say, 5 km/s. Express the result in tons of TNT equivalent so you can compare it to small nuclear explosives. $\endgroup$ Commented Nov 5, 2016 at 16:11

1 Answer 1


I am going to write an estimation of the effect of 8000 kg falling from LEO to solid ground .

As it's a homework question, I won't work out any figures, as I feel the OP should check my assumptions (please, please do :) and perform the calculation.

My assumptions are:

  1. The mass of the rods is 8000 kg.

  2. They will fall vertically, transitioning from horizontal orbital motion to vertical by appropriate weight distribution and with stabilising fins fitted at the far end of each rod.

  3. The rods are made of tungsten, which has a melting point of 3,400 degrees Celsius, comparing this to the maximum re-entry temperature of an Apollo space capsule (around 2700 degrees Celsius ), implies that air friction will not melt them or cause any loss of mass.

  4. They should fall with 0 kph initial vertical velocity and end with a ground velocity equal to their terminal velocity. I am assuming that, at the time of the drop, all spacecraft velocity is effectively horizontal.

  5. Assuming a diameter of 1 metre cross section (the OP says 1 foot, but to maintain stability during the drop, and to transition from horizontal to vertical motion effectively, I think the front of each rod will need to be biased with more mass to the front). This gives us a cross sectional area of $3.14 × .5^2 = 0.785 m^2$

  6. So what is the terminal velocity $V_t$ of a rod with a frontal cross sectional area of $0.785 m^2$?

From Terminal Velocity Wikipedia

$${\displaystyle V_{t}={\sqrt {\frac {2mg}{\rho AC_{d}}}}}$$


${\displaystyle V_{t}}$ is terminal velocity,

${\displaystyle m}$ is the mass of the falling object,

${\displaystyle g}$ is the acceleration due to gravity,

${\displaystyle C_{d}}$ is the drag coefficient,

${\displaystyle \rho }$ is the density of the fluid through which the object is falling,

${\displaystyle A}$ is the projected area of the object.

Taking $$m = 8000 kg, C_d = 0.005, g = 9.8 m/s^2, \rho = 1.225 kg/m^3, A = 0.785 m^2$$ I have assumed 0.005 as the drag coefficient of a flat plate parallel to the flow ($Re \ge 10^6$) and assumed average air density.

The K.E. is then $\frac {8000 × V_t^2}{2} $

This K.E. is then the energy available to act effectively as an explosive and can be compared to other explosions using Energy Released, Orders of Magnitude

  • $\begingroup$ I get 5716 m/s, or Mach 16.8; with 1.3e11 J of energy, or several times more powerful than the largest non-nuclear weapons developed to date. The Wikipedia article suggests an impact speed of Mach 10, which is 2.8x less kinetic energy, but still on the order of magnitude of the most powerful non-nuclear weapons. Thanks! $\endgroup$
    – adam.baker
    Commented Nov 7, 2016 at 4:35
  • $\begingroup$ Just check the Reynolds number, wikipedia says 0.001 for Re less than 6. But I can't remember Re calculation method. Either way. I think I would rather not be in the area when the rods from god arrived. Regards $\endgroup$
    – user108787
    Commented Nov 7, 2016 at 9:03

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