# Measurements of an asteroid replacement

Say I wanted to create an impact similar to an asteroid with a diameter of 1.5 kilometer/1 mile with an iron core. It would be going about 18km/s on a 90 degree angle. An impact to level cities but not disturb the Earth as a whole a lot.

Now iron is far from dense compared to some other materials. If I artificially produced a projectile from tungsten, what would the dimensions of the projectile be to cause the same impact damage?

My Original plan was to use asteroids farmed in the solar system but reading up on Project Thor it got me thinking. A species that can use asteroids for planetary bombardment, uses FTL. Such a projectile isn't an insane investment.

• I'm not really sure what you want to know. How big an asteroid-shaped lump of tungsten would need to be to cause the same level of damage as a lump of iron? You would just need the same total mass. Commented Mar 22, 2017 at 19:38
• So a tungsten call of about 430 meters would work? Commented Mar 22, 2017 at 19:42
• Is that what you're asking? You may want clarify that in the question, then. Commented Mar 22, 2017 at 19:43
• My question being if I want a projectile from something as dense as tungste, that creates an impact similar to the one discribed above. What material would work and what shape? Sphere, pole, etc. Commented Mar 22, 2017 at 19:45
• Ermm humans have thought a similar weapon to what you 'want' ... it iscalled called 'rods from gods' is a weapon using the 'kinetic bombardment' en.m.wikipedia.org/wiki/Kinetic_bombardment Commented Mar 26, 2017 at 18:27

As I understand your question, you want to know which size should the tungsten asteroid have to keep the same level of (low) damage on impact.

You want then to keep the same energy upon impact. The energy involved in the impact is the kinetic energy of the asteroid, which, keeping the same speed, we can only change by changing its mass.

Tungsten density is 19250 kg/m3, while iron density is 7960 kg/m3, about 2.4 times higher, then.

To keep the same mass we would need to reduce the volume of the same ratio, and thus the diameter would be scaled down by the cubic root of 2.4, which is about 1.34. This gives you a diameter of roughly 1100 m.

• Your iron density is off, by about a factor of 10. I think you missed a digit Commented Mar 22, 2017 at 19:57
• We all know that being less dense than water iron floats! Why else would they use iron to make ships? Commented Mar 22, 2017 at 21:41
• I fixed the error
– L.Dutch
Commented Mar 23, 2017 at 5:43