*how deep does an impact go*?

Today we are going to do momentum.

Take a look at the diagram below (click on it to enlarge).

The impactor (as yesterday) of mass

**m**hits the earth with velocity

**u**m/s and immediately becomes a plasma. As it meets earth material (we assume molten rock) it shares its momentum with the new material. The result is a growing cone of tunneling material as shown in the diagram below. The semi-angle of the cone is

**α,**its depth is

**d**metres and its base radius is

**r = d * α**.

Multiply the resulting volume by the density of molten rock

**ρ**which is 5.5 tonnes per cubic metre and decide the final velocity

**v**at which we believe further tunnelling will stop. Perhaps 10 metres per second?

**Click to enlarge**

Since γmu = ρα^{2}d^{3}v by conservation of momentum,

d = (γmu/ρα^{2}v)^{1/3}.

Now for some numbers. For the *1 kg impactor at 0.999c* considered yesterday, with α = half a degree and with a final plasma velocity of 10 m/s, the crater-depth **d = 1.2 km**.

This 1.2 km deep crater is *nothing like the journey to the centre of the earth* we speculated about yesterday. If we increase the cone angle for a wider dissipation - say 2.5 degrees - then the depth decreases to only 400 metres.

Suppose we increase the mass of the impactor to a cubic metre of ice weighing one tonne, at the original half-degree spread angle? The penetration depth **d** is now around 12 km. Notice that the increase of mass of a thousandfold has only increased the penetration depth by a factor of ten. This is because of the cube law we see in the equation above.

What would get us to the centre of the earth? A million tons (10^{9} kg) at 0.999999c would dig a crater 3,700 km deep. I reckon this would make a bit of a mess of the earth.

Here's the spreadsheet to play with (includes yesterday's).

**A relativistic impactor is like a nuclear detonation**

How does this compare with nuclear weapons? An approximate formula for the crater diameter d (km) of a nuclear explosion of M Megatons is simply:

d^{3} = M.

So a 27 Megaton detonation would produce a 3 km diameter crater. A rule of thumb states that the crater depth is around 1/5 of its diameter, so for the 27 Megaton device, the crater depth would be around 600 metres.

This is almost identical to the result given by the spreadsheet for a 1 kg impactor at 0.9c (kinetic energy = 27.84 Mt).

According to the Wikipedia article on the Orion spacecraft (powered by nuclear detonation) the initial plasma velocity of a 1 Megaton bomb is 10,000 km/sec. The velocity seems to scale linearly with the bomb yield suggesting that a 30 Megaton bomb might create a plasma shockfront expanding at a speed close to that of light. When this hits the ground, it might be indistinguishable from a relativistic impactor and therefore the cratering effects could be very similar.