# Computing the released energy of several nuclear explosions

1. Dec 16, 2005

### Wickerman

Hey, i'm new here. After looking around i think this is the right place to post this question thread.

I'm interested in a hypothetical question. I know that the EMT formula to calculate the energy released from a nuclear explosion is = Y to the power 2/3 where Y = megatons.

What i'm interested in is somewhat concerning MIRVs, and how to compute the energy released by many nuclear warheads at once in (for all means) the same spot.

For purposes of this discussion, let's assume the world wouldn't be completely annihilated by this explosion :p

So here it is:

There are 1 million nuclear explosions that take place in (for all means) the same spot. Each individual explosion is the result of a 10 Mt warhead.

I know the energy released by the explosion of a 1 Mt warhead = 4.18 x 10 to the power 15.

Initially i wanted to think that the resulting energy would be = 10 (number of Mt) x 1 million (the number of explosions) x 4.18 x 10 to the power 15.

But then i thought to myself that THAT would be the resulting energy of the explosion of a 10 million Mt warhead. Aka the initial quantity (10 Mt) x 1 million. BUT that would assume the entire quantity of TNT would be stored in one single warhead. And that's not what i'm interested in.

I'm interested in how to add up the resulting energy of 1 million explosions of 10 Mt warheads.

I'm willing to bet it adds up differently but i have no idea how to do it or what the result would be.

I'm sure it's greater than just 4.18 x 10 to the power 22.

So........what is it?

:D

2. Dec 18, 2005

### Staff: Mentor

1 Mt warhead = 4.18 x 1015J and 10 million or 107 would give about 4.18 x 1022J, more or less, because that is a somewhat arbitrary factor. It is effectively additive.

Now how that energy manifests itself, i.e. blast vs radiation is another matter.

I am not sure why one would be so concerned with such a hypothetical calculation.

3. Dec 19, 2005

### Wickerman

Yes, i understand that, but your calculations make it sound like there's a 1 million megaton warhead. It's not. It's 1 million 1 megaton (let's say) warheads. Wouldn't the strength of the energy increase?

Like for example if you drop a large pebble in the water you'd get a smaller effect than if you drop 10 pebbles that add up to the same weight as the initial pebble into the water.

Wouldn't the energy from the 1 million explosions somehow interact and be more potent than just a 1 million megaton warhead?

As for why i'm so concerned, let's just say i'm freaky :p

4. Dec 19, 2005

### Staff: Mentor

I did not imply that there is a 1 million MT warhead. I simply said the energy from one million 1 MT warheads would be additive.
Each detonation would not enhance the other detonations. In reality some warheads might release more, but that dependends entirely on the warhead geometry/design. The detonation of each warhead, assuming they all detonate simultaneously would not necessarily affect the others.

On the other hand, the clustering effect might provide a more efficient conversion of the innermost warheads, and so perhaps their yield might be effectively increased.

IF the 10 pebbles were bound, the mechanical response would be about the same as a single large mass, assuming the density would be about the same. If the 10 pebbles broke apart when they hit the water, they might actually produce a lower energy (lesser energy density) response than a single mass - in this case - it is a matter of transfer of momentum from the mass to the water.
Not necessarily, and in fact it might be less for geometrical reasons. One would have to look at the work generated by the expanding plasma/gases and the pressure waves to appreciate the difference. Fortunately, there does not exist, at the present time, the material to make 1 million 1 MT devices.

If one asks, would distributing 1 million 1 MT devices be more destructive than a single 1 million MT device, then the answer would be yes - again for geometrical reasons.

5. Dec 20, 2005

### Wickerman

That's what i was interested in. Cluster effect vs. the energy being additive. That's the way i saw it, that due to the clustering effect, the yield would be effectively increased.

Hmmm......well thanks for the answers :)

6. Jan 25, 2006

### Morbius

Wickerman,
The energy isn't going to be additive.

It would be additive IF you could get all those warheads to explode at EXACTLY the
same instant in time. However, any warhead that doesn't go off exactly when the first
warhead goes off - and that warhead is anywhere nearby when the first goes off - will
be destroyed by the first warhead that goes off. This is known to military planners as

No - military planners that intend to make multiple nuclear attacks on targets that are
close together, have to stagger the times at which those targets are hit.

That 2/3 power formula is easy to understand. It assumes that the volume of the
fireball or the volume of space that receives a given overpressure is proportional to
the energy of the device - or yield. The volume is assumed to be spherical - so the
radius will be the 1/3 power of the volume, thus the 1/3 power of the yield.

The area of "footprint" of the blast will be the circular disc which has the radius of
the sphere. The area of this disc will be proportional to the square of the radius,
which itself is the 1/3 power of the the yield. Therefore, the area is the 2/3 power of
the yield.

Dr. Gregory Greenman
Physicist