My crack team of engineers and scientists is building a new weapon; the Supernova Rifle. They assure me it's going to be very powerful but as Galactic Emperor I'd like to know that my space credits are being wisely spent.
The function of this rifle will be to take the energy output from a supernova in a given surface area relative to the collapsing star; and project that energy down range in a "bullet-like" manner.
The .50 BMG has a bullet diameter of .510 inches or 13mm; which gives an area of ≈0.81713 sq/inches or ≈5.30929 sq/cm.
So given the total "surface area" of a supernova how much energy of the supernova is found within the cross sectional area of 1 .50 BMG round?
I do not want the TOTAL value of a supernova, rather imagine the path of a bullet as a cylinder and that cylinder extending out of a spherical volume. Imagine sticking a Pin into a tennis ball where the tip of the pin is at the center of the ball and extends to the surface. That's the volume ratio I'm looking for. A bullet sized column of supernova energy. What I'm asking; Is such a tiny fraction of a supernova still that powerful?
Compared to a conventional .50 BMG is this better or worse?
Can something the size of a .50 BMG cartridge feasibly contain that much energy or would it weigh far to much to be practical?
Does my Nova Rifle need to be made of something better then steel to withstand firing these rounds like handwavium or scificilite? Because those are very expensive and I don't know if the Empire can devote the needed credits for exotic rifle designs.
My quick calculations:
- A star with 30 solar radii has a radius of 20871000000 meters (30 solar radii is the lower end for supergiant stars according to wikipedia) This gives the star a volume of 3.81x10^31 cubic meters
- I take a cylinder inside this sphere equal to the radius of the star in length and with a radius of .0065 meters (50 BMG bullet radius). This cylinder has 2.77x10^6 cubic meters
- This results in a small value 7.27x10^-26. Multiply this by the approximate power of a supernova 1x10^44 and we get 7.27x10^18 Joules.
(I'm of course assuming the energy distribution of a supernova is uniform throughout the sphere, and that it makes sense to use a solar radius prior to the supernova event)
According to wikipdia again this is somewhere in between South Korea's and the USA's total yearly energy usage.
Lacklub sees to have produced an answer in the same ballpark as well which is good.