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.

Since my weapons need to both do terrible things and look awesome doing it, my engineers are basing everything off the Barret M82A1 and as such the ammo will resemble the .50 BMG.


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.

Edit 2:
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.

  • 1
    $\begingroup$ I'm fairly confused as well. $\endgroup$
    – AndreiROM
    Feb 19, 2016 at 16:00
  • 1
    $\begingroup$ You might also want to clarify the 'surface area' comment. The amount of energy per square meter carried by the shock front of the supernova will lessen with time and distance from the original star... $\endgroup$
    – Joe Bloggs
    Feb 19, 2016 at 16:04
  • 3
    $\begingroup$ I am not understanding how this is relevant to WorldBuilding. Stripped of the rifle component, it's only a physics question - whether a certain amount of energy could be contained within a certain volume of space. From a physics standpoint - either that's a black hole or it's not. How does it relate to WorldBuilding? $\endgroup$ Feb 19, 2016 at 17:17
  • 1
    $\begingroup$ @TheAnathema Stripped of the reasons most of our questions are physics based, I'm also curious if that energy level is something a rifle type weapon is capable of safely firing that type of energy and if the ammo wouldn't weigh too much from a density perspective. In short could soldiers actually wield this type of weapon? $\endgroup$
    – Culyx
    Feb 19, 2016 at 17:24
  • 3
    $\begingroup$ Why would it be a rifle? Do you just mean a gun, or is the barrel actually rifled? $\endgroup$
    – Samuel
    Feb 19, 2016 at 17:35

2 Answers 2


This might be a physically possible gun, depending on what you mean by "the total 'surface area' of a supernova". Because the star changes size when it collapses to form the supernova, and then expands greatly (and quickly), it is hard to pinpoint an accurate area.

One of the possibilities is that if you put a supernova where the sun is, then what energy would go through a rifle-sized hole on earth? This ends up with a surprisingly relatable answer:

Energy of supernova: 1.5*1044 J = 1.669*1027 kg

At the distance that the earth is, through a .50 caliber hole, there is 7*1016 J = 0.788 kg

This means that you only need to carry about 400 grams of antimatter to fire the rifle. The difficulty is finding a way to focus it into a beam, but this seems like a physically possible device.

  • $\begingroup$ In a strange coincidence, I remember that a megaton is on the order of 10^-16 J... $\endgroup$
    – Joe Bloggs
    Feb 19, 2016 at 19:16
  • 1
    $\begingroup$ I was looking at my calculations and wondering when you posted up, makes me feel a bit more confident in my own math =D $\endgroup$
    – Culyx
    Feb 19, 2016 at 19:38
  • $\begingroup$ So the soldier may be able to carry the ammunition. At a distance of 5 miles from the target, he has a 50% chance of survival. $\endgroup$
    – Jim2B
    Feb 21, 2016 at 21:46

Supernovae. A supernova is an explosion of a massive supergiant star. It may shine with the brightness of 10 billion suns! The total energy output may be $10^{44}$ joules, as much as the total output of the sun during its 10 billion year lifetime. Using $E = mc^{2}$:

$10^{44}~\text{J} = m \cdot (300000000~\frac{\text{m}}{\text{s}})^{2}$

$m = \frac{10^{44}~\text{J}}{9 \cdot 10^{16}~\frac{\text{m}^{2}}{\text{s}^{2}}}$

$m = 1.11 \cdot 10^{27}~\text{kg}$

That is more than the mass of 100 earth-sized planets. I don't think anything other than a black hole can be dense enough to contain all that mass/energy in such small volume.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .