Escape
velocity of the moon is about 2.4 km/s. Essentially, this means that
your projectile needs to be traveling 2,400 m/s in order to escape the
moon's gravity. This is simplified a bit from the actual case (I don't
think this is the proper forum to get into an intense discussion on
Lagrangian orbits) but it's close enough for our approximations. ^{Source}

If we're using a traditional .50cal BMG slug, it's maximum speed is
928 m/s. It would need to be traveling a little less than three times
that speed in order to escape the moon's gravity. ^{Source}

Let's try something a little faster - the .223 Winchester Super Short Magnum slug. *That*
has a max muzzle velocity of around 1,325 m/s. So even with one of the
fastest slugs in the world, we're just over halfway to escape velocity. ^{Source}

In short, even with a zero-cost free-return trajectory to the Earth,
your bullet will not even come close to escaping the moon's gravity.
Actually, the bullet can't even make a full orbit of the moon at the
surface. Orbital speed at the surface of a body is just the escape
velocity divided by the square root of 2. For the moon, that's about
1,700 m/s. ^{Source}

So, that option is thoroughly exhausted. However, let's see what
happens when we consider alternate methods of accelerating a metal slug.
A few years ago, the US performed successful tests on an experimental
Railgun-type weapon, which accelerated a slug weighing about 3.2kg to
speeds of about 2,400 m/s. ^{Source}

Now, on Earth, this still wouldn't help much, because the slug would be
immediately slowed down by air resistance, friction, and heating
effects (some really crazy shit happens when you hit Mach 7). On the
moon, however, with no atmosphere, we're governed by nothing more than
Newton's 1st law of motion.

However, this *still* wouldn't even get you *close* to
getting back to the Earth. Escape velocity is the minimum speed you need
to escape the moon's gravitational pull. That means that the further
you get away from the moon, the slower and slower your projectile will
travel, until your velocity is essentially zero (relative to the moon).
What does this mean, exactly, in terms of orbits? Well, imagine the
projectile escapes the moon's gravitational pull. So it's no longer
falling towards the moon, which means that, in effect, it's in orbit
around the Earth. What does this orbit look like? If the slug's velocity
relative to the moon is effectively zero, that means its position
relative to the moon stays the same. There's only one way this could
happen - if the slug were in the same orbit as the moon, just ahead or
behind the moon enough to not get captured by its gravity.

Essentially, all we've managed to do is put a 3.2kg slug in a lunar orbit around Earth. ^{Hooray!}

Eh, fuck it. We've come this far. How fast do we need to fire this fucker to hit the Earth?

Somewhat counter-intuitively, the most efficient way to get the slug back to Earth is to *slow down*. However, we're slowing down relative to the *moon*,
which means that we're going to be firing our slug so that it leaves
the moon's gravitational influence in the direction opposite to the one
the moon is traveling in. Think of throwing a baseball out of the back
of a pickup truck. The baseball still has some forward velocity, but
because we threw it out of our truck, it's going to hit the ground
behind us. This is sort of the same concept.

So how fast (or slow) does the slug need to be traveling to hit the
Earth outright? I'm going to ignore fancy maneuvers like aerobraking
because we're already into the realm of the theoretical here. Best to
keep our wild machinations simple. To figure this out we need to
calculate some basic orbital parameters.

rp = periapsis radius = radius of orbit at its closest to Earth. This is the radius of the Earth (duh!). 6378 km.

ra = apoapsis radius = radius of orbit at its furthest from Earth.
This is the moon's orbital radius from Earth (also duh!). However, this
distance isn't consistent, because the moon's orbit is not perfectly
circular. Suffice it to say, it will require less effort on the part of
our Railgun if the moon is at its furthest point from the Earth. It will
be traveling the slowest here, so we have to throw our baseball less
hard to get it out of the truck bed. This is 405,400 km.

a = 1/2(rp + ra) = semimajor axis. The entire length of the ellipse
is twice this number. So you can think of it like an average orbital
distance (sort of). It comes in handy later. 205,889 km.

e = ra/a - 1 = eccentricity of orbit. It's a dimensionless measure of
how squashed the ellipse is. 0 is a circle, 1 is a parabola. In our
case, it's .969, which is a pretty stretched-out ellipse!

Now to find velocity at our apoapsis (which is the speed we need to be going when we leave the moon's sphere of influence). v^{2}
= 2u/r - u/a. u is something called a gravitational parameter, which is
just a constant related to the orbited body. In this case, it's Earth,
and u = 398,600.44. There are units associated with this number but
they're weird and nonsensical, so I'm not going to bother. In any case,
we find that our velocity when we leave the moon's gravitational pull
must be .175 km/s, or 175 m/s.

*But* we're not done yet. To get an orbital velocity around
the Earth of .175 km/s, we must subtract that from the moon's orbital
velocity, which is .965 km/s at apoapsis. So we must escape the moon's
gravitational pull with a velocity of .79 km/s. Let's make it .8 km/s,
just for clarity's sake.

Now, to escape the moon. Orbital mechanics is weird. I just want to
preface what I'm about to go through by saying that. It's really,
really, weird. There's a certain spot on the moon you'd need to place
your railgun, and a certain direction you'd need to fire to ensure that
your slug left the moon's gravitational pull. But for the sake of
argument, let's say we're standing in the correct spot and facing in the
correct direction. How fast does our slug need to travel at the
surface, so that when we escape the moon's gravitational influence, it's
traveling at .8 km/s?

Believe it or not, we just use the same formula for velocity that I
stated above. However, we don't know a. And since this is a hyperbolic
orbit, there's no apoapsis. How do we calculate a? Simple. We use the
fact that at a *very large distance* away from the moon, we have a
constant velocity (.8 km/s). If we assume that, r is essentially
infinite, which means that 2u/r reduces to zero.

We're left with v^{2} (infinite) = u/a. Remember, u for the
moon is different than u for Earth. a works out to be 7660.625 km. Now
we have all the missing pieces!

How fast do we need to fire our slug from the surface of the moon?
The answer is a surprising 2.51 km/s, or about 2,510 m/s. That's really
not that much faster than escape velocity! Come on, science! Step it up!
Just a little extra juice and we're in business.

Okay, now to hit a wasp's nest. I don't think a critical strike is
likely. At all. However, we may be packing enough energy that we can go
for an area of effect thing and just hope to vaporize a few of those
pesky critters.

Using the same calculations as before, and ignoring air resistance
effects (honestly the bullet is gonna burn up in the atmosphere anyhow,
but let's say for awesome's sake that it doesn't), the 3.2kg slug will
impact the earth at a velocity of 11.27 km/s. Mach 33. Yeowch. The slug
will be packing what seems to be a whopping 203,082 kJ of energy, until
you consider that this is about 5% of the energy of a ton of TNT. This
is about 43 kg of TNT worth of power, about as powerful as a car bomb.
Not exactly the nuclear strike we were hoping for (fuck you, wasps).

Okay, oceans cover 71% of the Earth's surface. Antarctica is about
3%, and we'll say another 5% is excluded for all the inhospitable wasp
climates. This leaves about 20% of the world's surface area where wasps
CAN live.

Okay, people, this last part is coming straight out of my ass. Here's
where we get really, really stupid. Like, really stupid. I'm going to
posit that 70% of the world's habitable surface is covered by
agricultural land ^{Source,} 15% is residential (suburban) and 5% is urban. This is a gross, gross, gross simplification but that's the best we've got.

On agricultural land, let's say a nest kill will happen 1 out of
every 40 shots. In residential areas, 1 out of 30 (wasps like people's
houses, don'tcha know), and in urban areas, it's not gonna happen. If I
remember my stats correctly, this leaves us with a .15% (remember,
that's a *percentage* chance) that we'll hit a wasp's nest. So,
about every 667 shots, we'll hit a nest. Considering that, by this
metric, there's a 1% chance your slug will impact a heavily populated
urban area, that's about 6 city blocks destroyed for every one wasp's
nest.

**TL;DR:** Just buy a fuckin' can of Raid, you pussy.