6
$\begingroup$

I know you need oxygen for the ignition, but presumably, if the bullet is impervious (or water-tight) and if there is a little air enclosed in the bullet case, wouldn't that be sufficient to fire it ?

And, giving the speed, would that be enough/too much to send it into orbit around the moon ? Would you need to fire it at a certain angle ?(which will presumably make it an elliptic orbit)

$\endgroup$
10
$\begingroup$

Almost.

The speed of an object in orbit depends on the radius of the orbit, and the mass of the body being orbited.

enter image description here

The lowest orbit possible is where the bullet is just grazing the surface, so r = the radius of the Moon. M is the mass of the Moon, and G is the gravitational constant. Let's do the math:

wolfram alpha calculation

The result is approx 1.7 km/s.

A .50 BMG Saboted Light Armor Penetrator (SLAP) reaches 1.2 km/s muzzle velocity. So even a very powerful round such as this one moves a bit too slowly.

It is conceivable that a tricked-out .50 round, with an extra load of powder, and possibly shot from a specially prepared rifle, could reach the 1.7 km/s speed needed; that would be a 40% increase in velocity over the SLAP round.

In that case, if fired at the horizon, it would do a circular orbit, would go round the Moon and hit the shooter in the back 1 hour and 47 minutes later, provided it doesn't hit a mountain or something en route, and provided the orbit is along the equator.

If you're not shooting along the equator, it will still do an orbit, but its motion relative to the surface of the Moon will be a weird spiral and it will not return to the starting point (the orbit itself will still be a closed loop, a circle, but the Moon spins underneath it, so seen from the Moon it appears like a spiral).

The circular orbit (standing on a hill, shooting at the horizon) requires the smallest muzzle speed to complete the loop. Any other orbit would require a greater muzzle velocity from the rifle. The more elongated the orbit, the higher the velocity.

Plus, if you shoot higher than the horizon, the bullet would hit the ground before it reaches you. You have to stand on an elevated point (a hill), and make sure the rifle does not point at the ground either forward or backward. This is because the return trajectory of the bullet after completing one loop is aligned with the rifle (the rifle's barrel is tangent to the trajectory), and if the butt of your rifle points at the ground, on the return trajectory the bullet would have to come out of the ground, which is impossible.

A laser mounted on the rifle must not point at the ground no matter whether the laser is pointing forward or backward, otherwise the orbit gets clipped by the ground.

TLDR: Stand on a tall hill or a mountain, shoot while the rifle is exactly horizontal. Use a top-tier extremely high velocity round, custom-loaded with extra powder, with a high caliber sniper rifle. Anything less would not work.

I know you need oxygen for the ignition

You don't. Gun powder contains its own oxygen. You can fire guns in a vacuum just fine. Primers also go off just fine in a vacuum.


It should be noted that the important number here, 1.7 km/s, is the orbital velocity near ground level on the Moon. It must not be confused with the Moon's escape velocity (the speed at which an object breaks away from orbit and keeps going away into space forever, never to return), which is higher.

$\endgroup$
  • $\begingroup$ (1.7/1.2)^2 ~ 2, so the energy released must twice that of the mist powerful gun. $\endgroup$ – Walter Feb 25 '17 at 10:33
  • 1
    $\begingroup$ Nice answer. Though I'd add: For this to work you wouldn't just have to stand on a hill, in fact you (rather: the barrell) has to be above the highest point (along the equator for the equatorial orbit, highest point of the whole moon for the non-equatorial orbit). $\endgroup$ – iraserd Mar 17 '17 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.