While I was watching this beautiful video, the absence of air friction pushed me to ask myself: While standing on the surface of the moon, what is the initial velocity by which you can fire a bullet to put it into orbit around the moon so it will hit you in your back. And how much time you should wait the bullet to hit you.

Let us assume that the moon has no mountains and is a perfect sphere, and your height is 2 meters.

  • $\begingroup$ I think a more interesting twist to this question is, how much energy would be required. I mean, if I replace the bullet with a baseball, would it be possible to "pitch" the ball such that it would then come back and hit me? Assuming you can't fire guns on the moon (only lasers! Pew! Pew!) $\endgroup$ Aug 13, 2014 at 11:22
  • $\begingroup$ At what angle would you need to aim the gun? $\endgroup$ Aug 13, 2014 at 13:37
  • $\begingroup$ @CoderDennis I would assume parallel to the ground would be optimal. $\endgroup$
    – Cruncher
    Aug 13, 2014 at 14:56
  • $\begingroup$ It would simplify the question if the shooter/target were to stand at one of the Moon's poles... $\endgroup$
    – DJohnM
    Aug 14, 2014 at 23:19
  • $\begingroup$ Re: Assuming you can't fire guns on the moon, There's no reason to assume that. But if you plan to try the experiment, I suggest that you thorougly de-grease all of the weapon's parts, and then lubricate it with a dry lubricant. Oil will boil away in vacuum, after which the phrase "vacuum welding" may become relevant. $\endgroup$ Apr 21, 2015 at 17:08

2 Answers 2



Velocity required: 1680 m/s
Time to hit you: 6500 seconds

Part 1: Velocity required

(Using Google search values)

Radius of moon = 1737.4 kilometers
Mass of moon = 7.34767309E22 kilograms

Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple Newtonian mechanics, we set the acceleration due to gravity equal to the centripetal acceleration required to move the bullet in a circle of the appropriate radius:

Acceleration due to gravity:

$$ F = m a = \frac{ G M m }{ r^2 }$$ $$ a = \frac{ G M }{ r^2 }$$

Where $m$ is mass of bullet, $a$ is acceleration of bullet, $G$ is gravitational constant, $M$ is mass of moon, and $r$ is radius of bullet's orbit.

Centripetal acceleration:

$$ a = \frac{ v^2 }{ r }$$

Where $a$ is acceleration of bullet, $v$ is tangential velocity of bullet, and $r$ is radius of bullet's orbit.

Setting these equal:

$$ \frac{ G M }{ r^2 } = \frac{ v^2 }{ r }$$ $$ v^2 = \frac{ G M }{ r }$$ $$ v = \sqrt{ \frac{ G M }{ r }}$$

Plugging in values: (note that if you fire the bullet 2 meters off the surface of the moon, this additional height is virtually negligible and thus I only plug in the radius of the moon here)

$$ v = \sqrt{\frac{ 6.67 \times 10^{-11} \text{ N} * 7.35 \times 10^{22} \text{ kg} }{ 1737.4 \times 10^3 \text{ m} } } = 1680 \text{ m/s} $$

(rounded to 3 significant figures)

Part 2: Time to wait

Simply divide the total circular distance traveled by the bullet by the tangential velocity of the bullet (which we found previously).

$$ d = 2 \pi ( 1737.4 \times 10^3 \text{ m} ) = 1.092\times 10^7 \text{ m} $$

To find time:

$$ t = \frac{ d }{ v } = \frac{ 1.092 \times 10^7 \text{ m} }{ 1680 \text{ m/s} } = 6498 \text{ s} $$

Thus, it would take around 6500 seconds to hit you in the back.

  • 3
    $\begingroup$ OP did not state that the orbit needed to be circular :P If you place the apoapsis on the diametrically opposite side of the moon, you will get lower numbers. $\endgroup$
    – Aron
    Aug 13, 2014 at 10:45
  • 3
    $\begingroup$ "this additional height is virtually negligible" - in particular it's less than the precision of your estimate of the moon's radius. If we had a sufficiently precise size of the "perfect sphere" assumed in the question, then we might as well throw in the 2m ;-) It would add 12.6m to the circumference, that's nearly another hundredth of a second! $\endgroup$ Aug 13, 2014 at 14:47
  • 2
    $\begingroup$ is 1680m/s achievable? Would this actually work? We should put a bunch of bullets in orbit as a moon defence system :) $\endgroup$
    – Cruncher
    Aug 13, 2014 at 15:00
  • 3
    $\begingroup$ @Cruncher There don't appear to be any commercial rifle rounds that fast. The fastest Wikipedia lists is the .223 WSSM @ 4520 feet/sec ~ 1380m/sec. The 120mm cannon used on the M1 Abrams can reach 1750m/sec. This suggests that a rifle capable or reaching that muzzle velocity should be possible in theory; however keeping the recoil to something the human body can handle could be problematic. Assuming it wouldn't result in a kaboom I'm sure there're hand loaders who'd be willing to try if you provided them transport. $\endgroup$ Aug 13, 2014 at 15:55
  • 2
    $\begingroup$ @DanNeely as paul kind of mentions, guns are designed based on earth-like restrictions. Sometimes on earth, heavier bullets travel faster. Simply because they blow through air resistance better. However, on the moon, we can drop a lot of the restrictions that we have on earth. Bullets can be very very light, and not just flutter to the ground like a feather. large low density bullets, should allow for more acceleration, and still have the surface area required to be properly pushed. The lighter bullet shouldn't have the same recoil either(we're not increasing force from gun) $\endgroup$
    – Cruncher
    Aug 13, 2014 at 16:01

An interesting way to answer Part 2:

Using the angular velocity version of the centripetal force equation:$$F_c=m\omega^2r=\frac{GMm}{r^2}$$If we assume that $r$ is both the orbital radius and the radius of the sphere being orbited and that the density of that sphere is $\rho$, then:$$M=\frac43\pi \rho r^3$$ then the equation becomes::$$m\omega^2r=\frac{G\frac43\pi \rho r^3m}{r^2}$$After cancelling $m$ and $r$ and taking a square root, we get:$$\omega = \sqrt{\frac{4\pi G}{3}}\times \sqrt{\rho}$$Note that the first root contains only universal constants, and the second contains only the density of the primary.

The orbital period of a surface-grazing satellite varies inversely as the square root of the density of the primary.

Thus, if we know the orbital period for a low earth orbit, and that the moon is about $60\%$ the density of earth, the time to shoot yourself on the moon is easily found...

  • $\begingroup$ My Kerbals tell me that your numbers are only correct for a polar orbit. $\endgroup$
    – Aron
    Aug 14, 2014 at 12:41
  • $\begingroup$ @anon: How does the orbital inclination affect the orbital period? $\endgroup$
    – DJohnM
    Aug 14, 2014 at 16:52
  • $\begingroup$ My Kerbals tell me that the Mun rotates once every 24 hours and that, from the rotating frame of reference that is the surface of the mun, a gun being fired east to west would have a different orbit than a gun fired west to east. $\endgroup$
    – Aron
    Aug 14, 2014 at 16:54
  • $\begingroup$ I doubt that the moon rotates once every 24 hours... $\endgroup$
    – DJohnM
    Aug 14, 2014 at 17:14
  • 1
    $\begingroup$ Your reference contradicts your position: "Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit Earth" $\endgroup$
    – DJohnM
    Aug 14, 2014 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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