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.
tl;dr:
Velocity required: 1680 m/s
Time to hit you: 6500 seconds
(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)
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.
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...
