So, I've been turning this over and over in my mind, and I've yet to puzzle it out:

Let's say I built a cannon on the trailing apex of the moon (i.e. the point that is on average right under the retrograde vector of the moon's orbit around the Earth).

Were I to fire a projectile from this point with a muzzle velocity of ~1km/s (i.e. ~ The Moon's average orbital velocity of 1020 m/s), would it hit the Earth?

My immediate reaction is "Yes, of course it would, c.f. truck going at 50 mph firing a soccer ball at 50 mph backwards leaving it standing still.", but is there some thing or other I've missed?

Edit: There was a thing I missed: The Lunar escape velocity. So let's say the cannon fires with a muzzle velocity of Lunar escape velocity + Lunar orbital velocity?


2 Answers 2


It turns out that the interaction of the earth and the moon's gravity produces some non-intuitive results. The short answer appears to be that you can indeed hit the earth from the moon using less than the escape velocity of the moon.

The key, of course, is to consider that the escape velocity of the moon is calculated "in isolation", when in the case under consideration the earth's gravity, when acting in opposition to the moon's, will reduce the velocity requirements.

I've done a simple model of the system, notable mainly for the assumption that the moon's orbit is circular, and used a fairly coarse time step of 1 second and Verlet integration. In all cases, the cannon is assumed to be at the equator of the moon, firing perfectly retrograde. The frame of reference is the barycenter of the earth-moon system, and the motion of the earth is included.

EDIT - The following bold section turns out not to be true, but was an arifact of a bug in my simulation. Rather, there is a single window for a "direct" impact path, spanning 2512 to 2660 m/sec. Transit time for a perfect hit (path travels through earth center) is 418 ksec at 2574 m/sec launch. My apologies. I'm sure I've got it right this time. Positive.

In this case, there are two windows of launch velocity. The first, 2253 to 2260 m/sec, produces something like a direct fall to earth, with a trip duration of 424 to 440 ksec (about 6 days). The second window is larger, from 2470 to 2603 m/sec, and the resulting paths take a longer path, looping outwards and then back in. Trip times run from 418 to 424 ksec. Intermediate velocities (2260 to 2470) are similar to the successful paths, but they don't actually get within the radius of the earth, due in part to the earth's motion around the barycenter.


This model stops at 1 million seconds, so it's entirely possible that large values of launch velocity miss the earth the first time, then interact with the moon to produce a later impact, but I assume that this is not what is intended.

EDIT 2 - Since the consequence of falling from the moon's orbit and missing the earth is an orbit with apogee equal approximately to the moon's orbital radius, it's clear that there will be any number of possibilities for further interaction with the moon, and possible second tries. Depending on the details of the orbit (whether it keeps the same rotation as the moon's orbit or not), there is also the possibility of a slinghot resulting in a 2 km/sec projectile velocity, which will not be adequate to eject the projectile from the system, but will certainly cause a much larger orbit.


How can the projectile escape the moon's gravity when the velocity is less than the escape velocity? Simple. For velocities near escape velocity, the projectile spends a long time at large distances (compared to the moon's diameter), gradually slowing down, and at exactly escape velocity will take infinite time to reach zero radial velocity. During this long period of low velocity the earth's gravity will attract the projectile, and when the radial (lunar) velocity is on the order of the (terrestial) orbital velocity the result is acceleration towards the earth, which increases radial distance much more rapidly than would occur without the earth.

  • $\begingroup$ Hah, this is pretty cool. Do faster paths become available around escape velocity + orbital velocity, or does it just escape the system altogether? $\endgroup$ Commented Sep 26, 2016 at 15:34
  • $\begingroup$ @WillihamTotland - See edit. Short answer is no. $\endgroup$ Commented Sep 26, 2016 at 19:02
  • $\begingroup$ Ah, okay, that makes more sense. (That gap was a bit weird, but you're the guy with the simulation. ;) Do you have any graphics for it? $\endgroup$ Commented Sep 26, 2016 at 19:06
  • $\begingroup$ I imagine there is a second solution, where you fire the cannon with sufficient velocity that the ball will orbit the sun in the opposite direction, and hit earth after about six months. Slightly harder to aim... $\endgroup$
    – Floris
    Commented Sep 26, 2016 at 19:06

No. Not with a muzzle velocity of 1km/s (the orbital velocity of the moon)

Yes it is possible if the muzzle velocity is larger.

I like the idea but the escape velocity of the moon is 2.38 km/s (NASA Moon Fact Sheet). So with a muzzle velocity of 1km/s the projectile will fall back to the moon.

A larger muzzle velocity would be needed to escape from the moons gravitational pull and cancel the initial orbital motion. Without crunching any numbers, I would guess that you need a muzzle velocity in the ballpark of 3-4km/s (basically 2.38kms + 1km/s).

Edit: clarified my answer since the question now includes the orbital and escape velocity of the moon.

  • $\begingroup$ Is escape velocity of the moon + orbital velocity of the moon right, tho? I mean, it seems right, no argument there, but is it? $\endgroup$ Commented Sep 23, 2016 at 19:44
  • $\begingroup$ @WillihamTotland It is closer to right than using the orbital velocity alone. There might be more to it but I can't think of something else that would come into play in a major way. I'll think about it a bit more. $\endgroup$ Commented Sep 23, 2016 at 20:15
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    $\begingroup$ The escape velocity is used to get it to infinity. You just need to get it to a point where the earth will take over. Which would be a little less. But this is a slight variation and your answer is a nice place to start. $\endgroup$
    – Michael
    Commented Sep 23, 2016 at 21:35
  • $\begingroup$ @Michael That's exactly the kind of math I -ne…- want, for fun. How best to do those calculations? $\endgroup$ Commented Sep 23, 2016 at 21:52
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    $\begingroup$ Also, the underlying assumption seems to be that with the two velocities producing a vector opposite the moon's orbital velocity the object will fall directly to the earth, with a path through the earth's center. But this is not necessary, as all you need to do is come within 4,000 miles of the center. So the required velocity is somewhat less. $\endgroup$ Commented Sep 24, 2016 at 1:49

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