So I’m an eccentric trillionaire, and I’ve made up my mind that I want to shoot cannonballs and hit the Moon. My gun is buried in a vertical hole at the equator, and shoots straight up, and thanks to handwavium I can fire cannonballs at whatever absurd velocity the math requires and the cannonballs don’t care about Earth’s atmosphere. I guess that I can only hit the Moon when it is passing across the plane of the equator. To be safe, I’d want to fire cannonballs at at least escape velocity, so my misses wouldn’t come crashing back down to Earth. While I could get away with just getting as far as the Moon’s sphere of influence and letting it pull the shots in, I don’t trust my aim that well.
Dumb intuition tells me that by the time my cannonballs climb to the height of the Moon, they will be moving Eastward slower than the Moon does in its orbit, such that the Moon will ‘overtake’ them. If I fired enough cannonballs, the Moon would lose orbital velocity and drop closer to Earth.
But is that correct? If I fire cannonballs fast enough, or time them right, can I have them ‘overtake’ the Moon in its orbit, and maybe boost the Moon into a higher orbit with enough cannonballs? Or does firing faster cannonballs only change the radial component of their velocity when they get to the Moon, and not their ‘eastward’ velocity?
There’s very little difference between the speed you’d need to just barely reach the Moon and escape speed, so sure, you might as well go for escape speed.
As for the rest, it depends on what you mean by a “higher orbit”. If you make your canonballs super-fast, you’ll be able to increase the Moon’s orbital energy, while making very little change to its angular momentum. The effect would be that the apogee distance (the furthest it gets away from Earth) and its average distance would both increase, but its perigee distance (the closest it gets to Earth) would decrease. Eventually, you could get the Moon itself into an escape orbit (I’m pretty sure that would happen before the perigee got close enough to impact).
I think he’s talking about shooting the balls so they pass in front of the moon–front being the side facing the direction the moon is traveling. The mass of the canon balls would have a tiny pull on the moon, increasing the moons orbital speed. Shouldn’t that increase its apogee without affecting the perigee? Eventually, with enough cannon balls fired, the orbital speed of the moon would increase enough to result in a higher orbit. Fire them at the right times and you could theoretically put the moon in a higher circular orbit. Right? I mean, I’m sure the mass of cannon balls required would be greater than the mass of usable iron on the Earth, and we’d have to negate all the effects this would have on the Earth itself, but I think I see what he’s getting at.
If you shoot at the moon straight up from the equator (when the moon is there) doesn’t that mean that you’re shooting at the orbital nodes?
Due to the precession of the apsides I think you are shooting at different orbital anomalies over time, i.e., the moon will be at perigee, apogee and everything in between during the shots at different times.
So you can raise/lower the perigee/apogee in any combination depending on the precise timing.
I think you could also change the orbital inclination, or make it spin forward and backward, or send it on a slingshot trajectory out of the solar system… wasn’t there a documentary about this some time ago?
I think you need another cannon on the Moon that shoots the cannonballs back to the Earth for reuse. Shoot them back and forth.
I was thinking about impacting the Moon rather than slingshotting the balls past it, and wondering if I could increase the Moon’s angular momentum at the same time.
I can imagine lobbing cannonballs just right so that they drop onto the ‘leading edge’ of the moon and rob it of a bit of angular momentum with each impact without contributing much radial velocity (ie the ball is at its own apogee right when the Moon plows into it). From the Moon’s POV, the cannonball would impact near vertically at the center of the leading hemisphere. To my mind, this would be a bit like a retrorocket firing to drop the Moon into an orbit with a closer perigee.
I can also see how hitting the Moon dead center in the middle of the Earth-facing side with a fast cannonball would force its orbit to become more elliptical, increasing apogee but decreasing perigee. It still seems like this would rob some angular momentum from the Moon since the cannonball isn’t moving east as fast as the Moon is at impact. Or is it?
I’m wondering if there’s a way to have the cannonballs plow into the trailing hemisphere of the Moon and add angular momentum, pushing the Moon into an orbit where both apogee and perigee are larger. Ideally they would be impacting vertically in the center of the trailing hemisphere, but intuition says that can’t be arranged and I’d have to settle for some other scenario. But intuition can be wrong, right?
All of your cannonballs will actually increase the Moon’s angular momentum, since they will themselves have some (small) amount of angular momentum in the same direction as the Moon’s. But importantly, they’ll also be increasing the Moon’s mass, enough that the total angular momentum per mass will be decreasing.
I was assuming that all of the cannonballs would physically collide with the Moon, and it’s hard to imagine how such a collision could be anything other than perfectly inelastic. The only thing that could change by striking it in different locations would be to change the amount of angular momentum that goes into the orbit, versus the amount that goes into rotation.
That might actually be something worth considering: If we can get the Moon spinning in the opposite direction from its orbit, that would free up more angular momentum for its orbit. But it’d also open up tidal effects (the same effects that synchronized its orbit and spin to begin with), and those would be a beast to calculate.
I’m not sure if I understand the question(s) correctly, but…
I’m not sure why you position your canon on the equator, but it is impossible to hit the moon in a straight line. Not because you have to overtake the moon, which stays comfortably more or less in the same place (at least long enough to take aim), but because of rotational friction of the earth (aka wind). Unless you hit the moon exactly dead center in the heart, you might give it spin rather than shift its orbit, depending on the energy of your cannonball.
I was thinking of inelastic collisions, too. Maybe I’m thinking about this all wrong.
Let me rephrase: when I fire the cannonball, it gets a velocity vector with a radial component from the gun and an easterly component from the rotation of the Earth. When the cannonball reaches the Moon, ISTM that the easterly component of the cannonball’s velocity will always be less than the easterly component of the Moon’s velocity vector (because intuition). So the Moon will always be overtaking the cannonballs along its orbit. Or will it? Can I launch the cannonballs so that their easterly velocity is greater than that of the Moon at impact? Can I get them to plow straight into the center of the trailing hemisphere, like I think I can get them to drop straight into the center of the leading hemisphere? If I fire the cannonballs faster, doesn’t that just increase the radial component of velocity at the time of impact, but not the easterly velocity component?
I confess I wasn’t thinking at all about imparting spin to the Moon as well.
I’m confused. Why would the easterly component of the cannonball’s velocity be greater than that of the moon? There’s a reason the moon appears to go from east to west in the sky, despite the fact that it’s actually traveling west to east.
BenedictusXIV, he expressly made his cannonballs impervious to the effects of wind in the OP.
The moon has an orbital velocity from the earth frame of about 1 km/s.
Perhaps if you visualize the Earth-moon system rotating around a common barycenter it will help.
From the barycenter perspective you will need to “lead” your cannon ball’s shot to hit the moon which will put the cannon ball path a path forward of the moons current position. The vector will be more “East”. When switching back to the earths reference frame, if you don’t have more than 1 km/s to the east you are going to miss it in the same way that if you try to toss a ball to a friend on a merry-go-round you will never hit the mark unless you lead.
It is the impact of being in a rotating frame of reference.
The orbital velocity of the Moon is roughly 1 km/sec. If you fire a cannonball from the Equator at just escape velocity, by the time it gets to the Moon its easterly velocity component is (looks at some math examples) about 0.01 km/sec, I think.
If your cannon balls are of a significant enough mass to affect the moon’s orbit, they will have an additional reactive effect on the earth itself, pushing it away from the moon.
Escape velocity is about 7mi/s, and the moon at apogee is 252,277 miles away, so it would take something like 12 hours to get there, so the shooter’s vantage would be on the wrong side of the earth to spot the impact.