I've got a lunar colony that struggles to obtain resources, so they're looking for ways to launch satellites into orbit without using up a bunch of rocket fuel.

One plucky engineer has suggested that trebuchets are the way to go, pointing out that the lack of a lunar atmosphere and the reduced gravity are clear advantages that make it possible on Luna. We've got plenty of materials, but nothing from the future. Our satellites are ~1 kg and have withstood all forces we've been able to apply to them so far.

Given current technology, could a trebuchet be used to launch a satellite into orbit around the Moon?

I've tagged this question hard-science but am open to some theoretical/unsupported conjecture.

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    $\begingroup$ Yes, absolutely positively yes. Cue the engineering pissing contest for who can get that satellite closest to the speed of light using only weights, pulleys and the moon's total mass. $\endgroup$
    – Cognisant
    Commented May 3, 2018 at 6:20
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    $\begingroup$ Yes, but since the lower gravity also applies to the counterweight I am pretty sure electromagnetic catapults will be better. $\endgroup$ Commented May 3, 2018 at 6:32
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    $\begingroup$ The hard-science tag isn't open to unsupported conjecture, science-based is though. $\endgroup$
    – Separatrix
    Commented May 3, 2018 at 7:03
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    $\begingroup$ @Dubukay, but there's plenty on escape velocities, material strengths, and probably design of trebuchets. You should in theory be ok on that count. $\endgroup$
    – Separatrix
    Commented May 3, 2018 at 7:38
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    $\begingroup$ Remember your KSP lessons! This will theoretically create an orbit which will be trebuchet height only at its lowest point. You must supply further force once at maximum height to round the orbit off else it will crash back to mun ... erm... moon on almost exactly the other side of the mun. $\endgroup$
    – Windlepon
    Commented May 3, 2018 at 8:56

7 Answers 7



No matter how much velocity you can give your satellite, its trajectory will either contain the point it was launched from or not be a closed curve.

an object is in orbit if it returns to its previous position in phase space (assuming we are talking about a close orbit).

if we assume the trebuchet to be the only source of speed, then it means the moment the projectile leaves the device it must already be in orbit, implying the same projectile will impact the machine after exactly 1 orbit. (give a look at Newton's cannonball for reference)

So your satellite will either fly away or hit the ground. You can provide most of the velocity with the trebuchet but your satellite needs at least some other source of thrust to correct the orbital trajectory.

Other issues make a trebuchet a poor choice to send a satellite into an orbit. They are powered by gravity, something there ain't much of on the moon (see L.Dutch answer for that).

Trebuchet have a tendency to impress an angular momentum to the projectile something that is undesirable for something that isn't spherical and has active components inside, not to mention it is all energy removed from the velocity you want to deliver to the satellite.

Finally, the trebuchet deliver the energy mechanically so if we assume that your machine can deliver orbital speed to the projectile, we must also assume some part of it will accelerate to the same speed and therefore be subject to a lot of stress.

Ultimately, you need to correct all of these problems, and while this might be possible the trebuchet is ultimately a poor choice especially when compared to simpler, cheaper and more effective alternatives like a coilgun/mass driver.

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    $\begingroup$ Nice edit, well deserved +1, although the amount of velocity and hence stress would vary a lot based on the actual orbit you want. Earth gravity makes trajectories bit complex. $\endgroup$ Commented May 3, 2018 at 10:17
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    $\begingroup$ Although it is correct that you can't launch anything into orbit with ballistic means only, this isn't really a relevant issue here: you can launch something into a surface-tangent pseudo-orbit that only requires a very small Δv to make it a low-apogee elliptical orbit. That can be done with a simple monopropellant booster. I'd still call such a satellite “trebuchet-launched”. $\endgroup$ Commented May 3, 2018 at 11:58
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    $\begingroup$ A trebuchet has the advantage over a cannon that because of the way the sling on the end of it's arm works, it can release the satellite relatively high off the ground at a flat trajectory and the trebuchet will be out of the way on the next orbit (assuming we aren't firing again at the exact wrong time) $\endgroup$ Commented May 3, 2018 at 13:01
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    $\begingroup$ So what you're saying is that we need is a trebuchet with an arm long enough to reach into orbit... $\endgroup$
    – ckersch
    Commented May 3, 2018 at 16:17
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    $\begingroup$ The correct answer is "YES", via a purely mechanical trick: have the trebuchet launch two masses that are connected by a tether---imagine a gaucho throwing a bola---with the two-mass "bola" twirling as rapidly as feasible. At apogee, explosively sever the bola-tether, with the timing such that the payload bola-mass continues in a circularized orbit, while the sacrificial bola-mass returns to the moon. $\endgroup$ Commented May 3, 2018 at 21:51

It is possible...eventually

To be clear @SilverCookie's answer is correct. In simple Kepler mechanics, when you give a single delta-v to an object there are two 'orbital' possibilities: the orbit returns to where it started, or it is not a closed orbit and your trebuchet payload is above escape velocity.

However, in this specific case, it is very possible for an object to reach lunar escape velocity (2.38 km/s) without getting anywhere close to escape velocity relative to Earth (11.2 km/s). Here is where we have the possibility to establish an orbit with various interactions of our payload with the Earth and the moon.

A trebuchet launch at greater than escape velocity relative to the moon will have at least two vector components relative to Earth: the 2.38 km/s velocity in the direction of the launch ($v_l$), and a ~1.02 km/s velocity in the direction of the moon's motion ($v_m$). The approximation is due to the fact that the moon itself has variable velocity as it pursues its elliptical orbit, in the range (0.970, 1.08) km/s. The final launch velocity $v$, dependent on angle between moon's motion and launch ($\theta$) would be (from the Law of Cosines):

$$ v = \sqrt{v_l^2 + v_m^2 + 2v_l v_m \cos\theta}$$

The codomain of this function, given the values above, is (1.36, 3.40) km/s. The minimum and maximum are found with launch angles of $\pi$ (directly away from the moon's motion for the minimum) and 0 (directly in line with the moon's motion for maximum). Note that this all involves only orbits in a 2-D plane.

The launch angle will in turn be the orbital angle of the object; so long as its orbit is prograde (in the same direction of the moon). If the launch angle is retrograde, you subtract $pi$. If the maximum speed is chosen, the angle is 0, which is the angle at perihelion. If the minimum speed is selected, launch angle is $pi$, but the orbit is retrograde so orbital angle is again 0; the object is still at perihelion. The angular velocity of the launched object is $$\omega = \frac{v\cos(\theta)}{r}$$ The minimum possible angular velocity is 0, the max is $8.85\times10^{-6}$.

Now we can use Kepler's equations to get some ideas of how to get back into a lunar orbit using the Earth.

Use the Earth's atmosphere to drop your velocity.

Since we have seen that it is possible to have zero angular velocity relative to the Earth, then there must be possible launch trajectories that intercept the Earth. If so, then it is possible to get one that brushes the Earth's atmosphere.

Use the Earth's atmosphere over a series of orbits to bleed off kinetic energy from our launched object. Once the object has lost enough energy, its velocity at the distance of the moon's orbit will be reduced to below the Moon' escape velocity. If the object then approaches within the Moon's Hill Sphere (about 60,000 km), it will start orbiting the Moon instead of the Earth.

Mission accomplished. I will admit that there is probably a lot of trial and error and lost payloads involved, but it is possible and that's all that matters.

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    $\begingroup$ Now, that's thinking. +1 $\endgroup$ Commented May 3, 2018 at 19:57
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    $\begingroup$ "...there is probably a lot of trial and error and lost payloads involved, but it is possible and that's all that matters." This guy Kerbals. $\endgroup$ Commented May 3, 2018 at 20:39
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    $\begingroup$ Speaking of KSP, aerobraking lowers the periapsis in addition to the apoapsis. Is this true in the real world as well? $\endgroup$
    – Fax
    Commented May 4, 2018 at 12:25
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    $\begingroup$ @Fax Yes! By Kepler's second law, orbital KE should be proportional to the square root of the area ellipse of orbit. So, when you decrease the orbital ellipse but one of the foci must stay the same (the Earth), the apoapsis and periapsis must both change (those are related to the semi-latus rectum, in conic terms). $\endgroup$
    – kingledion
    Commented May 4, 2018 at 13:01
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    $\begingroup$ Aerobraking at Earth is complete and utter overkill here. This is not necessary to leave the moon-surface-intersecting orbit. (If you did use aerobraking, you'd damn sure need boosters anyway to correct the path, because this is much harder to simulate with high accuracy.) $\endgroup$ Commented May 5, 2018 at 1:24

The average orbital velocity around a body of mass m at a distance r from its center (assuming the mass of the satellite is small with respect to the other body) is given by $$v_0 = \sqrt{\frac{mG}{r}}$$

Assuming you want to fly few meters above the surface you need to have a velocity of $1678\:\mathrm{\tfrac{m}s}$. This corresponds to $1.4\:\mathrm{MJ}$ of kinetic energy for $1\:\mathrm{kg}$ of mass.

Since a trebuchet works using gravity, you will need to have that much potential energy stored into the counterweight. This gives you the constrain of having a mass $M$ placed at height H in the moon gravity $g_\mathrm{m}$ so that $$HM = \frac{E_K}{g_\mathrm{m}} = 868530\:\mathrm{m\cdot kg}$$

Assuming your trebuchet has a counterweight moving $10\:\mathrm{m}$ vertically it means you need a mass of $86853\:\mathrm{kg}$ as counterweight.

Using lunar rocks as ballast this requires a cube with $7\:\mathrm{m}$ side (credits at Daniel for the hint. (Math corrected.)), and its center of mass will be located at its center, $3.5\:\mathrm{m}$ above the bottom.

The counterweight CoM will have then to swing from $+13.5\:\mathrm{m}$ to $+3.5\:\mathrm{m}$, which is the realm of possibility.

To be sure that the orbit of your satellite will not intersect any other point of the Moon, you are limited to place your trebuchet on the top of the highest lunar mountain, which according to Wikipedia is Mons Hadley. From there any plausible length of the trebuchet arm will be enough to fly above the peaks, including your trebuchet (unless you want to play catch and release with your satellite).

Nevertheless such a low orbit will be highly sensitive to deviation induced by minimum variation of local gravity or drag by solar wind, and thus unlikely to be stable on the long term.

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    $\begingroup$ That only works if the trebuchet has a 10km long arm. $\endgroup$ Commented May 3, 2018 at 9:29
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    $\begingroup$ You could use torsion or springs instead of a counterweight to make this workable $\endgroup$
    – Subbies
    Commented May 3, 2018 at 10:01
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    $\begingroup$ 86853 kg does not sound extraordinary. A lead sphere with 1.3 meters radius weights more... $\endgroup$
    – koalo
    Commented May 3, 2018 at 10:16
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    $\begingroup$ @koalo, last time somebody went on the Moon they found no lead, neither in balls or ore form... $\endgroup$
    – L.Dutch
    Commented May 3, 2018 at 10:29
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    $\begingroup$ Note that these calculations assume the mass of the trebuchet's arm as zero. In practice, an arm strong enough to accelerate 1kg to orbital velocity in a few dozen meters will weigh many kg itself; you'd need a counterweight at least an order of magnitude heavier. $\endgroup$
    – Skyler
    Commented May 3, 2018 at 21:24

As Silvercookies said, a (closed) Kepler orbit that starts at the surface will always intersect the surface again.

However, that's really irrelevant here, because Kepler orbits are an idealisation. They give a good approximation when you're orbiting a dominantly heavy object with good spherical symmetry (making it equivalent to a point mass). This is well fulfilled for the sun, which is why the planets have very Keplerian orbits. It is also fulfilled pretty well for the gas giants.

For anything else in the solar system, you don't actually get such orbits. The Earth-Moon and Pluto-Charon systems seem Keplerian, but that's just because they're two-body systems with no significant near other influence: this results in both bodies Kepler-orbiting their shared barycenter. But this doesn't work if there are more than two masses at work.

So, as kingledion remarked, you can basically just use the Earth to leave a Moon-tangent orbit. But it's not at all necessary to resort to aerobraking at the atmosphere here – it's sufficient to trebuchet into an orbit that gets close enough to Earth to be a bit disturbed by it.


Animation of an Earth-influenced lunar orbit

Conretely, you need to launch from the far side of the moon, with a bit of extra velocity in direction of the moon orbit. Seen from the moon, this orbit will look thus:

The unstable orbit, seem from the moon

Note that such an orbit isn't really stable – it may eventually crash. But that's the case even if you stay so close to the moon that the Earth's influence doesn't interfere, because the moon isn't completely homogeneous!

So: for a stable controlled long-term orbit, you'll need thrusters either way. But for quick experimental satellites, launching from the surface is not problem at all.


No. While it's possible to throw something into orbit (but you still need something to circularize it with), a trebuchet isn't capable of doing it. You're going to need a ridiculously heavy counterweight and a ridiculously long arm--and your arm is going to snap when you try it.

You need 1730 m/s to reach low orbit. That's 176 g-seconds (1g for one second.) Let's be pretty brutal and figure you launch at 10g (plus the lunar gravity.) You need to boost for 17.6 seconds to accomplish this. You moved 15km while doing this. (And reality is even worse--since you're not moving in a straight line you're actually experiencing more than 10g of acceleration.) You want to eject horizontal, without a gargantuan hole you can't start below the horizon. Thus you have 90 degrees of motion. Your trebuchet arm must be 9.5km long.

Out of what do you plan to build the arm that will stand up to the strain? I'm not up to calculating what it would take to keep the arm from snapping but my gut says it's going to be huge--which means it has a huge amount of momentum. Your counterweight has to be big enough to accelerate this whole thing. What's it's support cable made out of?

Note that the lower the acceleration of the cargo the worse the numbers become.

Rather than a trebuchet, what you want is a linear motor. Think of a maglev train, except it just keeps accelerating. The train itself is build so it won't fly off the tracks when it's weight goes negative. It accelerates to the right speed and releases the spacecraft, which is now in an orbit with a periapsis at zero and will have to circularize when it gets up there.

(Note that this scales pretty well. Limiting yourself to 5g but with a track wrapped all the way around the lunar equator you can eject anywhere from a sun-grazing orbit to a bit over solar escape.)

  • $\begingroup$ Orbital speed is directly related to the orbital height. For a trebuchet with a launch speed of 82 km/s (possible), then the orbital height above Earth would need to be 900000000 km. Or ~6 AU. But above the moon it would only need to be 8395 km! $\endgroup$ Commented May 5, 2018 at 4:48
  • $\begingroup$ @Draco18s While orbital speed is lower when you're higher you'll need more velocity at launch to climb out of the gravity well. As for your link, I can't find anything in it that supports your numbers--are you committing a unit error? (82 m/s instead of 82 km/s?) $\endgroup$ Commented May 5, 2018 at 12:43
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    $\begingroup$ *Looks at numbers* Shoot, you're right, I bollocks that part up. And the same about climbing out of the gravity well. I claim idiocy due to it having been 10:30 at night! $\endgroup$ Commented May 5, 2018 at 18:27

The two existing answers address difficulties in converting the energy imparted to the satellite by the trebuchet into an orbit.

You have limited resources, but don't clarify what resource are available. You need also consider that a trebuchet stores potential energy generated by humans over a long period of time and releases it in short period of time.

Counterweight trebuchets are powered by gravity; potential energy is stored by slowly raising an extremely heavy weight box (typically filled with stones, sand, or lead) attached by a hinged connection to the shorter end of the beam, and releasing it on command. Traction trebuchets are human powered; on command, men pull ropes attached to the shorter end of the trebuchet beam. The difficulties of coordinating the pull of many men together repeatedly and predictably makes counterweight trebuchets preferable for the larger machines


The humans will need to be fueled by organics (food) & Oxygen and will require significant people hours of time to store that energy.

You would likely get better results with the same volume of organics and less people hours, by converting the organics directly into combustible fuels. The only real advantage the trebuchet would provide is allowing the oxygen & carbon to remain on/in the lunar colony.

Organics used on moon to store energy in the trebuchet would be converted to Carbon Dioxide in the closed lunar environment (presumably) and be available to convert back to pure Oxygen and Carbon.

Organics used direct propellants (rockets) would either be used in space and lost or would need to be fired in a way that they could be recovered inside the closed system.

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    $\begingroup$ Whilst historically trebuchets were man-powered, I really doubt this particular implementation would have any reason to be. If anything it would draw direct from solar power to lift the counterweight, effectively acting as an incredibly large solar energy storage device capable of delivering all the stored power in a very short time-frame. It might take forever to reload.. but would only require the outlay of the solar harvesting device + maintenance rather than constantly burning a fuel. $\endgroup$
    – James T
    Commented May 3, 2018 at 15:19
  • $\begingroup$ I had similar thoughts, but if you are going to use solar energy there are more effective tools then trebuchets to store your energy. Springs or compressed gases could be better choices. $\endgroup$ Commented May 3, 2018 at 16:36
  • $\begingroup$ This is not a hard science answer. $\endgroup$
    – kingledion
    Commented May 3, 2018 at 17:18
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    $\begingroup$ @kingledion did you read the last line of the question? $\endgroup$ Commented May 3, 2018 at 21:47

I’m seeing your giant trebuchets modified with mangonel or scorpions to fire the satellite away from hte moon when the long arm of the trebuchet has provided them with their initial velocity tangential to the moon’s orbit.

Which ever form of torsion catapult adorns the tip of the trebuchet, they fire radially along the axis of the arm.

The release point is selected to maximize the gravity boost from both the moon’s and earth’s orbits.


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