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Similar to this question except not about using a machine: What payloads and launch speeds could a sling launcher get using modern materials on the Moon?

How fast could a stone be thrown using a sling while standing on the moon? Could you launch a rock into lunar orbit with a very long slingshot? Google has nothing about this.

Edit: I only said 'of the type whirled around above the head' to clarify that I didn't mean a catapult. Don't assume that that the solution needs to involve anything whirling above the head.

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    $\begingroup$ Orbital Velocity of the Moon is 1.03 km/s. There's no way a human could throw something at 1.03 km/s $\endgroup$
    – Star Man
    Apr 20 at 16:53
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    $\begingroup$ The same speed as it could be thrown anywhere else? There's no "g" in omega x r. $\endgroup$ Apr 20 at 18:03
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    $\begingroup$ @OrganicMarble The main limiting factor on Earth is air resistance. A stronger thrower could lob a heavier rock at similar speed, but the range of speeds is very limited by $v^2$ in air drag equation. On the Moon you could pump power into spin of the sling as long as you can control it and it doesn't break. $\endgroup$
    – SF.
    Apr 21 at 6:57
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    $\begingroup$ Your initial velocity depends only on your slingshot force and that will be the same as on earth. So the projectile should launch at the same speed, only difference being it does not decelerate due to drag and it arcs less due to gravity. But, if the slingshot is the same on the moon as on earth, I would expect to see absolutely no difference in your starting velocity. $\endgroup$
    – user39728
    Apr 21 at 16:48
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    $\begingroup$ @OrganicMarble So did I. If you use motors, bearings etc, air resistance stops being the ultimate limiting force, and the next threshold: that of your device ripping itself apart, will be mostly the same on Earth and on the Moon. $\endgroup$
    – SF.
    Apr 23 at 5:46
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Using a sling whirled around above the head, could someone on the moon throw a stone into orbit?

tl;dr: probably deeply suborbital; 1192 meters, 38 seconds, either straight up or downrange at 45 degrees.

And of course even if you had superhuman sling skills, your trajectory would either fly off into space (lunar C3 > 0) or return on an elliptical (C3 <0) trajectory and intersect the surface somewhere. Standing on the Moon a horizontal throw at 1680 m/s would theoretically be in circular orbit at an altitude of 2 meters, but it would just slam into some crater edge or mountain.


@OrganicMarble's

The same speed as it could be thrown anywhere else? There's no $g$ in $\omega \times r$

is insightful but @SF.'s

The main limiting factor on Earth is air resistance. A stronger thrower could lob a heavier rock at similar speed, but the range of speeds is very limited by $v^2$ in air drag equation. On the Moon you could pump power into spin of the sling as long as you can control it and it doesn't break.

is important as well, and the "helicopter method" shown below does just that!

Scientific American's Whistling Sling Bullets Were Roman Troops' Secret Weapon says:

In the hands of an expert, a heavy sling bullet or stone could reach speeds of up to 100 mph (160 km/h): "The biggest sling stones are very powerful — they could literally take off the top of your head," Reid said.

That's 44 m/sec, and the Tod's Workshop video Is a sling as powerful as a gun? demonstrates a 125 ft/sec ~ 38 m/s measurement without trying very hard.

$\sqrt{GM_M/r_M}$ is about 1680 m/s so were only a few percent of orbital velocity.

Okay, but how suborbital would 44 m/s get you?

If we already know we're profoundly suborbital in terms of speed, we can estimate the maximum altitude (shooting straight up, probably killing yourself when it falls back on you) from conservation of energy, and maximum distance from a parabolic trajectory at 45 degrees.

Max height:

$$mv^2 = mgh$$

$$h = v^2/g$$

for 44 m/s and $g=GM_M/r_M^2=$ 1.62 m/s^2 gives 1191 meters!

The likelihood that it hits you on the way down is therefore small.

Wikipedia's Kinematic quantities of projectile motion give us

$$x(t) = v_0 t \cos\theta$$

$$y(t) = v_0 t \sin\theta - \frac{1}{2}g t^2$$

and solving for the time $t$ of the 2nd zero of $y$ and putting that back to solve for $x$ we get

$$x_2 = \frac{2 v_0^2}{g} \cos \theta \sin \theta$$

and using

$$\cos \theta \sin \theta = \frac{1}{2} \sin 2 \theta$$

$$x_2 = \frac{v_0^2}{g}\sin 2\theta$$

we see the familiar result that the maximum range is at $\theta = $ 45° and it will be again! a distance of $v_0^2/g =$ 1191 meters!

In both cases the flight time is about 38 seconds.

This will be much further than Allan Shephard's golf ball shot 1, 2

Alan Shephard golfing on the Moon Alan Shephard golfing on the Moon

But seriously... in a Space Suit?

The truly stellar and fascinating video Slinging Target Practice - Six Techniques shows six slingshot techniques; Helicopter, Figure 8, Byzantine, Overhand, Underhand and Greek.

The "helicopter" technique seems best suited for suited astronauts. As long as the sling is not too long, you get it moving overhead with a smooth arm motion then just use small deflections to speed it up overhead.

You will be needing either some good wrist articulation from your suit, or have to make do with using elbow or shoulder to pump speed into the helicopter.

I think this could work, and will get you close to a 1 kilometer impact distance and 38 second flight time of your very suborbital projectile.

screenshot from Slinging Target Practice - Six Techniques https://www.youtube.com/watch?v=o6kdRs4x1fs

above: Helicopter technique, below: all six for comparison

screenshot from Slinging Target Practice - Six Techniques https://www.youtube.com/watch?v=o6kdRs4x1fs

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    $\begingroup$ If you were standing on the top of the highest mountain on the moon (maybe it would need to me near the equator of the moon?) then your stone's circular orbit (or larger elliptical orbit) would always have altitude at least 2 meters above the highest mountain, so I am not convinced it would crash into the ground. And what if the sling moved the stone in an inclined circle before it is released? Then the release point and hence perilune of the orbit could be much more than 2 meters above the top of the highest mountain. $\endgroup$ Aug 15 at 3:18
  • $\begingroup$ @MatthewChristopherBartsh I think you are referring to "Standing on the Moon a horizontal throw at 1680 m/s would theoretically be in circular orbit at an altitude of 2 meters, but it would just slam into some crater edge or mountain." The moment it starts moving at orbital velocity we can say it is "in orbit" so technically it doesn't matter how far it gets. Whether it makes it only a few kilometers or all the way around the Moon several times over, it will "just slam into some crater edge or mountain" fairly soon as the Moon's gravity is pretty lumpy especially at such a low altitude. $\endgroup$
    – uhoh
    Aug 15 at 3:24
  • $\begingroup$ @MatthewChristopherBartsh So while I certainly agree with you, what I wrote is still okay I think. If you need further convincing you can ask a new question about lifetimes of extremely low altitude lunar orbits. I talk about mascons and spatial harmonics of the lunar gravity field at large distance in this answer, but in the present case they would be particularly influential only kilometers above the surface. $\endgroup$
    – uhoh
    Aug 15 at 3:28
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    $\begingroup$ Good point uhoh, about it still being in orbit if it is moving at orbital velocity. I hadn't thought of that. Making it at least several times around the moon is what I would have in mind by "into orbit", though. The question asks only about getting into orbit, not about staying there for a long time. How high above the top of the highest mountain would the launch point need to be so that despite the lumpiness of the moon's gravity (another thing I hadn't thought of --well done) the stone could orbit for a long time? $\endgroup$ Aug 15 at 3:32
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    $\begingroup$ The lumpy gravity of the moon would make it unlikely that the stone would return precisely to its launch point, whether the orbit was circular or elliptical. So if the stone were launched from at (or even just near) the top of the highest mountain, the stone would probably miss the mountain completely having completed one orbit, and the make many more orbits before eventually hitting it. So the lumpiness of the gravity might help the stone make several orbits rather than just one orbit. If the direction of the stone were chosen carefully, there might not be any other high ground in its way. $\endgroup$ Aug 15 at 21:18
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Looking at the question you are linking I would be inclined to say that it is possible assuming the astronaut's suit is flexible enough and the thrower has enough endurance. (Although, remember, the low periapsis. Combine that with the instability of lunar orbits and you can't say that it will stay in orbit!)

I don't think spinning it above your head is the right approach. Rather, look at the hammer throw. Take your pebble (the mass limit is going to be pretty low), hook a cable to it. Most of the cable is on a spool on your suit. Stand on a peak as there will be some sag of the cable. Start spinning around, maintain a constant velocity but keep letting out cable. The pebble reaches orbital speed without exerting all that much force on the spinner.

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