this post was submitted on 12 Jul 2025
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No Stupid Questions

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The smart people have done their nerdy number-wizardry and said it's not going to work but if your huge slingshot was tall enough to reach into space, you could just put the spaceship on top and call it a day. How to get it there is an exercise for the reader.

[–] litchralee@sh.itjust.works 28 points 1 day ago* (last edited 1 day ago) (3 children)

Let's break this down into parts:

send a spaceship into space

Assuming we're launching from Earth's surface, we will need to: 1) get safely away from the ground (or else we'll crash first), and 2) achieve an orbital velocity of at least 11 km/s, which is needed to escape the influence of Earth's gravitational pull. If we launch from the equator and launch towards the east, we get the free benefit of the equatorial velocity, which is about 0.44 km/s, so that reduces our required speed to "only" 10.56 km/s.

huge slingshot

The thing with all machines that yeet an object into the air is that they're all subject to the trajectory calculations, again due to that pesky gravity thing. As much as we'd like things to be as easy as "point and shoot" in a straight line out into space, the downward force of gravity means we must aim upward to compensate.

Normally when one thinks of trajectory, it is to aim an artillery piece in such a way that it'll land upon a target in the distance, but typically at about the same altitude as where it was launched from. But if the artillery piece is perched upon a hill aiming down into a valley, then a smaller angle correction must be made because it would hit farther than intended. When aiming at a target located higher than the gun, the correction would be a slightly larger angle.

In this case, to aim into space -- assuming we mean something near the Karman line at 100,000 km above MSL -- that's a substantial height and we'll need to aim the slingshot with a substantial vertical component. The exact angle will depend on what horizontal component we need, which was discussed earlier.

how much rubber band

The relationship between the necessary vertical component (to overcome gravity) and the horizontal component (to reach escape velocity, which is caused by gravity) can be drawn as two orthogonal vectors, with the rubber band having to provide the angled thrust equal to the sum of those two vectors.

We've ignored air resistance, but with this simple relationship, it's clear that we can use basic trigonometry and the Pythagorean theorem to find that the rubber band vector is the sum of the square of those two vector magnitudes. Easy!

The only problem is that, on its own, the square of some 10.5 km/s is a huge number. Even 10.5 km/s without squaring is a huge number, already exceeding the speed of sound in air (0.343 km/s) many times over. I vaguely recall a rule somewhere that elastic deformation cannot exceed the speed of sound (EDIT: within the material -- see excellent comment below), for reasons having to do with shockwave propagation or something like that.

But I think it's all fairly intuitive that for a rubber band slingshot to accelerate an object, it too must be in contact with said object while accelerating. And while a rubber band contracting can reach air's speed of sound (barely remaining intact), it cannot go much beyond that nor accelerate another object to thoss speeds. To then ask for the slingshot to accelerate to 30x the (air) speed of sound would be asking too much.

For this reason, I don't think the rubber band slingshot to space will work, at least not for a typical linear slingshot. If you do something that rotates and builds velocity that way, then it becomes feasible.

[–] Treczoks@lemmy.world 1 points 5 hours ago

Of course a rubber band reaching to the edge of the atmosphere most likely exceeds its own tensile strength by far. And even if it was short enough just not to, it would simply be pre-stretched by it's own weight, you would not even need to hold it down.

[–] SirActionSack@aussie.zone 12 points 1 day ago (3 children)

I vaguely recall a rule somewhere that elastic deformation cannot exceed the speed of sound, for reasons having to do with shockwave propagation or something like that.

Can't exceed the speed of sound in the material, not speed of sound in the atmosphere.

But that nitpick probably doesn't change your assertion that it's not going to work.

[–] litchralee@sh.itjust.works 6 points 1 day ago* (last edited 1 day ago)

I've added this clarification to my comment. Thanks!

It seems that common types of rubber have a propagation speed in the range of 1.5-1.8 km/s, so we're still quite a bit away from 10.5 km/s.

[–] Successful_Try543@feddit.org 5 points 1 day ago* (last edited 1 day ago)

Actually, the (vertical) velocity of the deflected center point of the rubber band is faster than the axial contraction of the rubber band itself (at that point) which is limited by the speed of sound of the material.

Derivation: Pythagoras, chain rule

Based on that, knowing the speed of sound of rubber, one can obtain a minimal required length of the rubber band.

[–] Zwuzelmaus@feddit.org 4 points 1 day ago

Can't exceed the speed of sound in the material, not speed of sound in the atmosphere.

Speed of sound in the air is a serious problem as well. It would probably break most rubber bands.

[–] Zwuzelmaus@feddit.org 7 points 1 day ago (1 children)

velocity of at least 11 km/s, which is needed

free benefit of the equatorial velocity, which is about 0.44 km/h, so that reduces our required speed to "only" 10.56 km/h.

How did you go from seconds to hours so quickly? 🤔

[–] litchralee@sh.itjust.works 6 points 1 day ago* (last edited 1 day ago)

Whoops, force of habit. Fixed now to not be bicycle speeds lol.

[–] Toes@ani.social 16 points 1 day ago (2 children)

This feels like one of those xkcd questions.

Without involving maths, its fair to say it wouldn't be unlike having it launched by an explosion.

The closest thing I can recall related to this concept would be Project HARP.

Whatever you're launching would need to be capable of absorbing the unreal amount of G force and survive the resistance of the atmosphere. As you'll be required to launch the spaceship at escape velocity with a delta of near 0.

I don't believe rubber has the material strengths necessary to achieve that launch while still being able to store it on the face of the planet. So, imagine a ball of elastics the size of the sun and maybe there’s a chance because of the slow rate of acceleration achievable by not being on the earth any more.

[–] groet@feddit.org 11 points 1 day ago

Yeah I feel like this runs into the typical problem: you need so much rubber that the weight of the band itself is much much much higher than that of the rocket. So you need to add more rubber band for the rubber band, for which you need to add more rubber band again...

The same as why you can't build infinitely high buildings. The materials are just to heavy compared to their strength

[–] jol@discuss.tchncs.de 6 points 1 day ago

Zackdfilms posted a very relevant short a couple days ago: https://youtube.com/shorts/_VbmAWhjUuE

[–] solrize@lemmy.ml 7 points 1 day ago* (last edited 1 day ago) (1 children)

You couldn't do it with one slingshot, as several people have explained. But if the slingshot was big enough to launch a ship carrying someone with another slingshot, you'd do a little better. With the second ship carrying a third ship and so on, the slingshot round on the last ship might move fast enough. This is how multi-stage rockets work, by the way.

Added: the math you want is called the Tsiolkovsky rocket equation.

[–] Flax_vert@feddit.uk 3 points 1 day ago

Okay but how much rubber band