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/h, so that reduces our required speed to “only” 10.56 km/h.
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 (is velocity) 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 is a huge number, already exceeding the speed of sound (0.343 km/s) many times over. 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.
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 since a rubber band probably can’t reach the speed of sound (and remain intact), then it also cannot accelerate another object to the speed of sound. To then ask for the slingshot to accelerate to 30x the 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.
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.
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.
Let’s break this down into parts:
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/h, so that reduces our required speed to “only” 10.56 km/h.
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 (is velocity) we need, which was discussed earlier.
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 is a huge number, already exceeding the speed of sound (0.343 km/s) many times over. 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.
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 since a rubber band probably can’t reach the speed of sound (and remain intact), then it also cannot accelerate another object to the speed of sound. To then ask for the slingshot to accelerate to 30x the 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.
How did you go from seconds to hours so quickly? 🤔
Whoops, force of habit. Fixed now to not be sizable fractions of the speed of light.
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.
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.
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.
Speed of sound in the air is a serious problem as well. It would probably break most rubber bands.