If you jumped on a trampoline on the moon, could you bounce so high that you would leave the Moon’s gravitational field, leaving you to float off into space?
Lunar escape velocity is about 1.5 miles per second, and I don’t think you could bounce up from the trampoline faster than you fell on it.
I’ll go out on a limb and say “No.”
However, I think on the moons of Mars, you could throw a baseball into orbit, wait for it to come to you from the other direction, then hit it back again (at least that’s what Carl Sagan said one time).
The moon’s escape velocity is 2.38 km/s, or 5324 MPH. So I would think the answer to your question would be no.
Homeowners hates trampolines as well.
Lunar gravity is about 1/6th of that on Earth - I think that means you could bounce about 6 times higher - maybe a bit more, as there’s no air resistance, so no terminal velocity, but even if the trampoline is located over a pit, you’ll reach the elastic limit of the thing waaaay before you achieve escape velocity
Yup. If I did my math right, the orbital velocity at the surface of Phobos is a mere 8.03 meters/sec, well within the reach of any weekend baseball player.
If you had a really wicked follow-through, and put your whole body into it, pushing off the rubber like a sonofabitch, like maybe Sidd Finch, would you risk launching yourself into orbit as well?
8.03 meters/sec is a bit under 18 mph. You can’t do that from a standing start. You could probably launch yourself from a full run, assuming you could get your space suit to move that fast.
Would you be able to run?
One-sixth gravity allows you to jump higher, but it also means you would flex the springs less when you landed on the trampoline. You wouldn’t get that much higher than you would jumping from the ground.
Why do the math when XKCD have done it for you in exciting graphical form?
Best bit IMHO: “You could escape Deimos with a bike and a ramp.”
(Caveat: the figure under Earth in the main diagram should be under Venus - the correct Earth figure is in the inset)
On second thought…the springs will store the combined energy from previous jumps, so with multiple jumps you could eventually flex the trampoline all the way to the ground. Once the tramp hit the ground you’d be jumping as high as you could go. With a huge trampoline mounted way above the ground, loss-less springs, sturdy legs, and a lot of patience you could jump off the moon.
My cite suggests otherwise. The moon is pretty big and its gravity is not inconsiderable. That’d be a helluva trampoline - you need one that under Earth gravity (and excluding things like air friction) could launch you 288km up. “Sturdy legs” indeed.
Real-world springs will absorb energy with each stretch, preventing you from building up enough energy.
If you’re jumping six times as high on the moon (with its one-sixth-of-earth gravity) as on Earth, then your gravitational potential energy at zenith is the same as for an ordinary-height leap on earth. When you come down from that 6X height, the moon-trampoline is absorbing the same amount of kinetic energy that an earth-trampoline would (for an earthling’s 1X jump height). That means it will deflect just as far as the earth-trampoline would, and your legs would experience the same peak loading as they would on earth.
The jumper’s leg strength imposes a design constraint. For a trampoline with a given spring constant, it limits the maximum deflection of the trampoline, which limits the energy that can be imparted to the jumper on launch - and therefore limits height. To get around it, you could select a trampoline with a softer spring constant. This will deflect farther before reaching the limit of the jumper’s legs, and therefore have more stored energy to give the jumper, allowing greater heights. Since all of the energy involved comes from the jumper’s legs, it’ll take a while to achieve the reported escape velocity (~5K MPH).
Suppose you have a mass of 180 pounds (~80 kg). Lunary escape velocity is 5000 MPH (8000 KPH), and so you require a kinetic energy of 197.5 MJ. Equation for energy storage of a spring is 1/2KX^2, where K is spring constant and X is displacement. If your legs can comfortably put out a maximum force of 320 pounds (1420 N), then the required spring constant of the trampoline mat (assume it’s linear) is 0.0205 N/m, and the trampoline mat has to be displaced a whopping 69,540 meters (at which point the force on the jumper is the aforementioned 1420 N). Yeah, you need a 69-kilometer-deep pit under the trampoline.
If we had stronger legs we could use a stiffer trampoline and still get the needed energy storage, but alas, we are only human.
Somebody check my math…
[QUOTE=Joe Frickin Friday;12213488
Suppose you have a mass of 180 pounds (~80 kg). Lunary escape velocity is 5000 MPH (8000 KPH), and so you require a kinetic energy of 197.5 MJ. Equation for energy storage of a spring is 1/2KX^2, where K is spring constant and X is displacement. If your legs can comfortably put out a maximum force of 320 pounds (1420 N), then the required spring constant of the trampoline mat (assume it’s linear) is 0.0205 N/m, and the trampoline mat has to be displaced a whopping 69,540 meters (at which point the force on the jumper is the aforementioned 1420 N). Yeah, you need a 69-kilometer-deep pit under the trampoline.
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If your tramplonaut could jump half a meter high on earth he would be supplying about 400J with each jump. So getting to 200MJ would require 500,000 jumps. Each jump is going to involve a long decent into the pit, a long fling up, and a long rise and fall above the surface. He’d better pack a lunch.
More than 500K jumps. Starting from zero motion, the earliest bounces won’t displace the trampoline very far (since the jumper’s legs don’t move very far per jump, only a couple of feet), and so won’t exert very much force: our traveler won’t be able to pump much energy (per jump) into the system until later on when he’s achieving very large displacements and has something (a very displaced trampoline) against which to exert the full force of his leg muscles.
If he had a rocket pack with him - even one that exerted less force than his weight - he could get things done quicker. First, thrust yourself downward to load up the trampoline; when it comes to a stop and starts moving you upward, turn the rocket pack around and have it work together with the trampoline, pushing you skyward. When you reach zenith, turn it around and drive yourself down at the trampoline, then repeat the whole cycle. In this way you could escape the moon even with a rocket pack that can’t counter your full weight in thrust.
Trebuchet anyone?
Nope. The energy stored in a trebuchet is gravitational, which means that it’d be reduced by the same factor in lunar gravity. Identical trebuchets on the Moon and Earth would launch projectiles to the same height and the same distance from the device.
Kudos to you Sir.
Well, they’d go a bit further as there is no air resistance. As well as there being no air to slow the projectile while it is travelling, the trebuchet mechanism itself would be more efficient in a vacuum. That big basket can’t be very aerodynamic, the tip of the arm travels quite fast (air resistance is a cube of speed), and the sling must act like an air-brake.
Cross-threading, a trebuchet would make a good weapon on the moon, as you’d have plenty of moon rocks to use as ammo, and you wouldn’t have to worry about re-supply. Great for cracking open those pesky moon-bases.