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View Full Version : Putting a big net on the ISS and flying into it



Sum0
2009-Nov-17, 04:25 PM
Wow. It's been a long time since I last posted here (six years or so!).

A little thought experiment that's been nagging at me:

Say I drape a big net between two parallel trusses on the ISS (or indeed, any object in Earth orbit) - it's made of a sufficiently strong material to stop anything that flies into it. I get in my sub-orbital spaceship on Earth and fly straight upwards, so that my apogee intersects with the ISS. When I hit apogee, I fly straight into my specially-constructed net, and quick as a flash a robotic arm clamps on to my spaceship and docks me to the station.

What happens then? Have I just "cheated" my sub-orbital flight into an orbital one?

I suspect that what happens is that when I crash into the ISS the ISS also crashes into me, and my lack of orbital velocity slows the ISS down to the point where it falls out of orbit, but I'm not entirely sure.

slang
2009-Nov-17, 04:27 PM
Welcome back :)

You could also end up as a set of metal french fries, falling back to Earth.

Argos
2009-Nov-17, 04:30 PM
I suspect that what happens is that when I crash into the ISS the ISS also crashes into me, and my lack of orbital velocity slows the ISS down to the point where it

Is shredded to pieces.

Sum0
2009-Nov-17, 04:35 PM
I'm rapidly coming to think that this is technically infeasible for many reasons :D but assume my spaceship (and the ISS) are extremely resilient. Or at the very top of my flight, an insanely powerful magnet sticks me to the ISS as it silently whooshes past. Or I have an arrestor hook/cable arrangement, like on aircraft carriers. What then?

slang
2009-Nov-17, 04:42 PM
Hmm... to start with, a huge spin imparted to ISS, I would guess.

Amber Robot
2009-Nov-17, 04:45 PM
If your rocket does not have the same velocity in the direction of the orbit of ISS then you'll have a collision. If we look at a simplified version of the situation it might go like this: imagine the ISS is orbiting at the equator and you launch your rocket from the equator straight up. The ISS orbits about every 90 minutes, so there's about 16 orbits per day. Your rocket will have some velocity along the same direction because of the spin of the Earth, but that's only once per day, so the ISS velocity in the direction of its orbit will be 16 times as fast as your rocket's. The wikipedia page says that the ISS orbital velocity is around 17000 mph. If your rocket is then at about 1000 mph or so, then the collision happens at a relative velocity of 16000 mph. So, what do *you* think happens when two spaceships colllide at 16000 mph?

slang
2009-Nov-17, 04:52 PM
So, what do *you* think happens when two spaceships colllide at 16000 mph?

Seriously? Something (http://www.universetoday.com/2009/02/11/two-satellites-collide-in-earth-orbit/) like this (http://www.universetoday.com/2009/02/14/how-and-why-did-two-satellites-collide-this-week/), or even this (http://www.universetoday.com/2008/02/20/us-cruiser-strikes-dead-spy-satellite/).

jogleby
2009-Nov-17, 04:56 PM
If your rocket does not have the same velocity in the direction of the orbit of ISS then you'll have a collision. If we look at a simplified version of the situation it might go like this: imagine the ISS is orbiting at the equator and you launch your rocket from the equator straight up. The ISS orbits about every 90 minutes, so there's about 16 orbits per day. Your rocket will have some velocity along the same direction because of the spin of the Earth, but that's only once per day, so the ISS velocity in the direction of its orbit will be 16 times as fast as your rocket's. The wikipedia page says that the ISS orbital velocity is around 17000 mph. If your rocket is then at about 1000 mph or so, then the collision happens at a relative velocity of 16000 mph. So, what do *you* think happens when two spaceships colllide at 16000 mph?

Even if the spaceships were made of some magical material that allowed them to survive the impact, wouldn't the differences in angular momentum drag both ships out of orbit and cause them to fall back to earth?

grant hutchison
2009-Nov-17, 05:24 PM
Discarding all the technical difficulties that have been raised ...
We have a mass that is briefly stationary at the top of a vertical flight (call it "riser"), and we have another mass in a circular orbit (call it "orbiter"). The two stick together. What happens?
The momentum of the combined mass is not enough to maintain the previous circular orbit. It will enter an elliptical orbit, with the collision point as apogee. The altitude of the perigee will depend on the relative masses of orbiter and riser. A tiny riser will knock the orbiter into an orbit with a minutely lower perigee. An extremely massive riser will be knocked a little sideways by the orbiter, and the two will return to Earth on an elliptical orbit with such a low perigee that it looks like a near-vertical drop. Intermediate cases will result in ellipses which may or may not enter the atmosphere after half an orbit or so.

Grant Hutchison

jfribrg
2009-Nov-17, 05:28 PM
Lets say the IIS is about 300000 kg and traveling 17000 mph. Let's say that your payload is 1000 kg and traveling 1000mph. That means that the momentum of the IIS is about 2307795698 kg m/s. Your rocket is about 448028 kg m /s. Afterwards. the combined mass is about 301000 kg and the combined momentum is 2308243726 kg m /s, so the velocity of the combined masses must be 7592.9 m/s or 16920 mph. This would place it in a slightly lower orbit. The end result is that if your payload is not too great and if you can solve the engineering challenges of making a net and payload that can survive the impact, then I would say that your plan would work. Remember however that every time you do this, the ISS loses 80 mph (more than that if your payload is more than a ton), so your payload would probably need to bring along some fuel to lift the ISS back to its original orbit and this fuel would reduce the amount of other stuff that you can lift with your rocket. Remember the stardust probe? It had some kind of super elastic material to catch materials from a comet tail at super high impact velocities. Maybe you can adapt that material for your own purposes.

Ilya
2009-Nov-17, 07:31 PM
A variation on the theme (and SLIGHTLY more feasible :D ) is something called orbital pinwheel. It does not have to be an actual wheel, just a REALLY long bar spinning about its center is sufficient. This is how it works:

The bar orbits Earth clockwise. It has typical low orbital velocity of about 7.9 km/sec. But it also spins around it center of mass counterclockwise, in the same plane as it orbits Earth. So every time either end of the bar sweeps closest to Earth (the bar is instantaneously "vertical"), it moves "backward" compared to orbital velocity. And here is the kicker (almost literally) -- the bar spins so fast that each end travels around its center also at 7.9 km/sec. Which means that every time bar is vertical, the end nearest Earth is motionless with respect to Earth. If you happened to be above Earth's atmosphere at the spot where the pinwheel's arm is vertical, you would see it approach at terrifying speed but slowing down (from your viewpoint) at tens of gees, come to a complete stop, then instantly accelerate up and away again at tens of gees. If you grab onto the pinwheel at that instant, you'll find yourself in orbit. If you let go at the top of pinwheel, after one half-spin, you will sail away from Earth at 15.8 km/sec.

How long is the bar depends on how much acceleration you are willing to tolerate. The bar's end carries you away from your original standstill position at twice the centripetal acceleration. Say, maximum 30 gees with lungs filled with oxygenated fluid. Then pinwheel radius (or bar half-length) must be 7900^2/(15 * 9.8) = 420,000 meters. An 840 km bar. Make the bar 2000 km long, and you are subjected to more tolerable 12.6 gees.

If the bottom of pinwheel's sweep is at 100 km height (its center of mass is then on either 520 or 1100 km orbit, with above examples), it's well within reach of SpaceShipOne. Needless to say, law of energy conservation is not broken -- when the payload is grabbed at the bottom, the bar's spin slows down slightly, and remains slowed down when it is released on top.

WayneFrancis
2009-Nov-18, 01:54 AM
Even if the spaceships were made of some magical material that allowed them to survive the impact, wouldn't the differences in angular momentum drag both ships out of orbit and cause them to fall back to earth?

or be kicked into a different orbit.

novaderrik
2009-Nov-18, 02:48 AM
unless you have Star trek style inertial dampers, the ship might survive but the occupants will be turned to a gooey stain splattered against one side of the cabin.

Sum0
2009-Nov-18, 03:45 AM
These are some really good answers, thanks!

I suppose another way of looking at it would be if I launched a rocket directly towards the moon. When it hits the lunar surface, it's essentially entered orbit (although in lots of tiny pieces :P) and the moon is slowed down an infinitesimal amount.

Something related: I was reading up on space elevators recently. On a space elevator, you're only expending energy to go upwards, like my theoretical sub-orbital capsule. Where does the orbital energy come from to speed travellers into a geosynchronous orbit? Isn't it "stolen" from the counterweight at the top of the elevator?
I would presume that the drag of slow-orbital-speed elevators coming from the ground and moving up the cable would slow the cable down and eventually take it out of orbit, right?

WayneFrancis
2009-Nov-18, 05:33 AM
These are some really good answers, thanks!

I suppose another way of looking at it would be if I launched a rocket directly towards the moon. When it hits the lunar surface, it's essentially entered orbit (although in lots of tiny pieces :P) and the moon is slowed down an infinitesimal amount.

Something related: I was reading up on space elevators recently. On a space elevator, you're only expending energy to go upwards, like my theoretical sub-orbital capsule. Where does the orbital energy come from to speed travellers into a geosynchronous orbit? Isn't it "stolen" from the counterweight at the top of the elevator?
I would presume that the drag of slow-orbital-speed elevators coming from the ground and moving up the cable would slow the cable down and eventually take it out of orbit, right?

I might be mistaken but I think the energy is not stolen from the counter weight but the Earth. IE ever time you lift an object up you slow the rotation of the Earth down a bit.

Like the ice skater. When they push their hands out they lower the rotation rate of the entire system but the angular momentum remains the same.

mugaliens
2009-Nov-18, 08:10 AM
I'm rapidly coming to think that this is technically infeasible for many reasons :D but assume my spaceship (and the ISS) are extremely resilient. Or at the very top of my flight, an insanely powerful magnet sticks me to the ISS as it silently whooshes past. Or I have an arrestor hook/cable arrangement, like on aircraft carriers. What then?

First, welcome back! Wow - six years!

From what I gather, you're asking about the change in the orbit of the ISS, not whether you two would remain intact.

That depends on the relative masses of the ISS and your spacecraft. If your spacecraft were the same mass as the ISS, your combined velocities would be half of the ISS's original velocity, and you'd both plummet to Earth.

Even if your spacecraft were only a small fraction of the ISS's mass, it would still induce an eccentricity in the ISS's orbit that would dip it into enough upper atmosphere to drag it ever lower.

Yours is the concept of a rotorvator, which is actually sound, despite some engineering hurdles. The ISS simply doesn't have enough mass to swing it. Most rotorvator concepts include a small asteroid, warped into Earth's orbit via thruster, to serve as the anchor.