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A Question About Orbital Mechanics
Old 05-21-2015, 09:09 AM   #1
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A Question About Orbital Mechanics

An untethered astronaut pushes off from the International Space Station.

My understanding is that rather than just float away, he will go into a new orbit, and that orbit will intersect the orbit of the ISS on the other side of the Earth.

The reference I read (and can't find again) said that this happens if the push off is to one side of the ISS and not toward or away from Earth. But I'd guess that it would be the same for any direction. That is, push down toward Earth, and he/she would to into an elliptical orbit that would intersect the original orbit.

Does that sound right?
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Old 05-21-2015, 09:28 AM   #2
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No. At least not in an absolute sense. There has to be at least one vector that would send you straight to earth. Also, there would be some relationship based upon your mass vs the mass of the space station but I'm guessing you would model the space station as infinite compared to yourself. Same thing about how hard you can accelerate yourself. If you push hard/fast enough it would make a difference and if you push too little gravity pulls you back. And then you get to doing this in three dimensions! Not certain but I bet you would very likely do lots of orbits before coming back to the space station if drag didn't pull you to earth first!
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Old 05-21-2015, 09:41 AM   #3
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Really interesting question. I am certainly not a trained astrophysicist, but I did take some classical mechanics courses in college many years ago. My take on the situation where the astronaut pushes down from the shuttle directly towards Earth is that his new orbit would not intersect with the shuttle. This is because he is closer to the Earth and has less gravitational potential energy, yet has exactly the same tangential velocity with respect to the Earth. The reason that the shuttle stays in a stable orbit is because its tangential velocity is exactly enough to offset the gravitational force of Earth. When the astronaut moves down towards Earth, his tangential velocity will stay the same but the Earth's gravitational force on him will get larger (and larger), thus his orbital path will shrink. His orbits around the Earth will no longer form a complete circle, but would look like a corkscrew or spiral pattern from far away. And then, as inertia carries him closer to the Earth, his tangential speed will begin to slow due to frictional drag caused by the Earth's atmosphere, and the downward spiral will become more pronounced.

I think you could analyze the "pushing off to the side, not towards or away from Earth" situation in a similar way, but I haven't thought it through yet.

Of course, could be totally wrong in my analysis. Physics can be very tricky and sometimes quite counterintuitive, as I learned well in college.
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Old 05-21-2015, 09:50 AM   #4
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For every action there is an equal and opposite action and every object in motion remains in motion. My thinking would be that if one pushes off from the ISS, say toward the Earth, you may be in a new orbit but gravity (a function of mass) allows the ISS to "fall" to the Earth faster (because it's mass is bigger than the astronaut) and their orbits will eventually intersect. If one pushes away from the Earth, I don't think they would intersect because the mass of the ISS is greater therefore "falling" faster away from the astronaut, and the fact that the orbit of the astronaut is slightly further away from the Earth, kinda like running on the outside of a track instead of the pole position. The astronaut orbital speed is the same but the distance is longer and he has no means to accelerate to increase speed or shorten his orbital distance. Just my 2 cents.
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Old 05-21-2015, 09:52 AM   #5
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I don't have a clue, but this is an interesting thread.
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Old 05-21-2015, 10:02 AM   #6
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If you really want to find out, might want to look at:

Kerbal Space Program - Wikipedia, the free encyclopedia

In any case, atmospheric drag (which the ISS has to fight) will make any rendez-vous after just one push away not a given. Also pretty sure the direction of the push makes a difference. Basically like mr. ArtTinkerer said
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Old 05-21-2015, 10:02 AM   #7
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There is actually a game Orbiter Space Flight Simulator that allows you to manouver spaceships in orbit (the astronaut is a space ship for these purposes) There is also a tutorial on the game at : Go Play in Space (for download it is 150 pages).
In any case Slight velocity differences will result in slight changes in the orbit. You can use the simulator to try this out. 2 simple cases push off at apogee or perigee and the difference will maximize at the other end of the orbit (perigee or apogee). The orbits in the absence of drag interset at the point you push off, but as the periods differ the two bodies wont both be there. (this is pushing off in a prograde or retrograde direction). Pushing off perpendicular to the orbital direction (parallel to the earths surface) will change the plane of the orbit (its inclination).

So 1 orbits intersecting does not mean that the bodies will be at that point at the same time is the critical point. This is why the low energy interplantary transfer orbits only occur every so often so the space ship and the destination are close when the space ship intersects the destination planets orbit.
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Old 05-21-2015, 11:07 AM   #8
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Here's my thinking.

Yes there is a vector that gets you back to Earth, but only because you intersect the atmosphere or Earth, before you have a chance to complete your new orbit.

So, you push toward Earth. As you get closer to Earth, you go faster, and like a comet, when you go around the earth, the increased speed kicks you out further, resulting in an elliptical orbit.

When a spacecraft wants to get to a higher orbit (e.g. to rendezvous with another), it has to do two burns. It first does a burn to increase its speed, boosting the orbit, and then a second burn to stay in the new orbit. That's described in this video at 5:45:



This diagram shows that after one burn, the new orbit would come back to the same place:



The pushing-off astronaut would be doing only one burn.
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Old 05-21-2015, 11:23 AM   #9
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My question would be if you want the astronaut to come back to the space station just intersect the orbit


Let's say (and I have no idea if it is true) that the orbits will intersect.... but will he come back so he can get back on I would say it makes a difference on which way he pushes off..

In your example, he pushes off at the exact vector the space station is traveling... it looks like he would come back to the station after one revolution....

But what if he pushed off at a 90 degree.... he is not in a larger orbit, but has now changed his position relative to the ISS... so if he were able to push off at 10mph... in an hour he would be 10 miles from the ISS... eventually he would cover the full number of miles the orbit was at (say 30,000 miles), but he would be dead.... (30,000 / 10 = 3,000 hours / 24 = 125 days)....


So, I think it makes a big difference on which direction he jumps....
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Old 05-21-2015, 12:26 PM   #10
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Quote:
Originally Posted by TromboneAl View Post
Here's my thinking.

Yes there is a vector that gets you back to Earth, but only because you intersect the atmosphere or Earth, before you have a chance to complete your new orbit.

So, you push toward Earth. As you get closer to Earth, you go faster, and like a comet, when you go around the earth, the increased speed kicks you out further, resulting in an elliptical orbit.

When a spacecraft wants to get to a higher orbit (e.g. to rendezvous with another), it has to do two burns. It first does a burn to increase its speed, boosting the orbit, and then a second burn to stay in the new orbit. That's described in this video at 5:45:



This diagram shows that after one burn, the new orbit would come back to the same place:



The pushing-off astronaut would be doing only one burn.
However the space station would not be there at the time the astronaut returns since the orbital periods differ. This is why there is the large wait time at mars using the low energy orbits you have to wait until the earth and your ship will be in the same place at the same time to insert into the transfer orbit.
Pushing retrograde would shorten the orbit time, pushing prograde would
lengthen it (relative to the space station orbit).
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Old 05-21-2015, 01:38 PM   #11
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Quote:
Originally Posted by Winemaker View Post
For every action there is an equal and opposite action and every object in motion remains in motion. My thinking would be that if one pushes off from the ISS, say toward the Earth, you may be in a new orbit but gravity (a function of mass) allows the ISS to "fall" to the Earth faster (because it's mass is bigger than the astronaut) and their orbits will eventually intersect. If one pushes away from the Earth, I don't think they would intersect because the mass of the ISS is greater therefore "falling" faster away from the astronaut, and the fact that the orbit of the astronaut is slightly further away from the Earth, kinda like running on the outside of a track instead of the pole position. The astronaut orbital speed is the same but the distance is longer and he has no means to accelerate to increase speed or shorten his orbital distance. Just my 2 cents.
Without drag (present but greatly reduced at the altitude we are talking about) large and small masses fall at the same speed. In a perfect vacuum a feather and a bowling ball fall at the same speed. The mass matters because you push against the station. The stations orbital velocity changes relative to the force you apply but the speed difference is inversely proportional to the mass involved--you go faster cause you have less mass. The space station has much more mass than you so the speed change for it is very small. Again, the amount of force applied matters--you could shove yourself out of orbit if you pushed hard enough (but that might be beyond reason!). In fact, if you could push hard enough, you could shove yourself and the space station out of orbit!
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Old 05-21-2015, 04:54 PM   #12
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Quote:
Originally Posted by meierlde View Post
However the space station would not be there at the time the astronaut returns since the orbital periods differ. This is why there is the large wait time at mars using the low energy orbits you have to wait until the earth and your ship will be in the same place at the same time to insert into the transfer orbit.
Pushing retrograde would shorten the orbit time, pushing prograde would
lengthen it (relative to the space station orbit).
Ah, yes, I think that's key. So, if I want the orbits to intersect, the guy has to push off at 90 degrees to the direction of travel, and neither away nor toward Earth.

I really want that. NASA pulls off a risky, kludgy mission to rescue the two astronauts from a damaged ISS by sending up a SpaceX Dragon. But the ISS starts spinning uncontrollably. Long story short, they come around the earth and:
“Whoa!” Both Catherine and Ray said it at the same time.

“Say status, please.”

“Houston, we just caught sight of the station, and it’s spinning right toward us.”

In the view from Ray’s helmet cam, the ISS, now in daylight, was spinning like a huge propeller. Most of the components had been flung off, but the main truss and a few modules were intact.

“Ray, what is your ETA for Dragon?”

“Houston, I don’t think we’re going to make it.”

Ray’s camera showed Dragon with the ISS in the background. To me it looked like Dragon would pass safely by, but the astronauts were headed right for the spinning fan. In all the vastness of space these three objects, traveling at 17,000 MPH, were all going to get damn close to one another.

“Houston, using my thrusters to avoid the station. I’m low on GN2.”

“Copy, Ray.”

The camera view shifted wildly as Ray maneuvered. The spinning mess approached like an evil machine hellbent on destroying the vulnerable astronaut pair. I pressed my foot against the floor trying to put on the brakes.
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Old 05-21-2015, 11:19 PM   #13
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Originally Posted by TromboneAl View Post
Ah, yes, I think that's key. So, if I want the orbits to intersect, the guy has to push off at 90 degrees to the direction of travel, and neither away nor toward Earth.
To rendezvous in orbit the 6 orbital elements Orbital elements - Wikipedia, the free encyclopedia need to be identical. One thing is that it takes a lot of energy to change the orbital plane very much (This is why the space shuttle going to the space station had 10 min launch windows) So the plane change by pushing off normal to the orbit would be very small, and you likley would not get very far away. In any case if you recall the shuttle operations they would get within a few hundred feet and station keep for a while (In later flights post Columbia they did the somersault to check the tiles) In real life most likley they would station keep just close enough to see the station with (a kilometer or so) and keep station for a while.
Here is a like to a wikipedia article on the process and a couple of methods of rendezvous one you have station kept. Space rendezvous - Wikipedia, the free encyclopedia
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Old 05-22-2015, 09:59 AM   #14
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Okay, good. The Dragon and astronauts will actually be flung off the ISS, because it starts spinning just when they are evacuating. The Dragon will be on the Canadarm, the spin gets faster and faster, it can't be held and they, boink, the arm breaks and they get flung off. They can get pretty much energy that way.

The thing I'll get (well-deserved) grief over will be that they will have to be flung off at just the right angle for the rendezvous to happen.
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Old 05-22-2015, 12:36 PM   #15
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Here's a related article:

Spacewalking astronaut hurls rubbish out into space | Daily Mail Online

Relevant text:

Anderson threw the equipment in the opposite direction of the station's travel ...

There should be no danger of a collision between the free-floating tank and station before [it burns up in the atmosphere].
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