Would an astronaut be able to get back to the space station?

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Would an astronaut really bale able to make it back to the space station by throwing a wrench in the opposite direction if his cable broke and floated away? I heard this question as it relates to Newton's third law. Wouldn't the astronaut need to throw the wrench faster than he is moving before throwing wrench? Would the wrench even be able to give the astronaut enough force to get back to the space station since the astronaut has more mass than the wrench? He would not only have to throw the wrench fast enough to overcome his speed before throwing wrench but he would have to throw it superman fast to allow him to float back to the space station even if the astronaut is moving slow? No other forces would stop him after the wrench exerts a good force on him? How does being in orbit play a role on the situation? Thank you.
 

.Scott

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Yes, it is possible.
Let's say that the astronaut is drifting away at a speed of 5 feet per second - relative to the station. Also, let's assume that the astronaut is 20 times heavier than the wrench. The astronaut would have to fling the wrench at 20*5 feet per second to set his speed to that of the station. Any faster, and he will begin drifting towards it.

But that is not the whole story. The station is in orbit about the Earth - at roughly 1 orbit every 90 minutes. Flinging the wrench may not work even if it is thrown fast enough - because drifting towards the station may not be the correct thing to do to intercept the station.

Let's assume that the station is in a perfect circular orbit and let's say that the astronaut jumped directly towards the Earth from the station. He would then be in an orbit that would still be about 90 minutes long - but not circular. If he keeps the wrench attached to his belt, he will continue to drift downward from the station for only 22.5 minutes, then he will begin drifting back. At 90 minutes, he will strike the station at the same speed of his original jump.

On the other hand, if he jumps forward, he will enter a longer, slower orbit. He will drift forward, then up, then backwards. As the orbits go by, he will drift further and further back.
 
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Yes, it is possible.
Let's say that the astronaut is drifting away at a speed of 5 feet per second - relative to the station. Also, let's assume that the astronaut is 20 times heavier than the wrench. The astronaut would have to fling the wrench at 20*5 feet per second to set his speed to that of the station. Any faster, and he will begin drifting towards it.

But that is not the whole story. The station is in orbit about the Earth - at roughly 1 orbit every 90 minutes. Flinging the wrench may not work even if it is thrown fast enough - because drifting towards the station may not be the correct thing to do to intercept the station.

Let's assume that the station is in a perfect circular orbit and let's say that the astronaut jumped directly towards the Earth from the station. He would then be in an orbit that would still be about 90 minutes long - but not circular. If he keeps the wrench attached to his belt, he will continue to drift away from the station for only 22.5 minutes, then he will begin drifting back. At 45 minutes, he will strike the station at the same speed of his original jump.

On the other hand, if he jumps forward, he will enter a longer, slower orbit. He will drift forward, then up, then backwards. As the orbits go by, he will drift further and further back.
So if he flung his wrench at exactly 20*5 per second, wouldn't that make the net force zero, therefore he's moving at some kind of constant sequence with the space station although he's not making any progress towards it? You're also trying to say that if he threw his wrench more than 20*5 meters per second then there would be an acceleration towards the space station?
 

A.T.

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So if he flung his wrench at exactly 20*5 per second, wouldn't that make the net force zero, therefore he's moving at some kind of constant sequence with the space station although he's not making any progress towards it?
Yes, if you ignore orbital mechanics, and replace "net froce" with "relative velocity", and "constant sequence" with "constant distance".

You're also trying to say that if he threw his wrench more than 20*5 meters per second then there would be an acceleration towards the space station?
Yes, although Scott was using feet per second.
 

jbriggs444

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You're also trying to say that if he threw his wrench more than 20*5 meters per second then there would be an acceleration towards the space station?
In addition to the other minor corrections

Regardless of how fast he throws the wrench away from the space station, the result is an acceleration toward the space station during the throw. The important thing about throwing faster than 20*5 feet per second is that the result is a velocity toward the space station after the throw.

[If you are moving away from the station, merely slowing down still counts as an acceleration toward the station]
 

sophiecentaur

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Would an astronaut really bale able to make it back to the space station by throwing a wrench in the opposite direction if his cable broke and floated away?
It would have to depend on what caused the cable to break. If he is just "floating away" that suggests the cable just happened to come undone. In that case, and assuming he is not too far away from the ship he could forget about the orbital mechanics involved and fling the wrench directly away from the ship. I suspect that he could be allowed some additional smaller object on his person which he could use for a final fine adjustment. He could have a problem when he actually arrive at the ship because he would need to grip hold of something or he could just bounce off - like a space probe off a comet. Do they carry magnets? That could be handy.
Otoh, if the cable parted due to a significant force, there is no way the astronaut would be able to impart enough momentum to the wrench. Jumping off a massive 'service pod' could perhaps take the astronaut back to the ship.
I liked the jet thruster trick with the suit's oxygen supply - but it would require a fair bit of practice in EVA training courses.

Something (yet another thing) that annoyed me about the incident in Gravity that involved the astronaut pulling him (her?) self back with a rope. The angular momentum was ignored totally; If there had been any added tangential velocity due to the accident, he/she would have started to orbit fast round the ship and the rope would have wrapped itself round the ship faster and faster - like in that tethered tennis ball game. I could 'feel' it happening as I watched the sequence and I was very disappointed at the film makers fail.
 

jbriggs444

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rope would have wrapped itself round the ship faster and faster
Faster and faster angular velocity. Fixed tangential velocity since the "central" force is aligned with the target surface, not the target center.

Tetherball conserves energy.
 

sophiecentaur

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Faster and faster angular velocity. Fixed tangential velocity since the "central" force is aligned with the target surface, not the target center.

Tetherball conserves energy.
The astronaut was pulling himself in on the rope - doing work and adding to the rotational energy (his work must go somewhere.) You are right that the tetherball has no source of extra energy and therefore there would be no extra KE. It would seem that the technique would be to start the process off by hauling in a short way and then allow the rope to finish the job for you.
 

ZapperZ

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Would an astronaut really bale able to make it back to the space station by throwing a wrench in the opposite direction if his cable broke and floated away? I heard this question as it relates to Newton's third law. Wouldn't the astronaut need to throw the wrench faster than he is moving before throwing wrench? Would the wrench even be able to give the astronaut enough force to get back to the space station since the astronaut has more mass than the wrench? He would not only have to throw the wrench fast enough to overcome his speed before throwing wrench but he would have to throw it superman fast to allow him to float back to the space station even if the astronaut is moving slow? No other forces would stop him after the wrench exerts a good force on him? How does being in orbit play a role on the situation? Thank you.
It all depends on his initial velocity when the cable broke. If the cable broke and he's not moving much relative to the space station, then any amount of throwing of the wrench in the opposite direction will get him back. Otherwise, this question can't be answered without more info.

Zz.
 

.Scott

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... In that case, and assuming he is not too far away from the ship he could forget about the orbital mechanics involved and fling the wrench directly away from the ship.
The only situation when orbital mechanics can be ignored is when the maneuver to return to the ship occurs over a period of several minutes or less. If tossing the wrench results in a drift towards the station that would, ignoring orbital mechanics, get him back in say 15 minutes, then he is probably in big trouble.

For example, let's say that the tether loosened as he left the station in a forward direction (relative to the orbit). 90 minutes later he notices that the station is in front of him. He may not also notice that he is actually headed towards the station - but that is only temporary. So he attempts to correct by tossing the wrench backwards, directly away from the station. If he succeeds in propelling himself fast enough, it could work. But if he fails to reach the station within a few minutes, he is much worse off than he was to begin with. If he wasn't able to catch the station directly, he would have been better off tossing the wrench towards the station so that his orbital velocity (and therefore period) would better match the stations.
 

sophiecentaur

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he would have been better off tossing the wrench towards the station so that his orbital velocity (and therefore period) would better match the stations.
That's so counter-intuitive that you just have to explain further. The fifteen minutes you mention would be a long time for the poor gut to wait to do something about the problem.
I think this question cries out for a quick simulation. SOMEBODY !!!!
 

Filip Larsen

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That's so counter-intuitive that you just have to explain further.
If you linearize the relative orbit between two close objects themselves in orbit around a massive object you arrive at the Clohessy-Wilshire equations [1]. First thing to notice is that z motion (motion along h-bar) is a simple harmonic motion that is decoupled from x and y (v-bar and r-bar). Second thing is that there is are many interesting (and counter-intuitive) solutions to the x and y motion, e.g. closed 2:1 elliptic motion.

Regarding visualization, I have been looking for a good online resource for years without finding any, but today I found a GeoGebra page [2] that at least allows interested to get a rough idea of the x-y motion for different initial conditions.

[1] https://en.wikipedia.org/wiki/Clohessy-Wiltshire_equations
[2] https://www.geogebra.org/m/cdTAMXmv
 

A.T.

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That's so counter-intuitive that you just have to explain further.
Consider the rotating frame, where the station is at rest.

You have an effective potential similar to below in green, with the station at the lowest point (stable circular orbit). This is symmetrical all around, so the station rests in a circular potential groove. If you push off from it, you will be affected by that potential groove (pushing you towards its lowest area), but also by the Coriolis force (which tries to make you go in circles). These two effects generate the spiraling or elliptic motion seen here: https://www.geogebra.org/m/cdTAMXmv

central-force-motion.png


Plot from : https://www.wired.com/2016/06/way-solve-three-body-problem/
 

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.Scott

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That's so counter-intuitive that you just have to explain further. The fifteen minutes you mention would be a long time for the poor gut to wait to do something about the problem.
I think this question cries out for a quick simulation. SOMEBODY !!!!
We start with the spaceman at the station in a 90 minute circular orbit. We'll assume that the station stays exactly in that orbit even when the spaceman kicks off.

At this point, we can go to one of @sophiecentaur links ( https://www.geogebra.org/m/cdTAMXmv).
To create the situation where the spaceman is kicking off away from the front of the station, provide a vx0 of 1 and a vy0 of 0. In this simulation, an orbital period is about 6.3 time units. So we will take one of their time units to be a bit over 14 minutes to get our 90-minute orbit.

When the spaceman kicks off, he will be travelling faster than the station, but will have the same perigee. So his apogee will be higher than the stations - and since he is now in a higher orbit, his orbital period will be longer than 90 minutes. So, although he is going faster than the station, he will take longer to complete an orbit.

Looking at the geogebra plot, in about one-fifth of an orbit (18 minutes), our spaceman is no longer in front of the station but rather directly above it and loosing ground fast. After slightly more than 90 minutes (exactly one orbit for the spaceman), the spaceman is back to his perigee, directly behind the station, and travelling exactly as fast as he did when he kicked off. So he is actually heading towards the station. But not for long, 9 minutes later he will be ascending vertically relative to the station.

Although throwing the wrench directly at the station seems counter-intuitive, there is another was to look at it. If he throws that wrench exactly fast enough to not only cancel his initial velocity (vx0=1), but to reverse it (vx0=-1), he will suddenly be moving away from the station. But this time he will be looping down and after one orbit, he will rejoin the station.

Let's say that he can throw the wrench fast enough to get a change in velocity of 3. Assuming he didn't care how fast he struck the station, how fast could he get back?
Playing with that tool, it appears that throwing the wrench to change his orbital velocity from (1,0) to (-0.7,-2.5) will get him back to the station in about an hour.
 

.Scott

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As a further demonstration of how potent the orbital dynamics are - and what it takes to try to ignore them, let's revisit my last post using the https://www.geogebra.org/m/cdTAMXmv link.

Once again, we have our spaceman depart the station with a relative velocity vector of (1,0) - directly forward in orbit. Then he waits one orbit before deciding to return.
But when he returns, he thinks it is too counter-intuitive to throw his wrench towards the station to create a path towards it. So how much of a change in velocity will he need to reach the station if he limits himself to changes that are straight up, straight down, or generally towards the station.
Answer: He needs to go from v=(1,0) to v=(1,-6.6). With that delta-v (6.6), he will return to the station in about 36 minutes.
That's in contrast to the counter-intuitive delta-v of 2 that would get him there is 90 minutes.

In an earlier post, I said that if an attempt at a direct route took 15 minutes, our spaceman would be in big trouble. Playing with this simulator, he would nee to work much faster than that. At a delta-v of 40 towards the station (yielding v=(41,0)), in 10 minutes, he would have caught up with the station in the x direction, but would have have gained about as much altitude in the y as distance in the x - and he would be moving rapidly away from the station.

Even with a travel time of only 4 minutes, he would need a serious vy correction downward. Try v=(71,-20).

So, it would seem that as a practical matter, the orbital mechanics for a low Earth orbit cannot be ignored until travel times get below 4 minutes.
 

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