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Ok, this is a question from a student in my class that was inspired from us watching the ISS Live Earth Feed. He wanted to know if there was a point in space when you would start falling. He likened it to a swimming pool. He said if you were swimming in space was there a point when if you stepped over an imaginary barrier you would suddenly start falling towards Earth.

I was not able to answer him or explain what the transition would be like but thought I could get an answer here so over to you and thanks for anyone who takes the time to respond to an 14/15 year old student.

Thanks

Chris

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    $\begingroup$ Are you assuming you are orbiting? Just moving in space freely? Something else? I might suggest he try playing Kerbal Space Program, which gives an excellent idea of such things. $\endgroup$ – PearsonArtPhoto Apr 22 '16 at 13:29
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    $\begingroup$ @PearsonArtPhoto Yeah, it would be good if we could clarify if the student is aware that weightlessness isn't a product of being in space, but a product of orbit. $\endgroup$ – called2voyage Apr 22 '16 at 14:04
  • $\begingroup$ Another interesting property of orbital mechanics: if you left the station wearing a manoeuvring unit and started to slowly "fly" down towards Earth, you would in fact be moving to a lower orbit. This would mean you would begin to "drift" ahead of the station, then (probably) above and ahead of it in ever increasing circles. Very confusing! $\endgroup$ – Andy Apr 22 '16 at 14:18
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    $\begingroup$ To settle the discussion between Antzi and Andy, refer to Pushing down a projectile from LEO. $\endgroup$ – a CVn Apr 22 '16 at 14:56
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    $\begingroup$ Thank you. I don't know whether any answer is the right answer but it is the perfect answer for 14/15 year old to get them thinking. Top stuff! $\endgroup$ – Davies-Barnard Apr 22 '16 at 22:33
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Let's assume an astronaut on the ISS goes out for a swim.

When will he start falling ?

Actually he is already falling to earth. But he is also moving at the same speed as the ISS relative to earth (7.66km/s). To be specific, this is not called "falling" but orbiting. He never reaches the ground but just keeps turning around the earth.

However, our astronaut is very slowly losing speed, because of the very rarefied atmosphere. After months/years, he will reenter.

As our astronaut is moving slower and slower, he is also losing altitude. This is a very gradual process... Until he reaches the reentry interface, which varies by atmospheric conditions (meteo) but is about 128 km high. At this point, he starts losing speed MUCH more quickly, and also loses height much more quickly.

So from this point, yes, you can say he is falling back to earth.

PS: Of course, our astronaut will die from suffocation in his space suit and burn up during the reentry process.

TL;DR: Yes, at around 128 km high

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  • $\begingroup$ Apollo basically totally arbitrarily picked 400k ft as the entry interface altitude. 400,000 feet is 121.9 km. $\endgroup$ – a CVn Apr 22 '16 at 14:59
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For a far more theoretical perspective, I'll interpret the question as "can you swim through space (perfect vacuum, without reaction mass, i.e. rockets)?" and answer it like they might at Physics.SE

In flat space, no. If you were to move your arms forward, most of the rest of you would move backwards to keep your center of mass in the exact same point. Space, however, is not flat in the vicinity of things with mass (i.e. Earth), so you can swim. The math is beyond me to calculate how much an average human could move themselves in Earth's G field, but check out:

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There is no such point where the gravity is incredibly high and having low gravity areas beside it. The gravitational field of the earth (or any object) at outside points decrease with a parabolic path (in graph). And there is no such discontinuity. And hence, such points aren't possible. If the gravity free area is any spaceship , then the radial force component balances the gravity and in such case also, there is no such point. But there can be the point if you can deflect your velocity from 8kph approx. (satellite velocity) by large magnitudes {only in case of satellite}.

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  • $\begingroup$ Akash, your answer does not add much to the existing answers, and because of formatting and wording it is difficult to read. I suggest revising your answer according to the advice in How do I write a good answer. If you need assistance, you are welcome to seek help in chat once you accumulate more reputation points. $\endgroup$ – called2voyage Apr 26 '16 at 15:33

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