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Lets say our Spacecraft is traveling to a remote Galaxy at a constant speed of 1/X of the speed of light.

A brave Astronaut is leaving the spacecraft to a space walk, while not being attached to the spacecraft.

Will the astronaut

  1. hover near the spacecraft at the same speed as it (1/X of speed of light), or
  2. be quickly behind the spacecraft and will watch it disappear in the black horizon?

Is there any difference between such a situation when orbiting the Earth and when being in the deep space?

riorio
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9 Answers9

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As long as neither spacecraft nor the astronaut are accelerating or decelerating, the relative speed of the spacecraft and the astronaut remains the same. So the astronaut will hover near the spacecraft.

The actual velocity is irrelevant here, it's the same principle with every spacewalk: the ISS is moving at about 27,600 km/h, yet the astronauts do not "get left behind" when they exit for a space walk. They, too, move at about 27,600 km/h. They do move at a very slight relative velocity when they move along the spacecraft, though.

Things change if your spacecraft is accelerating or decelerating, though: in this case the astronaut needs to remain attached to the spacecraft to not get lost. As soon as they would let go, their current velocity would remain the same but the spacecraft would continue to change its velocity and the two would get further and further apart.

DarkDust
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  • In the case of ISS, isn't gravity producing acceleration to the station? So wouldn't the astronaut feel that the ISS is moving away from him because the earth pulls it more than the astronaut? – papakias Nov 08 '18 at 14:20
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    @papakias No, because gravity will pull on both the ISS and the astronaut the same. Drag will slow the ISS more rapidly due to surface area, but that effect will take hours or days to make itself obvious. – Saiboogu Nov 08 '18 at 14:28
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    I would elaborate that the gravitational force on ISS is greater than on an astronaut as it has more mass, but also this mass makes it accelerate slower that if the same amount of force was applied to the astronaut. Things basically cancel out, and the gravitational acceleration is the same for everyone. – NikoNyrh Nov 08 '18 at 16:32
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    @Saiboogu, agreed, although if the astronaut was not tethered and even just 1 metre away from the ISS and completed a full orbit, the astronauts orbit would be fractionally different and therefore the ISS would move away very slightly, and each orbit that distance would increase, and now that i think about it the lighter astronaut would be more susceptable to atmospheric drag, even with how thin the atmosphere is at that altitude – Blade Wraith Nov 08 '18 at 16:43
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    I would emphasize that there is no unaccelerated flight -- there's always some mass somewhere (stars, galaxies etc.). The key issue is that the masses are usually so far away that the resulting gravitational field can be considered homogeneous with very little error and thus affects bodies that are "close" to each other (close relative to the closest significant gravitational sources) equally. That would not be the case in low orbit, see https://en.wikipedia.org/wiki/Micro-g_environment#Tidal_and_inertial_acceleration. – Peter - Reinstate Monica Nov 08 '18 at 17:00
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    @BladeWraith, not quite. If the astronaut were 1 meter away, in a homogeneous gravity field, they would co-orbit each other, or, more likely, oscillate in distance to each other. Better visualization: https://www.youtube.com/watch?v=cxNJoaBLLNM -- And, atmospheric drag depends entirely on the relationship to surface area hitting the atmosphere relative to weight... which usually correlates to an object's density. Spacecraft being intentionally light, and usually quite hollow, a spacewalking human is more dense than the ISS, so will experience less atmospheric drag. – Ghedipunk Nov 08 '18 at 20:56
  • If the spacecraft and the astronaut is in an elliptic orbit, the speed is not constant. If the speed is not constant, acceleration could not be zero. But there is zero gravity during the complete orbit. – Uwe Nov 08 '18 at 21:37
  • Since OP asks this kind of question (apparently thinking that the faster you travel to more you stay behind regardless of any forces) saying that ISS moves with a specific speed is a bit risky and misleading. I would point out that it's a speed relative to Earth. You can also mention that the same applies for example to the speed relative to Sun or even Galaxy centre. In my opinion it actually strengthens the answer since otherwise you would have to ask "so to which move does the astronaut start falling behind?" leading to a conclusion we need to look from a relativistic perspective. – Ister Nov 09 '18 at 08:32
  • @Uwe Orbits occur due to acceleration caused by gravity. You cant have an orbit without gravity and you can't have an orbit without acceleration. There absolutely is gravity during a "complete orbit". – Matt Nov 09 '18 at 13:52
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It turns out that outer space is not a perfect vacuum: there are a few hydrogen atoms per cubic centimeter. (reference)

For large X, non-relativistic physics, the astronaut and spacecraft will stay close enough to each other.

Once X gets small, and you approach the speed of light, these hydrogen atoms could slow down your spacecraft. Therefore, to maintain constant speed against this "apparent headwind" you'd have to apply force to the spacecraft, and the space-walker would not be subjected to that same force.

My hypothesis is that the astronaut will slowly be left behind.

TomEberhard
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I feel this sort of question benefits from a series of thought experiments.

Imagine instead that you've got two astronauts, side by side, zipping through space at some constant speed.

They're kind of sweet on each other so they're holding hands. Awwwww.

But then they suffer a cruel change of heart and stop holding hands!

What do you imagine would happen?

Does anything change if one of the astronauts is much fatter than the other?

If we replace the very fat astronaut with a spacecraft, does that change anything?

(I'm asking these questions quasi-rhetorically, for the benefit of the original question-asker. No need to answer me in comments.)

Roger
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    What about the mutual gravity force? Until they hold their hands, they can remain at a fixed distance, but when they let go they slowly start approaching ;) – frarugi87 Nov 08 '18 at 16:06
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    @frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D – BlueCoder Nov 08 '18 at 16:10
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    If the partner of the very fat astronaut is replaced by a black hole the black hole gets to have vay fatty spaghetti for lunch. – Peter - Reinstate Monica Nov 08 '18 at 17:04
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    @PeterA.Schneider If the partner is replaced by a black hole, there are two options: (1) you've just created mass from nowhere, pushing this wweeelllll out of the realm of predictable, or (2) the black hole evaporates instantly, turning the fat astronaut into a Pittsburgh Rare and / or superheated gas. – wizzwizz4 Nov 08 '18 at 22:26
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    Re: gravity, note that the escape velocity of a 150kg astronaut+suit is only 56.06 nanometers / second. Good luck getting your velocity below that. – geometrian Nov 08 '18 at 22:31
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    @imallett, I seem to recall escape velocity is related to radius, so I take it we are assuming a spherical fat astronaut, in, for all practical purposes, a vacuum? – Jon P Nov 09 '18 at 05:44
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    While I see what your didactic goals are, I'd say this answer does not really work. Such a "think for yourself" approach would be worthwhile if the scenario would be closer to the knowledge of OP. But it is quite clear that he is not familiar with, to let more exotic things away, Newton's Laws, so any intuition he has on the two lover-astronauts would either be correct by pure chance (=> no gained knowledge), or incorrect (=> dito). Can you maybe find a thought experiment that is more close to our everyday experiences? – AnoE Nov 09 '18 at 14:17
  • @PeterA.Schneider That depends on the mass, radius, and spin of the new black hole – Gary Nov 09 '18 at 17:01
  • @Gary Good point, although my exclamation should still be in the ballpark. The energy is (independent of spin and radius) always roughly 10^19 J (the equivalent of 110kg w/ suit). IIUC spinning will slow evaporation, but since 100t evaporate in 1s I assume 100kg will stay under that even with some spin. Our fattie BH will just grow some love handles. – Peter - Reinstate Monica Nov 09 '18 at 18:00
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Another way to think is to consider two space walking astronauts; one inside the ship and one outside. Neither is touching the ship, both are moving at essentially the same speed in the same direction. All three pretty much stay together.

However, there could be a teeny tiny amount of acceleration experienced by each. For example, at an extremely high velocity, even the tiny impulse caused by each interstellar proton hitting a body can cause a bit of drag. The "indoor" space walker won't experience it, and so won't be slowed at all, but the ship will, and so will the "outdoor" space walker. It's not clear which one would be affected more, it depends on their cross-sectional areas and masses.

Then there are tidal effects. If there is a distant gravitational source, and there always is, that will accelerate all three the same. But if you are fairly close to a source of gravity, then it is possible that it affects them slightly differently because they will each have a very slightly different distance from the source.

For more on that see answers to Lowest ISS microgravity and for fun see How to get sunburned through the window of a General Products hull?

And before your ship does another neutron-star flyby to accelerate so fast, remember that what humans call UV is not the only thing that gets through a General Products Hull!

uhoh
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  • This is always the coolest way to explain this! – Fattie Nov 09 '18 at 03:22
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    The one inside would need to be in a vacuum for the situation to be equivalent. Otherwise, you have to account for air pressure. – jpmc26 Nov 09 '18 at 04:48
  • @jpmc26 Sounds good. In my mind's eye I pictured them both wearing suits for some reason, you've figured out why! There would be a small (order of part-per-thousand) buoyancy effect if there was air in there. – uhoh Nov 09 '18 at 04:55
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no, conservation of momentum is retained (an object in motion will remain in motion unless something acts upon it)...similar to being in an airplane and throwing a ball up in the air...seems like it should fly to the back of the airplane, but it won't...it'll act just like you were on the ground.

Joseph
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A brave Astronaut is leaving the spacecraft to a space walk, while not being attached to the spacecraft.

Will the astronaut

hover near the spacecraft at the same speed as it (1/X of speed of light), or

be quickly behind the spacecraft and will watch it disappear in the black horizon?

Newton's First Law of Motion ("an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force") means that the astronaut -- who is traveling at the same speed and direction as the ship while inside the ship -- will continue traveling at the same speed and direction as the ship when he steps out of it.

RonJohn
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The correct and complete answer is distributed among many previous posts. I try to condense them here, without attempting to reference all of you guys. All of the below information was provided in the previous answers.

The main point is that

  • neither the ship nor the astronaut tend to brake in empty space because of Newton's law.

Additionally, there are three very weak effects:

  1. Space is not completely empty. This depends on where you are, but there will be some countable amount of atoms (mostly hydrogen) per cubic meter. These take away your velocity, very very slowly. Whether the astronaut's or the ship's velocity decreases faster depends on the ratio of their mass to their cross-sectional area, respectively.
  2. Tidal effects also pull them apart. This is because they are located at slightly different distances to the surrounding sources of gravity. The closer you are to such a source, the stronger is the respective force, hence the astronaut and the ship experience different gravitational pulls.
  3. Mutual gravity pulls them together. Both the spacecraft and the astronaut have mass and hence attract each other.

Whether the astronaut will be able to measure a change of the distance between her or him and the spaceship (during her or his lifetime) depends on the exact initial conditions.

Everyday Astronaut
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Let's tackle this with a slightly different question:

Which falls faster? A bowling ball or a feather?

Now, everyone knows the feather will fall slowly, but that's because the feather has a massive surface area to catch the air around it. Without air resistance they fall at the same rate (see the video below for a most impressive display of that principle)

If an astronaut exits a spacecraft moving at 17,000 mph, they're still moving at a relative 17,000 mph because there's nothing to slow the astronaut down.

Machavity
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    Actually the hammer falls somewhat faster (if the objects fall not together but one after the other) because it pulls the earth towards itself just a tad stronger, so that the two collide a bit earlier. (My 10 year old remarked that when I suggested the feather/hammer experiment. Ouch.) – Peter - Reinstate Monica Nov 08 '18 at 17:06
  • @PeterA.Schneider The difference is negligible. The formula for gravitational attraction (when you account for the mass of the Earth) bears this out (I had a physics teacher make us do that math). Either way, the experiment shows how a vacuum changes things – Machavity Nov 08 '18 at 17:18
  • @PeterA.Schneider wow, your 10 year old kid just did a bit of extra hard science that is usually omitted but should be pointed out (of course with the remark that the difference is negligible as mentioned by Machavity). I really hope you've mentioned to them how great job they did? – Ister Nov 09 '18 at 08:39
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As others have explained, cosmic dust and orbital mechanics aside, the astronaut will cruise along with the ship. However, to make sure we cover all the aces, he'd better check the ship is not rotating before he leaves.

If it is, then while he is inside, he will find himself held to the outer walls by "centrifugal" force (really, it's the walls pushing him round in a circle). Once he exits, that pushing will send him drifting off at a tangent to the rotation. Since the craft will turn under him as he floats away, it will look like he is moving straight out from the door. At this point, a Wilhelm Scream might be appropriate.

Oscar Bravo
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