Assuming a spacecraft is traveling in a constant rate and our Astronaut will exit it to a space walk, will he...
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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
- 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?
Is there any difference between such a situation when orbiting the Earth and when being in the deep space?
spacecraft interplanetary spacewalk
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up vote
20
down vote
favorite
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
- 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?
Is there any difference between such a situation when orbiting the Earth and when being in the deep space?
spacecraft interplanetary spacewalk
New contributor
11
Is there a reason you think this would be different than a typical space-walk in Earth orbit?
– JPhi1618
2 days ago
8
Actually - he will be left behind. Jumping in to interstellar gas at near light speed, is just like a parachutist jumping out of an aircraft in to air. At high enough relativistic speeds the interstellar gas has "thickness", just as air does for an aircraft/parachutist. Only @TomEberhard 's answer is correct.
– Fattie
2 days ago
4
OP, note that in many scifi (and real prospective) scenarios, such interstellar spaceships would be continuously boosted. You continuously fire a ramjet or solar sail or some such half way, then turn around and fire it the other way for the other half of the journey. In this case the spacewalker would certainly fall behind. You'd have to "hang on" to the outside all the time, it would be "pulling away" from you.
– Fattie
yesterday
5
@Fattie (1) OP said constant speed so forget acceleration. (2) Speed is ${1over{x}}c$ so if $x$ is extremely large, forget relativistic collisions with cosmic dust. I walk to the shops at ${1over{x}}c$, just that $x=10^8$.
– Oscar Bravo
yesterday
5
Even at 0.1c, hitting a 1mg grain of dust is equivalent to detonating 250kg of TNT...
– Oscar Bravo
yesterday
|
show 7 more comments
up vote
20
down vote
favorite
up vote
20
down vote
favorite
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
- 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?
Is there any difference between such a situation when orbiting the Earth and when being in the deep space?
spacecraft interplanetary spacewalk
New contributor
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
- 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?
Is there any difference between such a situation when orbiting the Earth and when being in the deep space?
spacecraft interplanetary spacewalk
spacecraft interplanetary spacewalk
New contributor
New contributor
edited 2 days ago
RonJohn
239110
239110
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asked 2 days ago
riorio
20715
20715
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New contributor
11
Is there a reason you think this would be different than a typical space-walk in Earth orbit?
– JPhi1618
2 days ago
8
Actually - he will be left behind. Jumping in to interstellar gas at near light speed, is just like a parachutist jumping out of an aircraft in to air. At high enough relativistic speeds the interstellar gas has "thickness", just as air does for an aircraft/parachutist. Only @TomEberhard 's answer is correct.
– Fattie
2 days ago
4
OP, note that in many scifi (and real prospective) scenarios, such interstellar spaceships would be continuously boosted. You continuously fire a ramjet or solar sail or some such half way, then turn around and fire it the other way for the other half of the journey. In this case the spacewalker would certainly fall behind. You'd have to "hang on" to the outside all the time, it would be "pulling away" from you.
– Fattie
yesterday
5
@Fattie (1) OP said constant speed so forget acceleration. (2) Speed is ${1over{x}}c$ so if $x$ is extremely large, forget relativistic collisions with cosmic dust. I walk to the shops at ${1over{x}}c$, just that $x=10^8$.
– Oscar Bravo
yesterday
5
Even at 0.1c, hitting a 1mg grain of dust is equivalent to detonating 250kg of TNT...
– Oscar Bravo
yesterday
|
show 7 more comments
11
Is there a reason you think this would be different than a typical space-walk in Earth orbit?
– JPhi1618
2 days ago
8
Actually - he will be left behind. Jumping in to interstellar gas at near light speed, is just like a parachutist jumping out of an aircraft in to air. At high enough relativistic speeds the interstellar gas has "thickness", just as air does for an aircraft/parachutist. Only @TomEberhard 's answer is correct.
– Fattie
2 days ago
4
OP, note that in many scifi (and real prospective) scenarios, such interstellar spaceships would be continuously boosted. You continuously fire a ramjet or solar sail or some such half way, then turn around and fire it the other way for the other half of the journey. In this case the spacewalker would certainly fall behind. You'd have to "hang on" to the outside all the time, it would be "pulling away" from you.
– Fattie
yesterday
5
@Fattie (1) OP said constant speed so forget acceleration. (2) Speed is ${1over{x}}c$ so if $x$ is extremely large, forget relativistic collisions with cosmic dust. I walk to the shops at ${1over{x}}c$, just that $x=10^8$.
– Oscar Bravo
yesterday
5
Even at 0.1c, hitting a 1mg grain of dust is equivalent to detonating 250kg of TNT...
– Oscar Bravo
yesterday
11
11
Is there a reason you think this would be different than a typical space-walk in Earth orbit?
– JPhi1618
2 days ago
Is there a reason you think this would be different than a typical space-walk in Earth orbit?
– JPhi1618
2 days ago
8
8
Actually - he will be left behind. Jumping in to interstellar gas at near light speed, is just like a parachutist jumping out of an aircraft in to air. At high enough relativistic speeds the interstellar gas has "thickness", just as air does for an aircraft/parachutist. Only @TomEberhard 's answer is correct.
– Fattie
2 days ago
Actually - he will be left behind. Jumping in to interstellar gas at near light speed, is just like a parachutist jumping out of an aircraft in to air. At high enough relativistic speeds the interstellar gas has "thickness", just as air does for an aircraft/parachutist. Only @TomEberhard 's answer is correct.
– Fattie
2 days ago
4
4
OP, note that in many scifi (and real prospective) scenarios, such interstellar spaceships would be continuously boosted. You continuously fire a ramjet or solar sail or some such half way, then turn around and fire it the other way for the other half of the journey. In this case the spacewalker would certainly fall behind. You'd have to "hang on" to the outside all the time, it would be "pulling away" from you.
– Fattie
yesterday
OP, note that in many scifi (and real prospective) scenarios, such interstellar spaceships would be continuously boosted. You continuously fire a ramjet or solar sail or some such half way, then turn around and fire it the other way for the other half of the journey. In this case the spacewalker would certainly fall behind. You'd have to "hang on" to the outside all the time, it would be "pulling away" from you.
– Fattie
yesterday
5
5
@Fattie (1) OP said constant speed so forget acceleration. (2) Speed is ${1over{x}}c$ so if $x$ is extremely large, forget relativistic collisions with cosmic dust. I walk to the shops at ${1over{x}}c$, just that $x=10^8$.
– Oscar Bravo
yesterday
@Fattie (1) OP said constant speed so forget acceleration. (2) Speed is ${1over{x}}c$ so if $x$ is extremely large, forget relativistic collisions with cosmic dust. I walk to the shops at ${1over{x}}c$, just that $x=10^8$.
– Oscar Bravo
yesterday
5
5
Even at 0.1c, hitting a 1mg grain of dust is equivalent to detonating 250kg of TNT...
– Oscar Bravo
yesterday
Even at 0.1c, hitting a 1mg grain of dust is equivalent to detonating 250kg of TNT...
– Oscar Bravo
yesterday
|
show 7 more comments
9 Answers
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up vote
49
down vote
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.
13
@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
2 days ago
11
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
2 days ago
2
@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
2 days ago
6
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
1
@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: 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
2 days ago
|
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29
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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.
New contributor
1
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
2
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
2
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
4
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
1
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
|
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20
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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.)
8
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
2 days ago
4
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
3
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 A. Schneider
2 days ago
3
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.
– imallett
2 days ago
2
@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
yesterday
|
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9
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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!
This is always the coolest way to explain this!
– Fattie
2 days ago
1
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
yesterday
@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
yesterday
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5
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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.
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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.
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
add a comment |
up vote
3
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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.
2
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 A. Schneider
2 days ago
@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
2 days ago
That's a great link, @Machavity, thanks.
– Fattie
yesterday
@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
yesterday
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3
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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:
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.
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.
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.
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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.
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
add a comment |
9 Answers
9
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9 Answers
9
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oldest
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active
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active
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up vote
49
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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.
13
@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
2 days ago
11
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
2 days ago
2
@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
2 days ago
6
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
1
@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: 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
2 days ago
|
show 4 more comments
up vote
49
down vote
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.
13
@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
2 days ago
11
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
2 days ago
2
@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
2 days ago
6
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
1
@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: 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
2 days ago
|
show 4 more comments
up vote
49
down vote
up vote
49
down vote
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.
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.
edited 2 days ago
answered 2 days ago
DarkDust
5,09412044
5,09412044
13
@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
2 days ago
11
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
2 days ago
2
@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
2 days ago
6
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
1
@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: 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
2 days ago
|
show 4 more comments
13
@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
2 days ago
11
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
2 days ago
2
@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
2 days ago
6
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
1
@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: 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
2 days ago
13
13
@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
2 days ago
@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
2 days ago
11
11
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
2 days ago
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
2 days ago
2
2
@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
2 days ago
@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
2 days ago
6
6
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
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 en.wikipedia.org/wiki/….
– Peter A. Schneider
2 days ago
1
1
@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: 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
2 days ago
@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: 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
2 days ago
|
show 4 more comments
up vote
29
down vote
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.
New contributor
1
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
2
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
2
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
4
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
1
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
|
show 2 more comments
up vote
29
down vote
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.
New contributor
1
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
2
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
2
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
4
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
1
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
|
show 2 more comments
up vote
29
down vote
up vote
29
down vote
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.
New contributor
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.
New contributor
edited yesterday
New contributor
answered 2 days ago
TomEberhard
36114
36114
New contributor
New contributor
1
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
2
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
2
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
4
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
1
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
|
show 2 more comments
1
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
2
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
2
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
4
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
1
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
1
1
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
"Disclaimers: not a physicist" but your physics is absolutely spot-on! See this comment and also this answer
– uhoh
2 days ago
2
2
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
Why would you post an answer in a comment anyway? Please remove that rather bizarre note from the beginning of the post. Also, your reference says, "You have followed a link to a page that is not yet available for public viewing on the New World Encyclopedia." Question: galaxies and the space between them likely have different densities of matter; which one does your stat refer to?
– jpmc26
yesterday
2
2
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
In practice, a lot hinges on putting numbers on "slowly" here.
– gerrit
yesterday
4
4
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
And by "slow down your spacecraft" you surely mean "rip it to shreds"
– Lightness Races in Orbit
yesterday
1
1
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
@qazwsx IRTA "irritated", which may be true if there's time
– Lightness Races in Orbit
7 hours ago
|
show 2 more comments
up vote
20
down vote
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.)
8
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
2 days ago
4
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
3
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 A. Schneider
2 days ago
3
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.
– imallett
2 days ago
2
@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
yesterday
|
show 5 more comments
up vote
20
down vote
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.)
8
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
2 days ago
4
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
3
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 A. Schneider
2 days ago
3
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.
– imallett
2 days ago
2
@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
yesterday
|
show 5 more comments
up vote
20
down vote
up vote
20
down vote
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.)
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.)
answered 2 days ago
Roger
74916
74916
8
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
2 days ago
4
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
3
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 A. Schneider
2 days ago
3
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.
– imallett
2 days ago
2
@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
yesterday
|
show 5 more comments
8
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
2 days ago
4
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
3
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 A. Schneider
2 days ago
3
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.
– imallett
2 days ago
2
@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
yesterday
8
8
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
2 days ago
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
2 days ago
4
4
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
@frarugi87 how sweet! They can't avoid being nearer and nearer to each other...Awww :D
– BlueCoder
2 days ago
3
3
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 A. Schneider
2 days ago
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 A. Schneider
2 days ago
3
3
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.
– imallett
2 days ago
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.
– imallett
2 days ago
2
2
@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
yesterday
@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
yesterday
|
show 5 more comments
up vote
9
down vote
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!
This is always the coolest way to explain this!
– Fattie
2 days ago
1
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
yesterday
@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
yesterday
add a comment |
up vote
9
down vote
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!
This is always the coolest way to explain this!
– Fattie
2 days ago
1
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
yesterday
@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
yesterday
add a comment |
up vote
9
down vote
up vote
9
down vote
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!
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!
edited 2 days ago
answered 2 days ago
uhoh
31.8k15109392
31.8k15109392
This is always the coolest way to explain this!
– Fattie
2 days ago
1
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
yesterday
@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
yesterday
add a comment |
This is always the coolest way to explain this!
– Fattie
2 days ago
1
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
yesterday
@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
yesterday
This is always the coolest way to explain this!
– Fattie
2 days ago
This is always the coolest way to explain this!
– Fattie
2 days ago
1
1
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
yesterday
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
yesterday
@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
yesterday
@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
yesterday
add a comment |
up vote
5
down vote
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.
New contributor
add a comment |
up vote
5
down vote
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.
New contributor
add a comment |
up vote
5
down vote
up vote
5
down vote
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.
New contributor
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.
New contributor
New contributor
answered 2 days ago
Joseph
511
511
New contributor
New contributor
add a comment |
add a comment |
up vote
4
down vote
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.
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
add a comment |
up vote
4
down vote
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.
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
add a comment |
up vote
4
down vote
up vote
4
down vote
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.
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.
answered 2 days ago
RonJohn
239110
239110
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
add a comment |
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
Thanks for being the first one to mention Newton. OP's second case (does anything change if in Earth orbit) could be worth a further paragraph...
– AnoE
yesterday
add a comment |
up vote
3
down vote
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.
2
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 A. Schneider
2 days ago
@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
2 days ago
That's a great link, @Machavity, thanks.
– Fattie
yesterday
@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
yesterday
add a comment |
up vote
3
down vote
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.
2
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 A. Schneider
2 days ago
@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
2 days ago
That's a great link, @Machavity, thanks.
– Fattie
yesterday
@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
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
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.
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.
answered 2 days ago
Machavity
2,0911734
2,0911734
2
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 A. Schneider
2 days ago
@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
2 days ago
That's a great link, @Machavity, thanks.
– Fattie
yesterday
@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
yesterday
add a comment |
2
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 A. Schneider
2 days ago
@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
2 days ago
That's a great link, @Machavity, thanks.
– Fattie
yesterday
@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
yesterday
2
2
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 A. Schneider
2 days ago
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 A. Schneider
2 days ago
@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
2 days ago
@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
2 days ago
That's a great link, @Machavity, thanks.
– Fattie
yesterday
That's a great link, @Machavity, thanks.
– Fattie
yesterday
@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
yesterday
@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
yesterday
add a comment |
up vote
3
down vote
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:
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.
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.
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.
add a comment |
up vote
3
down vote
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:
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.
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.
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.
add a comment |
up vote
3
down vote
up vote
3
down vote
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:
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.
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.
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.
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:
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.
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.
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.
answered yesterday
rehctawrats
1,171522
1,171522
add a comment |
add a comment |
up vote
1
down vote
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.
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
add a comment |
up vote
1
down vote
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.
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
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.
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.
answered yesterday
Oscar Bravo
20124
20124
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
add a comment |
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
My understanding is that zero initial relative velocity is assumed.
– rehctawrats
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
@rehctawrats Even in a rotating ship, you could argue that the relative velocity between spaceman and spaceship is zero - angular velocity, that is! As for is assumed - on SE, nothing can be assumed...
– Oscar Bravo
yesterday
add a comment |
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11
Is there a reason you think this would be different than a typical space-walk in Earth orbit?
– JPhi1618
2 days ago
8
Actually - he will be left behind. Jumping in to interstellar gas at near light speed, is just like a parachutist jumping out of an aircraft in to air. At high enough relativistic speeds the interstellar gas has "thickness", just as air does for an aircraft/parachutist. Only @TomEberhard 's answer is correct.
– Fattie
2 days ago
4
OP, note that in many scifi (and real prospective) scenarios, such interstellar spaceships would be continuously boosted. You continuously fire a ramjet or solar sail or some such half way, then turn around and fire it the other way for the other half of the journey. In this case the spacewalker would certainly fall behind. You'd have to "hang on" to the outside all the time, it would be "pulling away" from you.
– Fattie
yesterday
5
@Fattie (1) OP said constant speed so forget acceleration. (2) Speed is ${1over{x}}c$ so if $x$ is extremely large, forget relativistic collisions with cosmic dust. I walk to the shops at ${1over{x}}c$, just that $x=10^8$.
– Oscar Bravo
yesterday
5
Even at 0.1c, hitting a 1mg grain of dust is equivalent to detonating 250kg of TNT...
– Oscar Bravo
yesterday