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anonguest

Random Scientific Musing of the Day

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One for the physics bods here.  Just one of those random no-reason oddball thoughts that go through ones mind from time to time, that you just gotta know the answer to.........

For a typical re-entry trajectory, at what point in a re-entry back to Earth would astronauts stop being weightless and fall to the floor (if they weren't already strapped into their seats, etc).

And YES, I have surfed Google for an answer but had no luck so far.

Anyone?

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There's no sudden point. In orbit the astronaut and the capsule are doing the same thing - take away the capsule and the astronaut would carry on in the same trajectory. Gravity is accelerating them both in the same direction, no sensation of weight. You feel weight on Earth because whilst gravity is trying to accelerate you the ground gets in the way and pushes back. The astronaut will get some sensation of weight as soon as a force acts on the capsule that isn't gravity alone. This will be a brief bit from the engines when they fire to de-orbit, and it'll start building up as atmospheric friction starts to act on the capsule.

Gravity at the altitude of the ISS is barely any different from on the surface, if you built a very tall tower up to that height and sat down on it it would feel like sitting on the Earth.

For all spaceflight curiosities I recommend playing Kerbal Space Program. Even though it'll mostly teach you how easy it is to make rockets with an alarming tendency to go out of control, or explode, or both.

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Thanks.  Yes, I realise there is no specific point when they suddenly become weightless. I also realise that the answer is also dependent (at least in part) on the specific trajectory that the re-entry vehicle takes?

That latter point in mind however raises the question as to whether the resumption to de facto full gravity (i.e. their falling to the floor IF they were not strapped in) would in fact be quite late into the re-entry process - and not very soon after undocking from the ISS.

Indeed in the case of the older style plummeting to Earth type descents (used by Russians and U.S in the Apollo era) surely tose astronaughts would be almost in free fall for much of the time??  Like the proverbial Einstein thought experiment of a man in an cut cables elevator hurtling towards ground.

Whereas in a more shallow trajectory descent, like the gliding path of the Shuttle, they would be firmly back on the floor very early on after dropping out of orbit - as their velocity drops below Earth escape velocity?

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Without knowing the exact descent profiles (and even then I probably wouldn't know how to interpret them!) I couldn't tell you for sure. My assumption is that forces >1g would be felt during re-entry as the slowdown from atmospheric friction is the dominant one in every spacecraft built. By the time those have settled down the speed would be so far below orbital speed that things would feel pretty normal. The descent profiles couldn't have been massively steeper for Apollo I think, otherwise it wouldn't lose enough speed before hitting the ground. All guesses there. I've not tried it but graphs of the descent profiles, including G-forces experienced, might be easily Googled.

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The weightlessness is an effect of being in a freefall trajectory - i.e. no air resistance.

Once there is atmospheric drag, then the braking force will be felt by the spacecraft - and that braking force will have to be transferred to the astronauts as a feeling of weight. 

Once a constant velocity is reached, then normal earth gravity would be felt. At some point, however, there would be a need to slow down to reach a stop, and therefore for some of the descent, there will be an effective weight greater than that at rest on earth.

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6 minutes ago, ChumpusRex said:

The weightlessness is an effect of being in a freefall trajectory - i.e. no air resistance.

Once there is atmospheric drag, then the braking force will be felt by the spacecraft - and that braking force will have to be transferred to the astronauts as a feeling of weight. 

Once a constant velocity is reached, then normal earth gravity would be felt. At some point, however, there would be a need to slow down to reach a stop, and therefore for some of the descent, there will be an effective weight greater than that at rest on earth.

Agree and understand about the air resistance issue.

Presumably, however, for the initial stages of re-entry the air resistance is negligible and so the spacecraft is in de facto free fall for at least a short while? (assuming the hypothetical case of a perfect vertical descent trajectory, as in a falling elevator). For that (short) period any astronaut not strapped in for ride home would remain floating about? But smoothly descend to the floor as the free fall effect disappeared due to air resistance kicking in?

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My first attempt was wrong, so this is my second (overwriting my first. Could still be wrong)!

I think the way to think of it is that when you are in free fall either around the earth or falling towards the earth, you will feel weightless. It is only once there is an upward component of force, say from an airfoil or parachute, that you will start to feel your weight.

If you are falling at a constant velocity inside the craft towards earth (no rotational component, could be terminal velocity), you will feel your full weight because a constant velocity is an inertial reference frame.

If in orbit, you will have to decelerate to start the decent towards earth, so you would feel a deceleration, but you'd only fall to the floor once you hit atmosphere because at that point you'd be effectively gliding and there would be a small upward force. You'd be decelerating quite a bit also, so you'd feel that force.

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Here's some information on G forces for Apollo re-entry. Depsite my earlier assumption that there wouldn't be all that much difference between different vehicles (if starting from the same orbit) the Shuttle apparently has a much gentler one than other spacecraft, for pretty much the reasons the OP gave. Not found a profile of that yet.

https://history.nasa.gov/SP-368/s2ch5.htm#135

 

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Thanks all for all that - and helping scratch that scientific trivia 'itch'.  😉

Alas, having now stirred up the dormant beast of nagging curiosity trivia my mind has now strayed into other long standing physics things that, like others, I am familiar with but never quite got 100% comfortable with/completely got my head around.

like the Cosmic Background Radiation from the big Bang...... IF the photons have lost energy (so now have longer wavelength than originally) then where has that energy gone?

 

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On 20/09/2018 at 19:40, Riedquat said:

Here's some information on G forces for Apollo re-entry. Depsite my earlier assumption that there wouldn't be all that much difference between different vehicles (if starting from the same orbit) the Shuttle apparently has a much gentler one than other spacecraft, for pretty much the reasons the OP gave. Not found a profile of that yet.

https://history.nasa.gov/SP-368/s2ch5.htm#135

 

Interesting that. I was originally thinking it could be a bit of a roller-coaster ride during reentry.

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On 22/09/2018 at 09:24, anonguest said:

like the Cosmic Background Radiation from the big Bang...... IF the photons have lost energy (so now have longer wavelength than originally) then where has that energy gone?

Dissipated into other forms of energy such as light emitted (including non-visible light)? As you know the sun has a time limit, albeit after billions of years of emitting heat and light.

What if you fire a bullet in space but there are no gravitational forces form nearby objects whatsoever? (planets etc.) Does the bullet slow down eventually? If so where has that kinetic energy gone?

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Thanks again to all those chipping in with answers, but........not so fast!

The issue of air resistance (on the falling/descending reentry vehicle) was mentioned as a contributing factor to the experience of weightlessness ceasing upon reentry, BUT.......

What about those modified airliners that are used to repeatedly create (albeit very temporarily!) weightlessness by virtue of the special flight trajectory they take?!  The air resistance on those aircraft plays no part in the considerations??

Thus the trajectory is solely the key?

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2 hours ago, anonguest said:

Thanks again to all those chipping in with answers, but........not so fast!

The issue of air resistance (on the falling/descending reentry vehicle) was mentioned as a contributing factor to the experience of weightlessness ceasing upon reentry, BUT.......

What about those modified airliners that are used to repeatedly create (albeit very temporarily!) weightlessness by virtue of the special flight trajectory they take?!  The air resistance on those aircraft plays no part in the considerations??

Thus the trajectory is solely the key?

The planes use their engines to counteract the air resistance. With the two cancelled out like that you feel weightless. The trajectory the pilot flies the plane is the one it would follow (given the same starting velocity) if it had no engines and was in a vacuum. It's only for a short time because, not being in orbit, that trajectory intersects the ground and it's a good idea to change from it before that happens...

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8 minutes ago, Riedquat said:

The planes use their engines to counteract the air resistance. With the two cancelled out like that you feel weightless. The trajectory the pilot flies the plane is the one it would follow (given the same starting velocity) if it had no engines and was in a vacuum. It's only for a short time because, not being in orbit, that trajectory intersects the ground and it's a good idea to change from it before that happens...

Hmmm........Still not conceptually clear on this.  There is no point during the flight when air resistance ON the plane is net zero? It is, after all, constntly flying through it at 500+ mph.

I'm gonna have to try and remember how to draw force vector diagram for this.

I never thought I would ever have to call upon my old A-level physics book ever again.  LOL .   Gotta go up to the loft to find it. Lord knows where it is.

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8 minutes ago, Riedquat said:

The planes use their engines to counteract the air resistance. With the two cancelled out like that you feel weightless. The trajectory the pilot flies the plane is the one it would follow (given the same starting velocity) if it had no engines and was in a vacuum. It's only for a short time because, not being in orbit, that trajectory intersects the ground and it's a good idea to change from it before that happens... 

I thought the plane use the engines to move the plane to match the speed at which gravity would pull it down, therefore mimicking conditions of 0 gravity, while air is a resistance factor to overcome in order to achieve that. If doing the same towards a planet with 0 atmosphere there would be no air resistance. Maybe I've got this wrong and it's based on being far enough away from the gravitation pull of earth?

Now......how about putting a bouncy castle in there too?

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9 minutes ago, Arpeggio said:

I thought the plane use the engines to move the plane to match the speed at which gravity would pull it down, therefore mimicking conditions of 0 gravity, while air is a resistance factor to overcome in order to achieve that. If doing the same towards a planet with 0 atmosphere there would be no air resistance. Maybe I've got this wrong and it's based on being far enough away from the gravitation pull of earth?

It's definitely in with the gravitational pull of the Earth - remember that's what keeps the Moon in orbit, and at the level things like the ISS orbit the strength of gravity is almost the same as at the surface. Planes obviously fly much lower than that.

You feel weightless when gravity is the only force acting on both you and your surroundings. Imagine a set of traditional scales, the sort where you put weights on one side and the thing you want to weight on the other. Sitting on the ground the scales tip when not in balance. If you held them horizontal and dropped them they wouldn't tip until hitting the floor, even if the masses on either side were quite different - hence weightless. Throw them in an arc and the same thing would happen, more or less. I say more or less because air resistance would play a part, although for a set of scales and the speed you could throw them at it wouldn't be noticeable unless they whole setup was made of paper. The plane is doing the same thing, only on that scale the air resistance is sufficient to deviate it from the arc it would otherwise follow, so instead it uses its engines to keep to that trajectory.

Being in orbit is the same thing as that arc, except that because the Earth is round by it curves out of the way and you never hit it. Hence being in orbit being described as falling towards the ground and continually missing.

The plane is doing the same thing, flying on the same sort of trajectory it would on the moon if it turned its engines off (and if it could fly on the moon, which of course it couldn't).

If the plane was matching the force of gravity then it's simply flying at a constant altitude.

I'm sure this would be easier if I drew some diagrams, that'll have to wait until I'm not at work.

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37 minutes ago, anonguest said:

Hmmm........Still not conceptually clear on this.  There is no point during the flight when air resistance ON the plane is net zero? It is, after all, constntly flying through it at 500+ mph.

The forces acting on the plane are air resistance, gravity, thrust from the engines, and lift from the wings. The trick is to use the lift and thrust to cancel out the air resistance, leaving just gravity. Then because the plane is falling at the same rate (or slowing down as it moves upwards) as everything inside it the things inside would appear to float inside the plane, at least until it has to deviate from that trajectory to avoid crashing. The air resistance is certainly still having its effect on the plane.

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12 minutes ago, Riedquat said:

The forces acting on the plane are air resistance, gravity, thrust from the engines, and lift from the wings. The trick is to use the lift and thrust to cancel out the air resistance, leaving just gravity. Then because the plane is falling at the same rate (or slowing down as it moves upwards) as everything inside it the things inside would appear to float inside the plane, at least until it has to deviate from that trajectory to avoid crashing. The air resistance is certainly still having its effect on the plane.

Okkkkkkkay.......  Off to get a pencil and paper now.......

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3 hours ago, Riedquat said:

The forces acting on the plane are air resistance, gravity, thrust from the engines, and lift from the wings. The trick is to use the lift and thrust to cancel out the air resistance, leaving just gravity. Then because the plane is falling at the same rate (or slowing down as it moves upwards) as everything inside it the things inside would appear to float inside the plane, at least until it has to deviate from that trajectory to avoid crashing. The air resistance is certainly still having its effect on the plane.

Yes, you have to be accelerating towards earth at 1g (relative to earth) to be weightless, so the plane has to overcome the air resistance and be accelerating toward the earth. Air resistance force is proportional to the square of the velocity (I think, but you can check it out) if you want to do some calculations.

Remember if the plane is traveling at a constant velocity in any direction(up, down, left, right, etc.), you'd feel 1g. On a rotational trajectory there would be a centrifugal force.

I've seen an experiment to test out movement in moon gravity in one of those planes (I think it was with Brian Cox), which would require an acceleration of less than 1g towards earth (about 0.8g I guess). You can probably work this out too.

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I've thought of a ridiculous analogy to the astronaut training plane.

Imagine you're bouncing along on a space hopper (on trampolines, let's keep you in the air for as long as possible). When it's in the air you aren't putting any pressure on it, it's not supporting you, you could wave is around, move it to just in front of you and let go and it would stay just in front of you until you hit the ground and it bounces away. As far as you're concerned you're weightless there. If we make the example even more implausible and put you inside the space hopper, so you couldn't even feel any air movement, whilst it's off the ground you there would be no way of telling if you're in orbit, in deep space, or a couple of feet off the ground, although in the latter case you'd find out pretty quickly when you hit it.

The air resistance effects on the space hopper are pretty negligable for that example, so stick a big parachute on it (don't worry, it'll get even sillier in a minute). That slows it down whilst you carry on inside it, until you hit the side. But put some engines on it that counteract the drag of the parachute and you're back to being apparently weightless. If you were on the Moon, so no air outside, there would be no drag and no need of the engines.

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Having tackled one (relatively basic) interesting scientific musing, I thought I would invite some of you (and newcomers too!) to chip in on this little one.

The Big Bang.  No need to explain it, BUT.....here's a novice/naive conceptual question that I have not yet found an adequate answer for (in layman's terms at least) anywhere on the internet......

IF, in the very earliest moments of the universe's existence it's density (albeit as energy), was de facto infinite then.....wouldn't that in effect make it just as a Black Hole?  In which case how did anything 'escape' from it?

Answers on a postcard to HPC Off-Topic board.  😉

 

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On 15/10/2018 at 19:24, Riedquat said:

My answer to this one is - damned if I know.

And I suspect that, at root, is the same for all cosmologists?!  There does seem to a lot of assumption in the whole very beginnings of things?

I'm not querying the stages of the subsequent evolution, based on properties of the various fundamental forces etc, but rather how could it all pop into existence at all?  IF, as in effect is claimed(?), there was a state of infinite density then what would drive it to not be at infinite density?

I recall a lecture given by an astronomer once who quipped, in jest, something to the effect that "the difference between the study of the solar system and study of cosmology is that we actually know some things about the solar system"

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I suppose it boils down to "no-one's come up with a better idea that explains what we see". There's also probably a bit of "can't say anything about infinite density, that's just the direction the bits we have modelled seem to be pointing in."

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