a feather?
is it possible to explain, in simply layman terms, why everything burns up on reentry? as opposed to dropping something from an aeroplane.
Clearly everything doesn’t burn up on reentry, since many astronauts have reentered the atmosphere without burning to death in a hellish blaze.[sup]1[/sup]
The reason things get hot when entering the atmosphere is not because they are entering the atmosphere, but because they are rubbing against it very fast. In order to achieve orbit, a spacecraft must accelerate to around seven or eight kilometers per second. Traveling that fast through the atmosphere generates a huge amount of friction, which makes heat.
By comparison, your fast jet airplane goes a piddling 0.2 kilometers per second or so. It also generates friction against the atmosphere, but not a lot of it.
[sup]1. However, some astronauts have burned to death in a hellish blaze without leaving the atmosphere at all.[/sup]
One way to look at the problem is with physics. An object in Earth orbit, or moving through the solar system, is going to have a great deal of kinetic energy. Before it can be at rest on the Earth’s surface, it must get rid of that energy. It does that by using friction with the Earth’s atmosphere to convert kinetic energy to heat. That same object, dropped from an airplane, is going to have much less energy that has to be dissipated before it can be at rest.
so if i gently throw a basketball from a standing/not in orbit/unmoving space shuttle towards the Earth, it might survive the trip to bounce on land?
The pillow still seems like a good question. A pillow has a pretty small amount of momentum compared with the amount of drag it has, so while it is up in the thin outer atmosphere, maybe the relatively small amount of atmospheric friction would be enough to decelerate it gradually without it experiencing any extreme temperatures. If it then slows down enough that it’s falling more or less straight down, it becomes a question of whether its terminal velocity once it gets into the denser atmosphere is enough to burn it up.
It could be that it’s no contest and a pillow may as well be a hunk of wood at those velocities, but I would think that given an object of low enough density relative to drag, it would be capable of slowing itself down in freefall from orbit without burning up. Maybe some real live physics dudes can weigh in on this idea.
Maybe this could be one of those scientific experiments they perform up in the space station.
Maybe someone with more credentials than me can correct me, buy my understanding of the heating during reentry is not as much as caused by the friction of the air molecules, but by the compression of the gases around the object.
I venture the WAG that a pillow is not, structuraly speaking, up to the task to compress anything; so if it´s going fast enough to be able to cause a shockwave in front of it, that in turn would heat up, that in turn will burn it up… the thing would have been shreded to bits before any of that could take place.
The latest New Scientist has an article on the future extreme sport of Spacediving. This involves lifting someone to near space (80km or so), and then skydiving back to earth. You need a pressure suit, and some thermal management (you end up very cold, then hot, then cool).
The point is, if you are not in orbit to start with but falling straight down, the atmospheric density increases pretty slowly, so your falling speed is self-managed and heat increases are low and manageable.
If you go higher, your speed increase outside of the atmosphere will give you too much energy to dump as you hit the atmosphere, and you would need heat shielding. Likewise for things in orbit - a re-entering spacecraft does not significantly decelerate to return to earth (it cannot carry enough fuel to do that). All it does is change the shape of its orbit from a near circle to an ellipse, whose closest approach passes through the atmosphere. The significant braking force comes from friction and the craft dumps orbital energy in the form of heat.
All this will be important as a way of rescuing astronauts and space tourists if a craft failure prevents re-entry.
So a pillow may survive reentry if dropped on the edge of the atmosphere from a suborbital flight path. It would not if dropped from an orbital path that intersected the atmosphere. If dropped from an orbital path that does not intersect the atmosphere, it will not reenter at all.
Si
This question has me wondering now too. Can’t the Space Shuttle just go into a geo-stationary orbit and then fire a rocket to move it straight downward and slowly descend into the earth’s atmosphere. At that point it should be able to glide right down without all that messy heat to deal with.
What am I missing here?
Down isn’t down, it’s actually “back”. You’re circling around the planet at X thousand miles per hour, so to go straight down would require braking to 0 miles per hour–something which is more feasible to leave to the atmosphere as opposed to taking up some more booster fuel to stop yourself in space. But even then you’ll be in free fall without a terminal velocity for quite a while. I couldn’t personally say how fast that is.
But the key point is that there’s a difference between dropping a pillow in mid-air, and accelerating a pillow to X thousand miles per hour and then releasing. If you strapped the pillow to the front of an airplane that could travel at satellite orbital speed one mile off the surface of the ocean with a pillow strapped to the nose, the pillow would destroy itself long before you even reached a speed at which it would burst into flame.
Well it depends what you mean by ‘fire a rocket’. Are you talking about a quick boost to start moving it downwards? Or are you talking about a long, sustained, fuel-guzzling burn to keep it moving directly downwards, by chucking out all that angular momentum?
If I’m in a shuttle in geostationary orbit and I throw a rock straight down, it’s not going to go straight down, it’s going to go down and forward, that is, in the direction I’m going in orbit.
ETA that was to DJMotorbike, and Sage Rat probably did a better job anyway!
But in this case, “forward” is stationary, relative to the earth, isn’t it?
Is there any real difference between these? I thought that the heat of gas compression is because of friction. Gas molecules try to move, but because of the compression they bump into other gas molecules more frequently that they would have, so, voila! Friction!
(PS: Note that the gas around a reentering space shuttle IS air. Prior to the shuttle, that air would have also contained some molecules which had been part of the heat shield, but even then, it was mostly air.)
The Shuttle cannot achieve geostationary orbit for a start - it does not have enough fuel to get there from launch, and could never launch with that amount of fuel. And then it would need the same amount of fuel while in orbit to stop (counter the tangential velocity of geostationary orbit = 11000km/h) to fall straight down to earth.
No. Geostationary orbits are just orbits that happen to keep the orbiting body directly over the same point of the rotating earths surface (a distance of 35000km, a tangential speed of 11000km/h). Relative to the earths centre of mass (the only point that actually matters at that distance), the satellite is travelling at 11k km/h at 90 degrees to the vector between the object and the CoM. No special rules apply. The earth could stop rotating and nothing would change. And to change orbit takes energy - to speed up or slow down into the new orbital height.
Thought question: If I throw something down from a geostationary satellite, and it did just keep going down, what is holding the satellite in orbit? What is the difference between the dropped object and the satellite?
Si
I accept that the thing is moving very fast relative to the centre of gravity of the earth - but still, the earth’s surface is stationery, relative to the object. The object’s speed is, I presume, an extension of the earth’s rotation.
Re. the thought experiment: if the earth wasn’t rotating, and had no atmosphere, and I was in orbit, the object I propelled directly vertically downwards would continue at the same rotational velocity as me, wouldn’t it, until it reached the ground? Or would it get flung into a different orbitational speed?
Re: geostationary orbits, you also have to remember conservation of angular momentum. Imagine the body in orbit as attached to the center of the earth by a string (gravitational force), and it is orbiting once every 24 hours. Draw in the string, and the body will begin to orbit faster and faster to conserve angular momentum.
Just like when a figure skater begins a spin with arms wide, then speeds up when the arms are brought in.
I suppose you could fire a rocket so that its trajectory is directly straight up from the center of the earth, but what purpose would that serve? Any payload you bring would just fall straight back down again.
No. The object’s speed is entirely dependent on gravitational attraction and altitude.
An orbit is a balance between being pulled straight down by the earths mass (gravity) and it’s tangential velocity. Too much velocity, and you move out into an orbit further away. Too little, and you move closer in. Every altitude has a specific orbital velocity (derived from the mass of the earth). The moon orbits at 384000 km and has a 28 day orbital period. Any other object orbiting the earth at 384000 km has a 28 day orbit. Any object orbiting the earth at 35000km has an orbital period of 24 hours (geostationary). Things closer go round faster - the Shuttle at about 200km has a pretty short orbital period (a few hours). The surface of the earth is not orbiting the earths CoM.
It would achieve a new orbit - as it gets closer to the earth, it’s tangential velocity would be wrong for the altitude, so the forces on the object would be unbalanced. These will eventually balance out into a new orbit, mostly similar to the original orbit, but probably more eccentric than the original, and a bit ahead or behind. It is incredibly counterintuitive and involves lots of vector maths. You certainly can’t just “eyeball” an orbital change - it really is rocket science in the ubergeek sense. In one story of the space programme (maybe Space, by James Mitchener) there is a scene where NASA rig up two jeeps on a special banked track, so that they replicate orbital mechanics (I can’t remember the exact details). This was to make a bunch of gungho test pilots who felt that they could fly anything realise that the one thing they could not do was “fly” a vehicle in orbit, let alone dock two of them manually. It all has to be done by computer. Even the manual “fine control” is mediated by computers that interpret what the pilot wants to do into the maths that make it actually happen without causing a collision.
Si
Now I getcha. :smack: Thanks for being patient enough to explain that.
If you were far enough out in space that the Earth’s gravity did not affect you greatly, and were at rest with respect to some reference frame outside the the Earth, the Earth would be screaming away from you at 66,000 mph, and a gentle drop of a basketball would simply result in a floating basketball.
You must remember that when the shuttle is in its standard orbit, it is moving at over 17,000 mph. It just looks like it’s blithely floating there because the nearest object that could give you a visual clue of its speed is the Earth, which is 200 miles away.
Once dropped, the pillow will begin to fall downward at about 9.8m/s[sup]2[/sup], slowly being counteracted by drag forces of ½ ρ*v[sup]2[/sup]*c[sub]D[/sub]*A. According to Wikipedia a “smooth brick” has a c[sub]D[/sub] of ~2.1 (but I’m not sure if I buy that). I don’t have numbers for a pillow’s cross sectional area, but maybe someone at home can weigh one and measure it for us? Density (ρ ) is determined by the 1976 Standard Atmosphere (or any other definition you care to use) but changes significantly with altitude. At work we assume no atmospheric drag above ~150km altitude, but you could conceivably go lower and still get accurate numbers.
Rather than set up a simulation program to integrate the velocity gains and losses, you might just look at Wikipedia’s Terminal Velocity entry and see what the V[sub]term[/sub] looks like at different altitudes. Once you’ve determined that, however, you still need to know what kind of speeds a pillow can survive… and I don’t have any kind of math that can help with that.
Someone once posted a little tip for this stuff like “forward is out, back is in, up is …” I don’t recall how it went.