Well? Could a person be lowered into the atmosphere from space (say, at the level the shuttle flies) without burning up upon re-entry if it were done slowly enough?
Clarify: What are you postulating as a platform to lower him from? The ISS? A shuttle in orbit? A shuttle either “on the way up” or “on the way down”? Something like a fighter jet or a 747, flying at X number of feet? A satellite?
Let’s say from a stable platform like the ISS. Assuming they’d have a tether a couple of hundred miles long.
Although I’m sure several people are going to jump in with concise answers, the issue here is that you don’t “burn up” just by moving into the atmosphere. The reason why the shuttle comes in as a fireball is because of the speed in which it travels through the air. If, for instance, you had someone “walk” from the ISS to the ground, nothing would happen to them (except they better be able to hold their breath for a while!).
The faster you move, the hotter you get. If that same person were to “run” at several dozen times the speed of sound on the ground, they would burn up too.
You might want to rethink that Mr. Blue Sky , if I’m understanding your statement. The ISS is moving along at a right pert clip, something like 22,000 MPH or so.
My understanding is that it is possible to reenter the atmospere without the glowing and burning we’ve come to know and love in our reentries. It would just take an extreme amount of fuel and or time.
An object in orbit is in “free fall” as it is. If you just drop something in the ISS, it will stay right there as ISS occupants see it.
Yeah, but he’d be given some kind of launch, maybe a shove between the shoulderblades. But he’d still be going too fast.
The orbiting shuttle is moving pretty fast, too. And so would be the shuttle on the way up or down. Even a 747 cruises at about 600 mph.
So, you’d have to postulate some kind of stationary platform? Why am I flashing on “secret missile defense platforms”? 
But they’d be in orbit, too…
Sorry, I know that’s not helpful…
Looking forward to seeing what the answer is to this…
Yes, it is theoretically possible, using the sky hook concept. This Wikipedia article on the space elevator makes for an interesting read.
Said platform from which said [hapless] astronaut would re-enter said atmosphere without said fireball would require that the aforementioned platform be in geosynchronous orbit. That would eliminate the lateral velocity that contributes to the shuttle heating. Thus, the only velocity left would be “down”, or toward earth, and if you could zip down at, say, 600 mph, you’d make it to the ground in due time without due heating.
thus said the aforementioned
Vlad/Igor
I highly recommend Arthur C Clarke’s Fountains of Paradise, mentioned in the link above. It is a science fiction novel about the construction of a space elevator from near (or on, I can’t remember) Sri Lanka to an orbiting geosynchronous platform. Very interesting speculations about physics and materials science.
The answer is probably yes, with the develpment of a whole lot more space technology.
Yes, absolutely. For the record, the altitude at which an orbit synchronizes with the rotation of the earth is a little under 36,000 kilometers, or a bit over 22,000 miles. The ISS is just a few hundred miles up, and the shuttle doesn’t get much further out than that.
It’s also worth noting that the ISS is not in a perfect orbit. From time to time, a booster launched from the ground is used to push the station back up; if left alone, the station would fall into the main part of the atmosphere and be destroyed. From that, I conclude that something dropped or thrown from the station would have as tough a re-entry as the station would. The object wouldn’t be moving straight toward the ground in an accelerating fall, which is how we instinctively but incorrectly visualize the event; rather, it would have approximately the same orbital velocity and trajectory as the platform from which it originated, which is thousands of miles an hour lateral to the ground.
Seems to me that if you put a human-comparable weight on a tether and shoved it out the bottom of the ISS, playing out the tether as the weight descends, it may actually start to bounce and skip off the top of the atmosphere (“top” being an arbitrary point where increasing atmospheric pressure creates sufficient drag on the object) like a tin can on a string rattling over the pavement behind a just-married-mobile. If you shot the object out the bottom with enough speed to punch through into the atmosphere, it would quickly get dragged out behind the ISS, requiring you to pay out more and more of the tether, thus creating an engineering problem similar to that facing the prospective builders of the space elevator. Even assuming you solved that problem and figured out how to keep the object more or less “below” the ISS, you’d have to figure out how to land it on earth without letting it gouge a huge trench in the landscape.
So, uh, I don’t see this working very well, even as a hypothetical.
IIRC the main issue, once they had incredibly strong lines (single chain molecules hundreds of miles long), was maintaining the platform/space station in the proper location. As cargo is hauled up and down the lines, the station has to absorb tons of force which tend to move it up or down, and once out of position vertically, the orbital speed would tend to very leading to leteral displacement. The issue was theoreically solved by slinging a counterweight out to a higher orbit than the station and somehow transfering the force up and down through thw whole complex in ways that tended to cancel out movement.
But this is not true. A geosynchronous satelite is still in orbit. That means it has to go very fast. It travels around the earth in exactly 24 hours, that’s fast. So your space traveler starts at that speed. True, he’s stationary with regards to the platform, he’s stationary with regards to the surface of the earth. Now, lets give him a push downward and see what happens. He still has the same sideways vector he had before, so he’s still moving as fast sideways as the space station. But now he’s in a lower orbit than the space station. If you are in a lower orbit than the space station but going the same speed, what happens? You don’t drift down to earth, you are instead put in an elliptical orbit and the high point of the elliptical orbit will be the exact same height you started out. The energy put into pushing you towards the earth doesn’t go away, you still have it, and it will carry you back away from the earth on your return.
What you are imagining is an elliptical orbit with one the high end up at Geosynchronous orbit, and the low end on the earth’s surface. And you’d be travelling at gigantic speeds relative to the earth’s atmosphere (not to mention the earth’s surface) when you’re at the low end of your orbit.
The thing to remember is that anyone in orbit is in orbit. That means they are falling towards the earth at a tremendous speed but constantly missing. You are NEVER “stationary” with regards to the earth.
Not from a platform orbiting the earth, but if you could somehow get yourself into an orbit about the sun that was slightly smaller (faster) than earth’s orbit, yet intersected the earth, you could come into the atmosphere at an arbitrarily slow velocity. The rotation of the earth would still leave you to deal with 1000mph winds if you came in over the equator, but if you came in at one of the poles, that wouldn’t be a problem.
Yes, but that object in orbit around the sun is going to be affected by earth’s gravity. After all, an object in orbit around the earth is in orbit around the sun too. You can’t just drift down to the earth’s surface because of that old 9 meters/sec^2.
Sure, but if you can get into atmosphere with around zero net velocity, as the orbit I described could come close to accomplishing, you’d be able to deploy a parachute and drift down.
IIRC NASA tested an emergency ‘bubble’ that could allow a astronaut to excape from orbit unpowered, though never used. It may have been back in the skylab days.
I also recommend Stanley Schmidt’s The Web Between the Worlds which came out at exactly the same time and was also about the building of a space elevator. In fact, it’s so similar to Clarke’s book in so many ways that Clarke actually wrote an introduction to it, pointing out the similarities and absolving Schmidt of plagiarism.
Since then, the Space Elevator has been used in lots of SF (Heinlein casually alludes to one as “the Beanstalk” in Friday. Larry Niven introduces an interesting one in Rainbow Mars, etc.), but most of these don’t go into the engineering as much. One interesting exception is Robert Forward’s Indistinguishable from Magic, which discusses some very interesting variations on the idea, some of which don’t involve non-existent super-strong materials.
No. A Geosynchronous orbit around the equator (a geostationary orbit, which I should have said in my first post) means that the satellite is “parked” over a geographical point on the earth’s surface. As seen from the moon, the satellite moves in an orbit, but seen from the earth, the satellite is stationary. Look at satellite TV antenna dishes: they are pointed at a satellite in a geostationary orbit and don’t move.
Let’s look at basic angular (circular) motion. Angular velocity is calculated: V ang = rotation in radians / time. One complete rotation = 2(pi) radians. We can say that the earth or a geostationary satellite has an angular velocity of 2(pi) radians / 24 hrs or 0.083 radians per hour. For any point anywhere on a straight line from the earth’s center out to a geostat satellite the angular velocity is the same: 0.083 radians per hour. the radius, or distance from the center of rotation is not in the equation.
However, if we want to calculate linear velocity from angular velocity, the equation is: V lin = V ang X radius. For a satellite 22 000 km from the earth’s rotational axis, the V lin = 0.083 rad/hr X 22 000 km = 5758 km/hr. Pretty fast. If an astronaut is lowered toward the roational axis for a distance of 10 000 km, then his V lin = 0.083 X (22 000 - 10 000) = 0.083 X 12 000 = 3141 km/hr. He is still moving at the same angular velocity, but his linear velocity is slower because his radius of rotation is smaller. Simply standing on the earth’s surface, you’re crawling along at only about 1600 km/hr. Standing at the north or south pole, where your distance from the axis is 0, your V lin is 0, even though your V ang is still 0.083 radians/hr.
Your analogy of orbital falling is correct, but misapplied. There are two vectors to a satellite’s movement: one is toward the earth’s center due to gravity (Vg), the other is parallel to the earth’s surface (Vp). Vg = Vp for a given stable orbit, meaning that the satellite moves forward (and away from the earth) the same distance it is pulled toward the earth. In a geostationary orbit, this takes place above the same point on the earth’s equator, without any change in position or distance.
Vlad/Igor
When did this come out in relation to Kim Stanley Robinson’s Mars Trilogy? They built a space elevator in that one, too.