Origami Plane to Fly From the Int. Space Station
SK writes "The University of Tokyo and the Japan folded paper (origami) plane society hopes to fly a paper airplane from the International Space Station to Earth. The plane will be 30-40cm long and weigh about 30 grams. A University of Tokyo research group has successfully designed a special paper plane model that was able to withstand a Mach 7 high velocity stream for 10 seconds. The experimental plane was about one-fifth the size and withstood temperatures as high as 300C without burning up." Unfortunately for most of us reading this, the original source is all in japanese.
Even though it's in Japanese, just use Google Translate to read it.
Japan wants to fly paper plane from International Space Station to earth:
http://mdn.mainichi.jp/national/news/20080118p2a00m0na025000c.html
Remember that the speed of sound changes with the properties of the air through which an object is travelling. The absolute speed of an object (i.e. in m/s) corresponding to a high Mach number deep in our atmosphere (say in the troposphere or stratosphere) would actually be much, much slower than the speed of sound in the mid-thermosphere (where the ISS is located).
Its a similar reason to why de-orbiting objects can travel faster than terminal velocity; they accelerated to that speed before the air resistance built up.
Aikon-
No. If you throw something from a satellite in a circular orbit, giving it a small 'downward' velocity component, the object will just end up in a slightly elliptical orbit.
One way of thinking about orbits is that a satellite is perpetually falling towards the earth, because of gravity, but also perpetually missing, because of the lateral velocity component.
To make the paper plane de-orbit, you could throw it in the opposite direction to the ISS at the same velocity as the ISS is travelling: 27 500 km/h. Then the plane won't have any lateral velocity component, and will fall straight down.
The site is bi-lingual, just click the "English" button.
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The force due to gravity of the ISS is so small it might as well be zero.
Out of modpoints but really liked a post? 1BDkF6TtmmeZ3yqXbz9yhdYVqRYnwFoXDj
As a rocket scientist, I'll take the reins here.
From the altitude the ISS is orbiting, there's no such thing as approaching the atmosphere "slowly". The ISS is traveling at about 17000 mph around the circumference of its circular orbit. In order to enter the atmosphere, a body in that orbit would have to slow down in order to enter an elliptical orbit which intersects the atmosphere. This requires a velocity change (delta v) of about 200-250 mph. Even with that change, you're still traveling at 16,750 mph, so that when you finally do hit atmosphere, the friction from the air will be very high, even if the air is thin. As the friction slows you down, you drop farther into the atmosphere, where the air is thicker and there is more friction. These two changes (air pressure and velocity change) work together to reach a point of maximum heating, and then taper off again as you reach subsonic speeds. The steeper the dive, the faster you reach thicker air, and the higher the max heating point will be.
Let's say for argument's sake that you wanted to drop straight down from where the ISS is orbiting, with no horizontal velocity. (That would require an instantaneous delta v of the whole 17000 mph, which is nigh impossible, but we'll assume we can for our thought experiment.) Since the ISS is orbiting at an altitude of about 225 miles, and the atmosphere is generally considered to start at the 62 mile mark, that's still 163 miles of vacuum free fall to contend with. Leaving out the square-of-the-distance effects of gravity fall off (which are close to negligible at these distances), we get a fall time of sqrt((163 miles)/(32 feet per second squared)) = 164 seconds. That gives us a velocity of (32 feet per second squared)*(164 seconds) = 5248 feet per second, or 3578 mph at the moment we hit the upper fringes of the atmosphere. The heating will certainly be less than the standard deorbit, but it is still a decent force to be reckoned with. Any angle larger than the vertical will require a smaller delta v but will result in a higher entry velocity and higher heating.
Now you might be thinking to yourself, "but AeroIllini! You totally contradicted yourself there!" I did. Except that as you vary the angle of entry from shallow to vertical, the graph of max heating reaches a peak and then falls off again. So for a very shallow entry, your heating will be lower than a steeper entry, but going even steeper the heating will taper off again until you reach vertical entry, which will have the lowest heating of all. Vertical entry also has the highest delta v requirement of all, and a shallow entry has the least delta v required.
I hope this answers your question.
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Unfortunately that suggests a roughly 70% chance it'd drop into the ocean.