Japan to launch origami planes into outer space

Times of London ^ | 02/07/08 | Leo Lewis

[KittyKares]

This is form of art is reaching new heights.

[dr_lew]

I expect the plane will get hotter than 300 C.

That's my first thought, but presumably they are doing their homework on this. Small objects have the advantage in reentry, since the kinetic energy is carried away ( or left behind ) by a layer of heated ( very thin ) air. For a large object, the size of this layer is comparatively small and the rate of energy loss requires a very high temperature. For a small object, not so much. You would have to do the math, which as I say, I have to believe they are doing.

[tlj18]

Also, the planes will have a high surface area to weight ratio, so they should descend less rapidly. Interesting experiment...

[dr_lew]

Also, the planes will have a high surface area to weight ratio, so they should descend less rapidly.

With a higher surface area to weight ratio, you should expect more rapid deceleration, and hence a more rapid descent, at least initially. It's only when it becomes a glider that it can "float to the earth" in the familiar paper airplane glide.

Interesting experiment...

It's got my attention.

[tlj18]

Well, it will descend more rapidly at first, because more of the rarified atmosphere will be striking it, and because of its low mass, it will have less momentum. However, once it enters the atmospheric regime, it will descend less rapidly because of the surface area. So basically its characteristics produce opposite results in the two different environments.

[TalonDJ]

Well, it will descend more rapidly at first

What? No it will not. It will descend slower the whole way. It will be descending at the same speed as anything else (just under orbital speed) initially and then slow down faster than something heavy would as soon as they encounter any air at all (which is pretty much instantly from low orbit). So the no, it will no descend 'more rapidly' than anything else at any point in it's fall.

[TalonDJ]

Oh, did you mean vertical speed not airspeed? Well maybe. Would have to plot it out to see if there is much difference there.

[tlj18]

It will lose speed more quickly due to its interactions with the upper atmosphere. However, it will experience more lift than typical space debris as it encounters substantial atmospheric concentrations, slowing its descent. So, initially, it will reach an altitude closer to the Earth than more massive objects, but then the situation flips.

Agree?

[TalonDJ]

Right it will start to fall out of orbit faster at first so it’s trajectory will be much steeper. It might be that it will loose speed enough to trade off that steeper entry angle, but yeah, I see what you mean now.

[dr_lew]

Orbital decay is kind of tricky. Paradoxically, an object in a lower circular orbit has a higher velocity than one in a higher circular orbit. How then can a frictional force acting antiparallel to the velocity move an object from a higher circular orbit to a lower one?

To solve this paradox, you have to account for the deviation of the direction of the velocity. If it tilts downward, then a component of the gravitational force acts parallel to the velocity, and this must be greater than the frictional force. Now the lower total energy, kinetic plus potential, is accounted for by the work done on the atmosphere by friction, even as the object moves faster in the lower orbit due to the net force, frictional plus gravitational, acting with a component parallel to the velocity.

[SunkenCiv]

It’ll take a long time for each reentry; it’ll be somewhat akin to Rutan’s “shuttlecock” model for reentry. But anyway, reminds me of the Japanese balloon bombs during WWII.