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Throwing paper airplane towards space question

  1. Oct 10, 2008 #1
    If you haven't heard yet, Japan is planning to throw paper airplanes from the ISS. Reportedly the airplanes will be able to survive re-entry:
    http://news.bbc.co.uk/2/hi/asia-pacific/7230949.stm" [Broken]

    Me and my friend were talking about it earlier, and we're at odds as to how exactly Japan will accomplish the throw. Basically, will a normal human powered throw from the ISS provide enough force for the airplane to successfully make it to Earth? Or will the paper airplane simply enter a slightly lower orbit around Earth?

    There's a few confusing bits in the scenario-
    1. Can it be accomplished by throwing the plane straight towards Earth or does it HAVE to be thrown in the opposite direction of motion (thrown backwards)? In either direction, could the throw from a human arm accomplish it?
    2. The ISS itself slowly falls to Earth and has to maintain its altitude with regular boosts back up, so I believe the airplane should fall back to Earth no matter how it's thrown or how much force is put in the throw. But what if the ISS was in a perfect geosynchronous orbit, could the plane be successfully thrown to Earth then?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 10, 2008 #2
    Sorry, I'd like to add a 3rd thing: Say the ISS or airplane weren't affected by conditions such as the slight atmospheric drag that slowed their velocity which would cause a continual drop in altitude. I really would just like to know if based solely on force applied from a human and after choosing a particular direction, towards Earth or backwards from orbit direction, would the plane make it to Earth?
  4. Oct 11, 2008 #3

    Jonathan Scott

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    If there were no atmospheric friction involved at all, I don't think it would be possible to throw a paper plane from the space station down into the atmosphere.

    Just for an order of magnitude estimate, suppose that you could throw the paper plane downwards from the space station at something like 30mph. A quarter of an orbit later, that velocity would now be tangential to the orbit, so that would be the minimum height reached. Given that an orbit is approximately 90 minutes, the plane would only be travelling downwards for about a quarter of that, and only straight down for a few minutes, so its orbit would only dip a few miles below that of the space station, which would not be enough to cause enough additional friction to trigger reentry. After a total of half an orbit, the plane would pass the space station again going up, and would continue passing in the downwards and upwards directions once per orbit, only gradually moving away.

    I don't think that throwing it backwards along the orbit would make much difference except that it would probably move away from the space station a bit faster.
  5. Oct 11, 2008 #4


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    Hi Yoduh,
    You can figure it out the answer to your third question using Kepler’s laws:
    The velocity in low earth orbit is about 8 Km/sec or 18,000 mph according to Wikipedia.
    A good baseball pitcher throws at around 100 mph.
    The radius of the earth is 4000 miles, that of the orbit of the space station, approximately 4200 miles.
    You need to decrease the orbit to one a perigee of 4000 miles and an apogee of 4200 miles so the paper airplane will hit the earth in half an orbit. This makes an average radius (or more correctly, a semi-major axis of 4100 miles or 2.5% less than the original orbit. Using the square-cube law the square of the period is inversely proportional to the cube of the period, hence in an approximate perturbative calculation, the period of the lower, smaller orbit is three halves times 2.5% less, or 3.75% less. But the circumference is 2.5% less, so the average speed is only 1.25% faster. (In a nonperturbative calculation you get a square root law.} But via the areal vector law, equal areas are swept out in equal times. The area is proportional to the radius squared, and the radius at apogee is 4200/4100 or again about 2.5% greater than the average radius, so the speed at perihelion is about 5% less than the average speed, which we already computed to be about 1.25% faster than the average speed of the higher circular orbit. Therefore you need to reduce the average speed by about 3.75% of 18,000 mph in order to de-orbit frictionlessly. This is about 675 mph, if I have not made any mistakes. (I should really say about 700 mph to better reflect the lack of accuracy in this perturbative calculation.)

    Looks like you need atmospheric friction to make this work.
    Reversing the calculation, a 100mph backwards throw should lower the perigee by about thirty miles.

    Jim Graber
  6. Oct 11, 2008 #5

    D H

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    You don't need to target the surface of the Earth to make a spacecraft hit the Earth in half an orbit or so. All you need to do is target the top of the thick part of the atmosphere, or about 75 miles above the surface. The atmosphere will do the rest. From a circular orbit 200 miles above the Earth, this corresponds to a delta V of about 130 mph.

    The astronauts do not need to throw the paper airplanes at 130 mph to make them reenter. All they need to do is let them go. The paper airplanes will eventually reenter because the ISS is well within the Earth's atmosphere. The troposphere extends up to 300 to 600 miles above the surface (the boundary depends on solar activity) and the exosphere, up to 10,000 miles. The ISS needs to raise its altitude on occasion (how often depends on solar activity) to keep itself in orbit. A paper airplane of course carries no thrusters; it will reentry eventually.

    No. A 100 mph backwards throw will lower the perigee by about 95 miles. Use the vis viva equation,

    [tex]v^2 = G\,M_e\left(\frac 2 r - \frac 1 a\right)[/tex]

    Suppose that from a circular orbit at distance [itex]r[/itex] from the center of the Earth you throw something backwards, decreasing the velocity by [itex]\Delta v[/itex]. The goal is to solve for the decrease in altitude [itex]\Delta r[/itex]. The semi-major axis is the average of the apogee (the current radial distance) and the perigee: [itex]a=1/2\,(r+(r-\Delta r)) = r - 1/2\,\Delta r[/itex]. Applying the vis viva equation,

    v^2 &= \frac {G\,M_e} r \\
    (v-\Delta v)^2 &= G\,M_e\left(\frac 2 r - \frac 1 a\right)
    = G\,M_e\left(\frac 2 r - \frac 1 {r - \frac 1 2\,\Delta r}\right)

    After a bit of grinding,

    \Delta r &= 2r\left(1-\frac 1 {2 - \left(1-\frac{\Delta v}v\right)^2}\right) \\
    &\approx 4r\frac {\Delta v} v
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