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Homework Help: Space Shuttle Problem

  1. Nov 16, 2006 #1
    While you were under the impression that you would be working at the international space station on a project involving biomechanics in a weightless environment, your boss has other ideas. Apparently, Spacetronics has developed a new material to be used for a heat shield on the newest version of the space shuttle. They are very proud of this material and hope to convince NASA to use it (there’s lots of money to be made)! Since it is way too expensive to test out the material, your boss has asked you to determine whether the material will work for the heat shield. She did note that your next project would be at the space station and involving biomechanics. She also pointed out that you would be taking the newly outfitted space shuttle home. So it is important that you get this right, or you would be toast, literally.

    The space shuttle is cylindrical with a diameter of 8.7 meters and a length of 56.14 m. The mass of the shuttle is two million kilograms. The shuttle would be coated with the material, 2 cm thick. The new material has a specific heat of 1000 J/kg K, and a density of 2500 kg/m3. If the temperature of the material exceeds 4000K, it will spontaneously combust, resulting in the shuttle exploding. You already know that as the shuttle enters the atmosphere at 100 km, it will experience a drag force, so while some of the potential energy will be converted to kinetic energy, some will also be converted to thermal energy. Hence, you need to be able to determine the speed of the shuttle as it falls to the earth, so that you can determine how much heat the shield will absorb, hence its temperature. The density of the atmosphere is given by the following expression:
    p= p0exp* -(MWgz/RT)

    where r0 is the density of air at the surface of the earth, MW is the average molecular weight of air in kilograms per mole (not grams per mole), T is the temperature in Kelvin and R = 8.314 J/mol K.

    Well I know:
    m= 2 million kg
    specific heat = 1000J/kg K
    density = 2500kg/m3
    Tempmax = 4000K

    where should I get started?
  2. jcsd
  3. Nov 17, 2006 #2


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    What are you trying to find? The speed of the shuttle? The termperature? What are you to assume about the path of the shuttle? Does the coating retain all of the heat created by collisions with the air? Does it conduct heat well (hopefully not) so that all of the mass of coating is warmed, including the part on the back side of the shuttle?

    Too many things left to speculation I think.
  4. Nov 17, 2006 #3


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    One can calculate its speed at the given height (assume circular orbit). This gives the kinetic energy at the start. You need to get the landing speed to determine the kinetic energy at the end (200 - 300 km/h?). It is also possible to calculate the potential energy difference between entry and landing. The difference between the mechanical energies at these two points is converted to heat via the drag on the shuttle (risk of getting fried at both ends of the trip!).

    I think one should assume that very little heat will be conducted to the shuttle so that the heat shield absorbs all of the heat.

    I think the dimensions of the shuttle serve to calculate the mass of the shield (quite dense material!) and would not bother with drag calculations.
    Last edited: Nov 17, 2006
  5. Nov 17, 2006 #4
    From reading it, I need to find the speed of the shuttle and once I find that, I can figure out whether the amount of heat absorbed is too much that it will cause the shuttle to combust. My whole problem is, how do I start to find the speed of the shuttle.
  6. Nov 18, 2006 #5


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    Use the centripetal acceleration.

    [tex]a_c = \frac{v^2}{r}[/tex]


    [tex]a_c = F_G/m[/tex]

    Newton's universal gravitational attractive force supplies the centripetal force.
  7. Nov 18, 2006 #6
    well the space shuttle is supposed to be descending to earth linearly so from what I already know, which is

    a= 9.8 m/s^2
    v initial = 0
    position final = 0 km
    position initial = 100km

    I am pretty sure I can calculate velocity now with the given information, but I don't know what equation to use. Any help?
  8. Nov 18, 2006 #7


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    You can't use a constant acceleration of 9.80 m/s^2 since this is is only a good approximation if one stays relatively close to the surface of the planet. You have to use conservation of energy but with the formula [tex] -G {m M_{earth} \over r}[/tex] for the gravitational potential energy (instead of the usual mgy, or mgh).
  9. Nov 18, 2006 #8
    well if the shuttle is entering earth's atmosphere and descending closer to earth, why can't I use that?
  10. Nov 18, 2006 #9
    any more ideas?
  11. Nov 19, 2006 #10


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    at 100 km up

    [tex]g_{100} = \frac{GM}{R_{100} ^2} = 9.53 m/s^2[/tex]

    that plus the drag on the shuttle will increase as its downwards speed increases.

    for the shuttle to start its downwards descent it is neccessary to slow its orbital speed (tangential speed) down by firing retro thrusters. The gravity of the earth will then start to pull it inwards.
    Last edited: Nov 19, 2006
  12. Nov 19, 2006 #11
    ok so if i use that for the accleration, i'm going to use a spreadsheet to calculate the acceleration as i get closer to earth, then i can calculate the velocity with the equation

    vf^2=v0^2 + 2 * acceleration * (distance final - distance initial)

    At 100km I get 6174.14m/s. This doesn't include the drag force though. So would I add the force of drag to that? That's not a velocity though. How would you calculate that in?
  13. Nov 21, 2006 #12


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    I think it is too much of a simplification to assume that the shuttle is falling like a stone to earth.

    Before the shuttle starts its downward descent back to the earth it is in orbit around it. Then they apply the brakes a bit (or let the drag take over by not vectoring the thrusters every now and then) so that their orbital velocity decreases.

    This cause the shuttle to move closer to earth since its orbital speed is not large enough to keep it from falling around the earth any more.

    On their way downwards their speed will increase drastically as they spiral around the earth (conversion of potential energy to kinetic energy).

    In order to decrease their forward speed they raise the nose of the shuttle slightly. This will increase the drag and keep their speed in control. Unfortunately this heats the shuttle up due to the friction of the atmosphere on its shell. A similar situation arises with the brakes on formula one cars. The brakes heat up when they are applied. In this case air flow over the discs cools them down again. With the shuttle the region that can be cooled down is not exposed so much to the air current, which leads to serious heat build-up (heat is radiated away by the tiles though, so according to my logic the tiles should be black on top in order to increase the radiation loss as much as possible).

    You could look at it this way: The shuttle is experiencing two forces one towards the centre of the earth and one opposing its direction of motion (velocity). The shuttle will therefore have two acceleration components - one caused by the drag and one by gravitation attraction of the earth. The drag force is not necessarily perpendicular to the gravitational force (the shuttle is falling down while it is moving forward). This means that the drag can be dissolved into two components - one upwards (opposing the downwards fall) and one decreasing its forward motion.

    So try the same trick as with projectile motion. Split the motion into two components - one towards the centre of the earth (falling downwards) and the other perpendicular to this direction and think of the eath as a flat plane with the shuttle approaching it at a downwards angle. The complication is just that both forces are not constant, so the acceleration in both directions changes during the motion.
    Last edited: Nov 21, 2006
  14. Nov 21, 2006 #13
    Ok now that makes a lot more sense. Thanks! :)
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