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Space Shuttle launch physics question

  1. Jul 15, 2004 #1
    Hello all.

    I am doing some research on the space shuttle for a class project, and I am having some problems with my data. I was hoping someone could show me the error of my ways.

    I got all of my data from this site: http://spaceflight1.nasa.gov/shuttle/reference/basics

    It says that 2 minutes (120s) into the launch the shuttle is moving at 4828km/h (1341.11m/s) and is at a height of 45km (45000m).

    Using V = at I calculate the acceleration to be:

    1341.11 / 120 = 11.18 m/s2 , or 1.14g

    This seems pretty low to me. I thought 3g was the norm?

    Using s = .5(a)(t^2) I calculate the distance to be:

    (.5)(11.18)(14400) = 80496m

    Which is a lot more than the 45km they claim. Even accounting for the curved path the shuttle takes on launch, I can't by that it travels 80km to reach a 45km height.

    The numbers get even worse later on. The website says at 8.5 min (510s) it is at 28968km/h (8046.67m/s) at a height of 109.26km.

    a = 8046.67 / 510 = 15.78 m/s2
    s = (.5)(15.78)(510^2) = (.5)(15.78)(260100) = 2052189m = 2052km

    Something just isn't adding up. Is there something I am missing in my calculations, or is there some variable I am missing? Any help would be greatly appreciated.
  2. jcsd
  3. Jul 15, 2004 #2


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    You're missing some stuff:

    1. The shuttle does not have constant acceleration.
    2. If the shuttle is accelerating upwards at 1 g, the pasengers are experiencing 2gs, since it's still overcoming gravity.
    3. The shuttle doesn't just accelerate upwards, it orbits about once every 90 minutes so it's got a pretty fast sideral speed.
  4. Jul 15, 2004 #3

    1. I know, the calculations for t=120 and t=510 would be average accelerations.
    2. Ah, forgot about that. That makes the 15.78m/s2 number seem better.
    3. I'm not talking about orbit here, only launch/ascent to "edge of space" at ~100km.
  5. Jul 15, 2004 #4
    Some more data

    Looks like I also left out the 1472km/h acceleration imparted by the Earth's rotation. I'll have to factor that in and recompute.

    Also, I have been trying to figure out the actual distance travelled by the shuttle during launch. I used altitude as "s" in my previous equations, which I knew were wrong (since "s" is total displacement) but gave me something to start with. I have been unable to find any accurate displacement data for the shuttle launch. I did get this info:

    At SRB seperation, 2min4s into launch, the shuttle is at an altitude of ~136000ft and is 32 statute miles downrange from launchpad.

    At External tank sep, the shuttle is 1000 statute miles downrange.

    Maybe I can use this rough data to calculate acutal distance travelled until MECO.
  6. Jul 15, 2004 #5


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    Actually, you're talking orbit immediately. Except the path of the 'initial' orbit is very elliptical - in fact, the 'back' part of the ellipse is below the surface of the Earth. At apogee (now out in space), the 'final' orbit is achieved by adding enough speed to circularize the ellipse.

    Every launch achieves some kind of an orbit - some satellites just have a 'wet' perigee.
  7. Jul 15, 2004 #6
    Might I suggest you download this app: http://www.medphys.ucl.ac.uk/~martins/orbit/orbit.html
    It taught me more about space flight, orbital mechanics and how the shuttle, Apollo, Gemini and Mercury spacecrafts fly in about a week of using it than I've learned in several years. It's fun too!
  8. Jul 15, 2004 #7


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    Cool topic.

    The way you find the instantaneous accelerations for a rocket is this:

    The thrust of the engine is determined by two things: The mass flow of the rocket and the specific impulse of the engine. The specific impulse, Isp, is a parameter which is unique to each engine, and is mostly based on the propellants used. It doesn't look like the site you linked has it for the SSME, so I'll recommend this site. Search by engine (on the right) for the SSME (it is your research after all :smile: ) The number listed is for Isp in vacuum, because performance is worse in an atmosphere. That's probably more detail than you need to worry about.

    To find the thrust, you take

    [tex]Th=I_{sp} * g_0 * \frac{dm}{dt}[/itex]

    Where dm/dt is the mass flow rate, and g0 is the acceleration of gravity at sea level.

    From F=ma, the actual acceleration will depend on how much fuel remains unburnt at the time in question.

    All rockets start their turn almost immediately after they lift away from the tower. An orbit is attained by going ~7.75 km/sec sideways. The reason they go up is only to get out of the Earth's atmosphere which would slow them down via drag. There is nothing to gain by going straight up. In fact, it really hurts you if you do it. The reason you can stay in orbit is because the Earth is curving away from under you at the same rate that gravity is pulling you in. Effectively, when solving for the acceleration, you can subtract a rotational V^2/R term away from the gravity to obtain the 'up'-'down' acceleration. Staying laterally slow during ascent means you need to fruitlessly burn fuel to overcome the unbalanced gravity, and is called a 'gravity loss'.

    Another formula you may make use of for your project is the ideal rocket equation

    [tex] \Delta V = -I_{sp}g_0ln(\frac{m_f}{m_i})[/itex]

    The change in velocity is -1 times the specific impulse times the acceleration of gravity at sea level times the natural logarithm of the final mass over the initial mass.

    If you know what the fueled and empty weights are, you can find what the theoretical speed of the shuttle will be. However much that is above 7.75 is the gravity & drag losses.

    Another factor would be the fact that the Shuttle drops off its SRBs and tank during the ascent. I don't know if you want to go into that level of detail for your project. If you do, you just do the rocket equation stepwise, first taking the fuel burnt from the SRB ascent, and then doing the next step with a smaller initial mass (because the dead weight from the SRBs aren't in the mix anymore). You then add the changes in velocity up.

    Hopefully my ramblings make sense. I love this stuff, and can geek out for hours on it.
    Last edited: Jul 15, 2004
  9. Jul 16, 2004 #8
    Yea, I really love this stuff. But my confusion continues.

    Regardless of any other data, the simple fact is that the space shuttle reaches a speed of 8046.67m/s in 510s...with an initial velocity imparted to it by the Earth's spin of 408.89m/s. To me, that means an average acceleration of 14.98m/s^2 using V = Vo + at. Sometimes it may be accelerating higher, sometimes slower. But this should be the average over the given time.

    Thus, if it has an average acceleration over those 510s of 14.98m/s^2, and still using the initial velocity of 408.89m/s, then it had to travel 2156682.9m (2156.68km) using s = (Vo)(t) + (.5)(a)(t^2).

    Even accounting for its parabolic trajectory, I doubt very much that it travels over 2000km before it reaches escape velocity.

    So again, what am I missing? Are you saying that I cannot use these equations?

    Thanks a lot for the discussion. It is incredibly hard to find people to "geek" over this stuff with. ;)
  10. Jul 16, 2004 #9


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    First off, it doesn't reach escape velocity; it reaches orbital velocity. The difference is this:

    Orbital velocity is the speed needed to get into an orbit which doesn't crash into the Earth at some point. For low altitude orbits, it is ~7.75 - 8 km/sec.

    Escape velocity is sqrt(2) times orbital velocity and that is the speed at which you will leave the Earth's system and never come back (assuming an empty universe aside from the Earth).

    Regarding the distance travelled:

    The Earth's radius is ~6378 km. That means that if you orbit once at an altitude of 200km you will have travelled ~41300 km. The ascent distance of 2000km is about 5% of a single orbit.

    Again, think about the magnitudes involved. At the end of the ascent, the Shuttle is going ~7.75 km / sec. It'll take a little over 4 minutes to go 2000 km at that speed. If you consider that a LEO orbit takes about 90 minutes, the magnitudes work out fine.

    The linear equations aren't any good for these calculations except to get rough ballpark calculations like you've done. Any real calculations need to be done in at least polar coordinates (for an arbitrary 2d orbital plane) to take into account the curvature of the Earth.
  11. Jul 16, 2004 #10
    Arg, of course I meant orbital velocity. Darn sausage fingers won't do what I tell em. ;)

    Well, my numbers finally make a little sense. At the point where the shuttle reaches "orbital" velocity of 8046.67m/s it is at an altitude of 109.26km and has moved downrange from the launch point roughly 1600km (1000 statute miles).

    Now the s=2052km number makes some sense.

    I didn't realize how far downrange the shuttle moved during launch/ascent. I assumed a much steeper launch angle with maybe a downrange distance of 700-800km. But it does seem as though the shuttle needs over 2000km to accelerate to 8km/s...just as the equations show. It is just very deceptive that nearly every resource I checked notes the 109km altitude and doesn't mention the actual travel distance.

    I saw one website that actually said "after travelling over 100km the shuttle is moving at 28000km/h" which is completely incorrect.

    Thanks for the help. The universe makes sense again!
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