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This is killing me

  1. Oct 3, 2007 #1
    This is killing me!!

    A 150kg rocket moving radially outward from Earth has a speed 0f 3.7km/s when it shuts off its engine 200km above Earths surface.

    (a) Assuming negligible air drag, what is the rocket's kinetic energy when it is 1000km above Earth?

    (b)What is the Max height the rocket can reach?

    Okay, my main issue is not solving the variables correctly. Every time I put them in my calculator I get a different answer. But maybe someone could varify my algebra and that I am using the correct values for the variables.

    (a)[tex]U_i+K_i=U_f+K_f[/tex]

    implies[tex]K_f=-\frac{GMm}{r_i}+\frac{GMm}{r_f}+\frac{mv^2}{2}[/tex]
    Using:
    [tex]r_i=6.37*10^6+2.0*10^5[/tex]
    [tex]r_f=6.37*10^6+1.0*10^6[/tex]
    [tex]m=150[/tex]
    [tex]M=5.98*10^{24}[/tex]
    [tex] v_i=3700[/tex]
    then
    [tex]K_f=3.8*10^7 J[/tex]



    (b) I did this without relying on my answer from (a)...

    [tex]-\frac{GMm}{r_i}+\frac{mv_i^2}{2}=-\frac{GMm}{r_f}[/tex]
    implies[tex]\frac{2r_i}{-2GMm+r_imv^2}*(-GMm)=r_f=r_f=7.4*10^3 km[/tex]
     
    Last edited: Oct 3, 2007
  2. jcsd
  3. Oct 3, 2007 #2

    Dick

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    Your solution to a) is so hashed up I don't think anyone could check anything. r_f is in Joules? It looks like a random string of letters and numbers to me. Can you clean it up?
     
  4. Oct 3, 2007 #3
    righto...
     
  5. Oct 3, 2007 #4

    dynamicsolo

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    The KE equation looks right here; you can save yourself a little work by stripping out the m, since the kinetic energy per unit mass of the rocket is the velocity squared. I'd say to first simplify the equation to

    (vf^2) = -(2GM/ri) + (2GM/rf) + (vi^2) ,

    then calculate 2GM = 7.97732x10^14 and hold onto it so you don't have to re-calculate it (since you'll use it often). (I'm wondering if you're getting different results each time because you're using this number to different levels of precision each time; I ran into that...)

    It looks like you converted everything consistently into SI. I'm gettng vf = 735 m/sec for part (a).

    I get something close to your answer for part (b), but you should work to higher precision to see the difference between this answer and the 1000 km. altitude that was used in part (a). Two sig-figs often doesn't cut it for celestial mechanics problems...
     
    Last edited: Oct 3, 2007
  6. Oct 3, 2007 #5
    But I am solving for K_f not V_f...but I can use your v_f to check K... it comes close at 4.05*10^7...I am sure you rounded somewhere. I did not....I carried all sig figures throughout.
     
  7. Oct 3, 2007 #6

    dynamicsolo

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    Ah, so you are! But it's really not hard for me to go from the vf^2 I got to Kf -- I have 4.053*10^7 J, so you're OK so far (besides, I'm not s'poseta tell you the answer to the question ;-) ).

    If you got different results each time keeping all sig-figs, I'm guessing there were input differences or errors on different trials. (I did keep all the sig-figs, too, but I hadn't used the same precision on 2GM every time -- once I did, the result was reproducible.)
     
    Last edited: Oct 3, 2007
  8. Oct 3, 2007 #7

    Dick

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    Hey, I get 4.10*10^7J. But I'm using slightly different values for mass and radius of the earth.
     
  9. Oct 3, 2007 #8
    Cool. I was using the givens....

    Any luck on Max height?
     
  10. Oct 3, 2007 #9

    Dick

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    Is that one coming up wrong too?
     
  11. Oct 3, 2007 #10

    dynamicsolo

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    I have a value, but you first... ;)

    (A suggestion there -- this is why I worked with vf^2: you already found vi^2 and
    (-2GM/ri), so just hold onto them; set vf appropriately and you can quickly extract rf.
     
    Last edited: Oct 3, 2007
  12. Oct 3, 2007 #11

    Dick

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    Used your givens and got 3.78*10^7J. Do I really have to check the second one too??
     
  13. Oct 3, 2007 #12

    Dick

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    Ah, dynamicsolo will do it for me. Won't you????
     
  14. Oct 3, 2007 #13
    You don't have to do anything. Like I said, I wanted to solve it independentl of part a. And I don't have the solutions to these...so how would I know if they're wrong?! That is the point of a check:rolleyes:

    See post #1...it's been there from the start.

    I held GMm....it was exact.
     
    Last edited: Oct 3, 2007
  15. Oct 3, 2007 #14

    dynamicsolo

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    Sorry, my ancient, creaky computer at home suddenly stopped accessing the 'Net and I had to re-boot (twice). I used R_E = 6.378*10^6 m , M_E = 5.98*10^24 kg , and
    G = 6.67*10^-11. I find in these types of problems that results can be pretty sensitive to small differences in precision, especially at the low-energy end of trajectories.

    What are you using for G? (I'm not seeing it on your list...)
     
  16. Oct 3, 2007 #15

    dynamicsolo

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    I meant your full value for r_f before you rounded it to two sig-figs.
    I have 7.415*10^6 m, giving a peak altitute of h_f = 1037 km.
     
  17. Oct 3, 2007 #16
    Ahhhhh...tricky, it did not occur to me to remove the earth radius...


    r_f=7.376*10^6

    so
    r_f-r_e=1006km
     
  18. Oct 3, 2007 #17

    dynamicsolo

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    All right, just to show the sensitivity to levels of precision, I redid the calculations using four sig-figs for all the constants (instead of only 3 for some in my earlier results):

    R_E = 6.378*10^6 m (equatorial radius; mean radius is 6371 km -- should we prefer this?);
    M_E = 5.974*10^24 kg;
    G = 6.673*10^-11 (in SI units) ,

    giving

    2GM = 7.9729*10^14 ;
    v_f = 740.0 m/sec , so K_f = 4.107*10^7 J (part a); and
    r_f = 7.416*10^6 m or h_f = 1038 km (part b).

    So your numbers were all right, but one really has to be careful about chosen values and precision.
     
    Last edited: Oct 3, 2007
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