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Missile's trajectory problem

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A missile's trajectory takes it to a maximum altitude of 1200 km.
    If its launch speed is 6.2 km/s , how fast is it moving at the peak of its trajectory?

    2. Relevant equations

    K+U = K_o+U_o

    U(r) = -GMm/r

    3. The attempt at a solution

    Alright some people have been telling me its zero. Intuitively thats what I thought, but
    I guess its wrong. Nevertheless, here is my attempt :

    K+U = K_o +U_o;

    =

    1/2mV^2 - GM_eM/r_d = 1/2mV_o^2 - GM_em / r_e

    and solving for
    V^2 = V_o^2 - 2GM_e/r_e + GM_e/r_d

    where,
    V_o = 6.2km -> 6.2*1000m
    G = universal graity = 6.67*10^-11
    M_e = mass of earth = 5.98*10^24
    r_e = radius of earth = 6.37 * 10^6;
    r_d = distance between the object and the center of earth
    = 1200km -> 1200*1000m

    and my answer is terms of meters
    ~2400m

    so dividing by 1000

    and

    v_p ~ 24m/s
     
  2. jcsd
  3. Mar 14, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Gravity

    Why would you think that? (It may be moving horizontally at the peak.)

    Looks good.

    You left off a factor of 2 in that last term.

    1200km is just the altitude, not the distance to the center of the earth.
     
  4. Mar 14, 2009 #3
    Re: Gravity

    so is this correct?

    Although my formula missed a factor of 2, I still did the calculation with the 2 factored in.
    Can you check my calculations?
     
  5. Mar 14, 2009 #4

    Doc Al

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    Staff: Mentor

    Re: Gravity

    Did you correct your value of r_d?
     
  6. Mar 14, 2009 #5
    Re: Gravity

    no, I am not sure why I should. The units wont match.

    I converted everything into meter and the final answer into km
     
  7. Mar 14, 2009 #6

    Doc Al

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    Staff: Mentor

    Re: Gravity

    The issue is not units, but that you are using r_d = 1200 km. That's the altitude; you need the distance to the center of the earth.
     
  8. Mar 14, 2009 #7
    Re: Gravity

    so is the distance 1200km + earth's radius?
     
  9. Mar 14, 2009 #8
    Re: Gravity

    how do I get the distance to the center of the earth of the object?
     
  10. Mar 15, 2009 #9

    Doc Al

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    Staff: Mentor

    Re: Gravity

    Yes, that's r_d.
    See above.
     
  11. Mar 15, 2009 #10
    Re: Gravity

    So the distance is 1200 * 1000m + earth's radius
    because
    1) It they are in both meters.

    2) Because the projectory is launched from the earth's surface, or top of the earth?
     
  12. Mar 15, 2009 #11

    Doc Al

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    Staff: Mentor

    Re: Gravity

    Good
     
  13. Mar 15, 2009 #12
    Re: Gravity

    Cool, thanks!
     
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