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Stuck on projectile motion problem using gravitational force

  1. Apr 30, 2017 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired vertically from the Earth's surface with an initial speed of 11.4 km/s. Neglecting air drag, how far (in meters) above the surface of the Earth will it go?

    2. Relevant equations
    (1/2)mv^2
    -GmM/R

    3. The attempt at a solution
    0083083ec8.png
    KEi+PEi = KEf+PEf
    (1/2)mv^2 - GmM/R = -GmM/(R+h)
    R+h = -(GM)/((1/2)v^2-GM/R)
    h = -(GM)/((1/2)v^2-GM/R) - R
    and plugging everything in, I got -1.744x10^8
    So I tried using kinetic and potential energy to solve for h, but keep getting a negative number. Am I doing this wrong?
     
  2. jcsd
  3. Apr 30, 2017 #2

    gneill

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    Staff: Mentor

    Have you learned about the implications of the sign of the total mechanical energy value?

    Try this in your conservation of energy equation: Rather than assuming that the final KE goes to zero, suppose that the value of h grows large enough that the PE tends to zero. What value does that yield for the final KE?
     
  4. Apr 30, 2017 #3
    I don't think we've learned about that yet, and how would I leave final KE in? I don't think I'm supposed to find the final velocity
     
  5. Apr 30, 2017 #4

    gneill

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    The idea is to investigate another possible scenario.

    When you set the final KE to zero you are making the assumption that the object will eventually slow to zero before falling back. What if the velocity never never reaches zero? What will be the maximum height then? So instead of assuming that the KE goes to zero, assume that the PE goes to zero instead. How will your equation look?
     
  6. Apr 30, 2017 #5
    would have
    (1/2)(11400^2)-((6.67*10^-11)(5.98*10^24))/(6.371*10^6)=(1/2)vf^2
    which would make vf = 2178.762
    but I don't see where I can get the h variable from if i set potential final energy to zero
     
  7. Apr 30, 2017 #6

    gneill

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    Good.
    Well, if the velocity never, ever, goes to zero, what can you say about h?
     
  8. Apr 30, 2017 #7
    It never stops, infinity? but the question is looking for a specific number
     
  9. Apr 30, 2017 #8

    gneill

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    Right, it never stops. There's no specific (finite) value for h. h tends to infinity. That's the best answer you can give.

    Some things to investigate:
    Look up the concept of escape velocity. You should be able to find or compute a value for the escape velocity from the surface of the Earth. Compare it to your given value for the initial velocity.

    Compute the specific mechanical energy for the object: That's your energy formula without the mass of the object in question so that you end up with the units joules per kg: ##ξ = \frac{v_i^2}{2} - \frac{GM_e}{r_o}##. There are three cases:

    1) If ξ is negative then the object is "bound": it will never escape. Its speed will eventually reach zero at some finite distance then fall back to Earth.

    2) If ξ is precisely zero it has exactly escape velocity. Its speed will never reach exactly zero, but it will approach zero in the limit. Again, it will never fall back and the "final" distance is infinity.

    3) If ξ is a positive value then again the object will escape, and moreover, its velocity will approach some value greater than zero as its distance goes to infinity (this "final" velocity is sometimes called the excess velocity).
     
  10. Apr 30, 2017 #9
    so i did sqrt(2GM/R) and plugged in to v12 and got

    ξ = 63816956.64
     
  11. Apr 30, 2017 #10

    gneill

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    You need to include the potential energy, too.
     
  12. Apr 30, 2017 #11
    got 99716437393501.805
     
  13. Apr 30, 2017 #12

    gneill

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    Staff: Mentor

    Okay, first, it's a bit silly to quote so many digits. Use exponential notation when the numbers go over 103.
    Second, what calculation did you do to include the PE? If you started with escape velocity I would expect your ξ to end up at or very close to zero.
     
  14. Apr 30, 2017 #13
    I used the ξ equation and plugged the escape velocity in vf2
     
  15. Apr 30, 2017 #14

    gneill

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    But there's no ##v_f## in:

    ##ξ = \frac{v_i^2}{2} - \frac{GM_e}{r_i}##
     
  16. Apr 30, 2017 #15
    i meant V initial
     
  17. Apr 30, 2017 #16

    gneill

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    Something is amiss. Can you write out the values you're using for each of the variables?
     
  18. Apr 30, 2017 #17
    ξ=(sqrt(2(6.67*10-11 * 5.98*1024))^2/2 - (6.67*10-11 * 5.98*1024)/6.371*106
     
  19. Apr 30, 2017 #18

    gneill

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    You forgot to divide by R in the first term. The escape velocity is given by

    ##v_{esc} = \sqrt{\frac{2 G M}{R}}##
     
  20. Apr 30, 2017 #19
    i got -6.26*10^7
     
  21. Apr 30, 2017 #20
    actually nvm forgot to square it, i got 0
     
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