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A projectile is shot straight up from the Earth's surface

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A projectile is shot straight up from the earth's surface at a speed of 1.10×10^4 km/hr



    2. Relevant equations
    mgh=.5mv^2


    3. The attempt at a solution
    I converted the speed to m/s and 3055.5556 m/s
    masses cancel out so i get 9.8h=.5v^2
    I plugged in v and solved for H and got 476347.95m, which was wrong.
    I'm not sure if this is how you are supposed to solve this problem, but i can't think of any other ways.
     
  2. jcsd
  3. Apr 16, 2013 #2
    "mgh" works only for low altitudes.
     
  4. Apr 16, 2013 #3
    Ok, what would I use then?
    What other gravitational equations have height in them?
     
  5. Apr 16, 2013 #4
    Newton's law of gravitation?
     
  6. Apr 16, 2013 #5
    Newton's law of gravitation is Fg=GmM/r^2. or Ug= GmM/r
    how would I get height out of that?
     
  7. Apr 16, 2013 #6
    What is "r"?
     
  8. Apr 16, 2013 #7
    radius and I'm not given any masses. the m would cancel out if I do KE=PE, but the M wouldnt
     
  9. Apr 16, 2013 #8
    Radius from where?
     
  10. Apr 16, 2013 #9
    r would be radius from center of the earth to the projectile.
    so would I do GmM/R=.5mv^2?
    G=6.673x10^-11
    M=5.972x10^24
    solve for R and subtract by radius of earth?
     
  11. Apr 16, 2013 #10
    That would be correct. Note that you could find out what GM is by considering the magnitude of the force of gravity right at the surface of the Earth.
     
  12. Apr 16, 2013 #11
    I converted 1.10x10^4 km/hr to 3055.55555m/s and solved for R and got 85328725.7m and then I subtracted by the Earth Radius which is 6371000m and got 78955725.7 as the final answer but that is still wrong
     
  13. Apr 16, 2013 #12

    gneill

    User Avatar

    Staff: Mentor

    The projectile begins at a certain radius, so it has some initial gravitational potential. It ends up at some new distance with another gravitational potential. The KE from the launch is exchanged for this change in gravitational PE.

    So you should write an expression for the change in gravitational PE and relate it to the change in KE.
     
  14. Apr 16, 2013 #13
    Excuse me for butting in but isn't there a relationship something like E = G(m1)(m2)(1/(r1) -1/(r2))

    or you can use calculus to integrate E = Int(G(m1)(m2)/r^2) from the surface of the earth r1 to r2
     
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