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Gravitational Potential Energy of a projectile

  1. Apr 10, 2009 #1
    A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.

    (a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is one-third of the escape speed from Earth?

    (b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is two-fifths of the kinetic energy required to escape Earth





    U = -GMm/r
    K = 1/2mv^2
    Radius of earth = 6.38e6
    Mass of earth = 5.98e24




    (a) Escape speed of earth = 11.2 km/s
    =11200 m/s

    U1 + K1 = U2
    -GMm/r + 1/2mv^2 = -GMm/(r+h)

    m cancels out
    11200/3 = 3733

    (-6.67e-11 * 5.98e24) / 6.38e6 + 1/2 * 3733^2 = (-6.67e-11 * 5.98e24) / (6.38e6 + h)
    -55550537.32 = -3.98866e14 / (6.38e6 + h)
    h = 800236.5781

    h / Re = 800236.5781 / 6.38e6 = 0.125 times the radius of the earth

    (b)
    U1 + K1 = U2
    -GMm/r + 1/2mv^2 = -GMm/(r+h)

    but it asks for 2/5 the KE, so

    -GMm/r + (2/5)1/2mv^2 = -GMm/(r+h)
    m cancels out

    -GM/r + 1/5 v^2 = -GM / (r+h)
    Plugged in the same numbers as part (a), except for v, which is 11200 m/s

    I got h, divided by Radius of earth (6.38e6) and get 0.670 times radius of earth


    Did I make any mistakes? Because the answers are wrong, apparently
     
  2. jcsd
  3. Apr 10, 2009 #2
    Well, at least the eq looks right for me. The only thing I could think of, is that h is counted also from the center of earth. So instead of .125, h is 1.125?
     
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