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Satellites again

  1. Apr 3, 2005 #1
    another problem bout satellites...

    to exit the solar system, the Pioneer spacecraft used a gravitational assist from jupiter, which increased its kinetic energy at the expense of Jupiter's kinetic energy. If the spacecraft did not have assist, how far out in the solar system would it travel? when it left Earth's vicinity, the spacecraft speed's relative to the sun, was 38km/s.
    r=2GM/v^2
    =2(6.673x10^-11N m^2/kg^2)(5.98x10^24kg)/(38000m/s)^2
    r=552694.4598m

    shoudld i use the mass of the sun coz the answer in my book is 7.9x10^11m
    i dont know wat to do, im so confused...
     
    Last edited: Apr 3, 2005
  2. jcsd
  3. Apr 3, 2005 #2

    xanthym

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    TWO (2) changes are required (assuming spacecraft of mass "m" applies no additional engine power & isn't significantly influenced by other astronomical entities):
    a) Use Sun's mass for "M".
    b) Use change in Potential Energy from {Earth's distance from Sun "Dearth"} to the {Unknown final distance from Sun "Dfinal"}:
    (1/2)*m*(vinitial)2 = -G*m*M*{(1/Dfinal) - (1/Dearth)} = G*m*M*{(1/Dearth) - (1/Dfinal)}
    Solve for "Dfinal".


    ~~
     
    Last edited: Apr 3, 2005
  4. Apr 3, 2005 #3
    xanthym, i dont know the {Earth's distance from sun "Dearth"}. can u give me the value for that?
     
    Last edited: Apr 3, 2005
  5. Apr 3, 2005 #4

    xanthym

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    If your textbook doesn't provide {Earth's distance from Sun "Dearth"}, you can use the following:
    Dearth = 1.496e(11) meters


    ~~
     
  6. Apr 3, 2005 #5
    pioneer

    u know that the Pioneer has no given mass?? so what will i put to the m in 1/2 mv^2 nad GmM(1/r - 1/r)?
     
    Last edited: Apr 3, 2005
  7. Apr 3, 2005 #6

    xanthym

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    Spacecraft's mass "m" cancels out from both sides of equation.


    ~~
     
  8. Apr 3, 2005 #7
    ohh yeahhhh... i forgot
     
  9. Apr 3, 2005 #8
    thanks so much for the help, xanthym!! hope u can help me again... heheh =)
     
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