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Geosynchronous Satellite

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A synchronous satellite, which always remains above the same point on a planet's equator, is put in orbit around Jupiter to study that planet's famous red spot. Jupiter rotates once every 9.84 h. Use the following data to find the altitude of the satellite above the surface of the planet. Jupiter has a mass of [itex]1.90 \cdot 10^{27}~kg[/itex], and a mean radius of [itex]6.99 \cdot 10^{7}~m[/itex].


    2. Relevant equations
    [itex]v= \large \sqrt{ \frac{GM_j}{R_j + h}}[/itex] [itex]M_j[/itex] mass of Jupiter; [itex]R_j[/itex] average radius of Jupiter.

    [itex]v= \large \sqrt{ \frac{GM_j}{r}}[/itex]

    [itex]v_{tan}=r\omega[/itex]

    3. The attempt at a solution

    I know that in order for the satellite to continually be suspended above the same spot on Jupiter, they have to be rotating through the same angles and take the amount of time to rotate through those angles. Hence, [itex]\omega_j=\omega_s= \frac{2\pi}{35424~s}[/itex] (I converted the hours to seconds).

    I thought of using the first forumula; but when I substituted
    [itex]v_{tan}=r\omega[/itex] and tried to solve for r, it became rather difficult. And so, I opted to use the second equation and perform the same steps. I solved for [itex]r[/itex], [itex]r= (\large \frac{GM_j}{\omega^2})^{1/3}[/itex]

    After the final substitution, [itex]r= 1.59 \cdot 10^8 m [/itex]. This, however, isn't the correct answer. What did I do wrong?
     
  2. jcsd
  3. Jan 27, 2013 #2

    Doc Al

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    Staff: Mentor

    You solved for r. What altitude does that correspond to?
     
  4. Jan 27, 2013 #3
    Oh, I need to subtract the average radius from the value of r I found. Also, I just noticed that the mass of Jupiter was given in kg, should I have converted it into grams?
     
  5. Jan 27, 2013 #4

    Doc Al

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    Staff: Mentor

    Right.
    No. Kg is the standard unit for mass.
     
  6. Jan 27, 2013 #5
    You are asked about the altitude above the surface of Jupiter.
     
  7. Jan 27, 2013 #6
    I got the proper answer. Thanks be to you both.
     
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