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Homework Help: Synchronous orbit with Saturn

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Satellite in synchronous orbit with equator of Saturn. Find altitude in km.

    rotational period = 38826 s
    radius of Saturn = 6.03E7 m
    mass of Saturn = 5.68E26 kg
    Gravity constant = 6.673E-11

    2. Relevant equations

    T^2 = (4pi^2)(r^3)/(Gm)

    3. The attempt at a solution

    I plugged and chugged and arrived at an answer of 113114516 m. Then I subtracted the planet's radius from this answer to find the altitude of 52797562 m. I then concluded that the synchronous orbit had to have an altitude of 5.28E4 km. I also solved the problem using F(g) = F(c) and arrived at the same solution. What am I doing wrong? Are my constants incorrect?
  2. jcsd
  3. Jun 4, 2009 #2


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    There is an interesting version of Keplers law (the equation you have) based on the density of the body.
    this is possible since you have r^3 and the mass in the equation anyway.
    One interesting result of this is that the orbital period at the surface of a body only depends on the density - for water it's a period of 3.3hours.

    What's also interesting about the density of Saturn?
  4. Jun 4, 2009 #3

    D H

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    The only things you are doing wrong are (a) expressing your answer with too much precision, and (b) writing G without units (G is not just a number. Change your units and you get a different numerical value for the speed of light -- and for G.)

    That said, your answers are correct. What makes you think you they are wrong?
  5. Jun 4, 2009 #4
    Vol of sphere = 4/3[tex]\pi[/tex]r[tex]^{3}[/tex]

    so density of sphere = vol(mass)

    Orbital period at surface? Isn't that just the rotational period?
  6. Jun 4, 2009 #5
    My "incorrect" answer was actually correct. I switched the rotational period to 10.23 hrs and came up with an answer that the online homework accepted. I certainly wasn't using significant digits...thanks for the reminder! BTW, I just left out the units for G to ease the typing requirements...lazy on my part, my bad!

    The density of Saturn should be less than water, right? It's a gaseous planet...
  7. Jun 4, 2009 #6


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    Sorry - I thought that for Saturn the geostationary orbit would be below the surface.
    I had used GM for Earth (400,000km - a useful figure to know) and got an orbital radius of 24,000km - which is below the surface.

    That was the reason for the point about density and water, I assumed that if the rotation speed at the surface of water is 3.3hr and Saturn is less dense than water it would be less, and so less than the 10hrs needed. I had the reasoning the wrong way around - sorry.
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