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Planetary Motion with satellite

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data

    A 20 kg satellite has a circular orbit with a period 2.4 h and radius 8.0×106m around a planet of unknown mass. If the magnitude of the gravitational
    acceleration on the surface of the planet is 8.0 m/s2, what is the radius of the
    planet?

    2. Relevant equations

    F = Gm1m2/r2
    Fc = m1v2/r
    v=D/T, D=[itex]\pi[/itex]r

    3. The attempt at a solution

    So I got the velocity of the the satellite by using speed=distance/time where the distance is the circumference of the orbit or [itex]\pi[/itex]r and T is the period in seconds. I'm having trouble because there is an unknown mass of the planet and its radius. How do I go about finding this values? Any help is much appreciated.
     
  2. jcsd
  3. Apr 30, 2014 #2

    patrickmoloney

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    Gold Member

    This is from the 2011 paper, it's actually a lot easier than it looks.

    T = period ## T = (2.4)(3600) = 8640 s ##

    Okay so I did this question using Kepler's second law (law of periods). It's a very useful law and is worth learning.

    <content deleted -- complete worked solutions are not permitted> -- gneill, PF Mentor
     
    Last edited by a moderator: Apr 30, 2014
  4. Apr 30, 2014 #3
    I thought the radius value was RTotal = RSatellite + RPlanet.

    Also using your method I got r = 5.3x106. Did you use g=9.8 or is there some other difference between or values? Also thanks for the help.
     
  5. Apr 30, 2014 #4

    patrickmoloney

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    Gold Member

    It's not the acceleration of the earth its the acceleration of the planet ## a_g = 8.0 m/s^2 ## So lets break this down, we use Kepler's second law, using a little bit of algebra this relation provides is with an approximation of the mass of the planet. Having the mass of the planet and the acceleration of the planet we have the radius of the planet.

    ## ∑F = ma_g = \frac{GMm}{r^2} ## which implies ## a_g = \frac{GM}{r^2} ##
     
  6. Apr 30, 2014 #5
    What is the radial acceleration of the satellite at 8x106 m? You know the radial acceleration at this location, you know radial acceleration at the surface, and you know that the radial acceleration is inversely proportional to the square of the radius.

    Chet
     
  7. Apr 30, 2014 #6
    You're not supposed to solve the problem. Just explain how the solution works.
     
    Last edited by a moderator: Apr 30, 2014
  8. Apr 30, 2014 #7
    Ok thanks for clearing that up and all the help guys
     
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