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Low altitude satellites orbiting around planet with twice radius

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Two remote planets consist of identical material, but one has a radius twice as large as the other. IF the shortest possible period for a low altitude satellite orbiting the smaller planet is 40 minutes, what is the shortest possible period for a similar low altitude satellite orbiting the larger one? Answer in minutes.


    2. Relevant equations
    volume of sphere=4/3 ∏r3
    speed=distance/time
    Ek=-1/2 Eg


    3. The attempt at a solution
    rB=2rA

    timeA=2∏r/VA
    40=2∏r/VA

    Don't really know how to move on from here? I know I have to find the time taken for the satellite to orbit planet B.

    TB=2∏(2rA)/VB

    I don't know VB, how would I find that?
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2013 #2

    PeterO

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    The larger planet will have a much larger mass [same material presumably means same density]

    Radius of orbit around the larger planet is larger [just a little larger than the planet itself]

    I would be looking at the effect of doubling the radius of orbit of the satellite around the smaller planet, then the effect of having a larger mass attracting the satellite.
     
  4. Apr 10, 2013 #3
    So I doubled the radius. I didn't notice their densities would be the same since they're made of the same material. Anyways:

    EKA=-1/2 EGA
    1/2 mvA2=-1/2(-GmMA/rA)
    vA2=GMA/rA

    and

    EKB=-1/2 EGB
    1/2 mvB2=-1/2(-GmMB/2rA)
    vB2=GMB/2rA
     
    Last edited: Apr 10, 2013
  5. Apr 10, 2013 #4

    PeterO

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    I was looking for an answer like: If you double the radius (while retaining the planet mass), you double/halve/quadruple/quarter the period.

    You would then look at: When you double/treble/quadruple/etc the mass of the planet (independent of the radius), you (some change) the Period.

    You then combine those effects:

    eg if one change means halving, and the other means increasing by a factor of 12; then net result is an increase by a factor of 6.
     
  6. Apr 10, 2013 #5
    Volume of B=4/3 ∏r3
    =4/3∏(2rA)3
    =4/3∏(8rA3)
    =8(4/3∏rA3)
    =8 volume of A

    So, planet B's volume is 8 times of A's....
     
  7. Apr 10, 2013 #6

    PeterO

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    I would be addressing the expression a = 4π2.R / T2
    one of the more useful expressions relating to the acceleration of a body moving in a circle.
     
  8. Apr 10, 2013 #7

    PeterO

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    Given they are made of the same material (imagine if you were using two steel balls to model the situation) what effect would that have on the mass.
     
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