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Satellite in Orbit (Quiz#4, 26)

  1. Oct 2, 2014 #1
    • Warning: Infraction points awarded to gcombina for not using the homework template.
    A satellite in orbit around the earth has a period of one hour. An identical satellite is placed in an orbit having a radius that is nine times larger than that of the first satellite. What is the period of the second satellite?

    (a) 27 h (c) 4 h (e) 0.04 h

    (b) 3 h (d) 9 h


    ** I truly don't know where to start.
    I have seen (online) people using equation for force of gravity, and centripetal acceleration but they dont explain why

    **Please do not use Keplen laws cause I havent studied that

    My attempt:
    The period is the distance of the orbit which is 2piR
    The first satellite moves because it has centripetal force which is Fc= mv^2/R
    The first satellite also has acceleration which is ac = v^2/R
     
  2. jcsd
  3. Oct 2, 2014 #2

    BvU

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    What is causing the Fc ? You need an expression for that, preferably one that also has R in it.
    Use the template. Read the guidelines. Don't write "I don't know where to start".
     
  4. Oct 2, 2014 #3

    SteamKing

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    It's Kepler's Laws, not Keplen.
     
  5. Oct 3, 2014 #4
    the Fc is caused by gravity
     
  6. Oct 3, 2014 #5

    ehild

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    And what is the expression for the force of gravity?

    ehild
     
  7. Oct 3, 2014 #6
    it is GMn/R^2

    but what should I do with it if I am looking for the TIME
     
    Last edited: Oct 3, 2014
  8. Oct 3, 2014 #7

    ehild

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    How is time related to speed and distance travelled ?

    What distance travels a satellite along its whole orbit?
     
  9. Oct 3, 2014 #8

    BvU

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    Ah, good to see you now have an expression for the force that functions as centripetal force. And it has an R in it too! Splendid! You can work it around to get something for v. And travelling a circular orbit of radius R brings you back to the same point in a time ....

    By the way, doesn't anyone of the helpers worry about the 1 hour period of the first satellite ? I got something of an underground orbit... ? Fortunately, the exercise can be continued nevertheless: IF the R satellite circles in 1 hour, THEN the 9R satellite does so in ... hours.​
     
  10. Oct 5, 2014 #9
    sorry i am lost

    you asked what is the relation between STD? speed time and distance?

    T = d/s
     
  11. Oct 5, 2014 #10

    BvU

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    We're getting there! What is d for one revolution in a circular orbit of radius R?

    [edit] Ah, I see you answered that already in post # 1: 'the distance of the orbit which is 2piR'

    @ehild: can't we assume circular orbits? It's already difficult enough for GC I would guess...
     
    Last edited: Oct 5, 2014
  12. Oct 5, 2014 #11

    ehild

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  13. Oct 5, 2014 #12
    yeah i read Kepler which is more simple
    t^2/R^3 = T^2/(9R)^3
    T = 27

    BUT, they dont want me to use Kepler's law
     
  14. Oct 5, 2014 #13

    BvU

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    Goodness. Post 1 said no Kepler. And it has R and 9R, so I take it they want you to derive this for circular orbits. You have ##GMm/R^2=mv^2/R## , you have ##T = d/v## and you have ##d = 2\pi R##. Play around until you have ##T=...##!

    @ehild: what about this 1 hour period infeasibility ?
     
  15. Oct 5, 2014 #14

    ehild

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    Read the last paragraph. It solves the problem with circular orbits.
     
  16. Oct 5, 2014 #15
    ok sorry, i suck at physics but need to pass this only course which is a prereq for my major which has nothing to do with physics!. Please be patient
     
  17. Oct 5, 2014 #16

    BvU

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  18. Oct 5, 2014 #17
    ill read that
     
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