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Speed of the satellite

  1. Nov 14, 2016 #1

    Kajan thana

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    1. The problem statement, all variables and given/known data
    A satellite orbiting the Earth moves to an orbit which is closer to the earth.
    What happens to the speed and the time it takes for the one orbit of the Earth.

    2. Relevant equations

    It is circular motion so I will use

    a=ω^2(r) a=(2π/T)^2/r

    3. The attempt at a solution
    If the r is decreasing, then the value of a is also decreasing this means that period will increase as it is inversely proportional.

    Can someone help me out?
     
  2. jcsd
  3. Nov 14, 2016 #2

    TSny

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    In the first equation, r is in the numerator on the right-hand side. But in the second equation you have moved r into the denominator. Did you mean to do that?
    Can you justify this statement?
     
  4. Nov 14, 2016 #3

    Kajan thana

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    a is proportional to r by the omega constant. If I increase a then r will also increase.
     
  5. Nov 14, 2016 #4

    TSny

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    Is it correct to assume that ω is constant when you switch orbits?
     
  6. Nov 14, 2016 #5

    Kajan thana

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    yes, the omega will change also but I am still clueless.
     
  7. Nov 14, 2016 #6

    TSny

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    It will help to consider where the acceleration is coming from. When the satellite is in a circular orbit, what force acts on the satellite?
     
  8. Nov 14, 2016 #7

    Kajan thana

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    Gravity is acting in between them two.
     
  9. Nov 14, 2016 #8

    Kajan thana

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    I don't know if I am on the right line.
     
  10. Nov 14, 2016 #9

    TSny

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    Yes, the force acting on the satellite is the force of gravity. If the satellite is moved closer to the earth, does the force of gravity on the satellite increase or decrease?
     
  11. Nov 14, 2016 #10

    Kajan thana

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    Increases, is ma is proportional to 1/r^2
     
  12. Nov 14, 2016 #11

    TSny

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    Yes. Good.
     
  13. Nov 14, 2016 #12

    Kajan thana

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    how will I still find out if the period is increasing or decreasing?
     
  14. Nov 14, 2016 #13

    TSny

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    Look at your equation a = ω r. [EDIT: Meant to write a = ω2 r] If you switch to an orbit closer to the earth, what happens to r? What happens to a? What happens to ω?
     
    Last edited: Nov 14, 2016
  15. Nov 14, 2016 #14

    Kajan thana

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    yes but like what you said the r is not a constant.
     
  16. Nov 14, 2016 #15

    Kajan thana

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    or is the R constant after it has moved closer.
     
  17. Nov 14, 2016 #16

    TSny

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    Using the formula a = ω2 r, you should be able to figure out if ω increases or decreases when the satellite is moved into an orbit closer to the earth.
    Does the left side of the equation get larger or smaller?
     
  18. Nov 14, 2016 #17

    Kajan thana

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    Left hand side get larger, I thought in order to look at proportionality something need to remain constant, isn't that the case?

    And also I want make sure if we are allowed to relate MA with gravity because my teacher told me MA can be only used when it is contact with something else.
     
  19. Nov 14, 2016 #18

    TSny

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    In the equation a = ω2 r, all three quantities a, ω, and r change when moving to an orbit closer to the earth. Nevertheless, you can still use the equation to deduce what happens to ω.

    Gravity is an "action-at-a-distance" force. It can act on an object even when the object is not in contact with any other object. The satellite is a good example. When it is in orbit, it is not in contact with any other object. Yet, there is the force of gravity acting on the satellite.
    F = ma will apply to this situation.
     
    Last edited: Nov 14, 2016
  20. Nov 14, 2016 #19

    Kajan thana

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    It makes sense.
    Thanks for the help.
     
  21. Nov 15, 2016 #20
    If it helps then try thinking about the extremes
    Try thinking about or solving the velocity required to orbit the Earth a meter above the surface and the velocity required to orbit the Earth far out like 1000 km away
    Compare the two results and see what relationship they hold
     
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