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Velocity of a satellite.

  1. Aug 19, 2011 #1
    I have a question that says the velocity of an orbiting satellite is 28620 km/hr. Calculate the height in km of the orbit above the earths surface. Gravity is 9.81, and earths radius is 6370000. What formulas do I need and how do I work out the height?
     
  2. jcsd
  3. Aug 19, 2011 #2

    Doc Al

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    What force acts on the satellite?
    What is the acceleration of the satellite?

    Apply Newton's 2nd law.
     
  4. Aug 19, 2011 #3

    PeterO

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    That 9.81 is the gravitational field strength at the surface, but what is its value up where the satellite is orbiting.

    If you knew that you could use the standard expressions for centripetal acceleration.
     
  5. Aug 19, 2011 #4
    That was all the information I was given on the question. To know the gravitational pull up on the satellite, I need to know the height. So Im stuck.
     
  6. Aug 19, 2011 #5

    PeterO

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    I didn't say there was a gravitational pull up on the satellite, I was referring to the gravitational field strength when you get up there.

    I suppose you could assume it is actually still 9.81 [even though it won't be] and see what you get for R, or you could use the relationship between Gravitational Potential Energy vs kinetic energy of a satellite in stable circular orbit.

    This page could contain some useful information and formulae

    http://www.sparknotes.com/testprep/books/sat2/physics/chapter11section3.rhtml
     
  7. Aug 19, 2011 #6

    Doc Al

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    I suspect that it was meant as a generic comment, like use g = 9.81 m/s^2. Of course, it's irrelevant for this particular problem.
     
  8. Aug 19, 2011 #7

    gneill

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    You may want to verify the given values in the problem statement, the velocity of the satellite in particular. The value looks a bit high to me for a satellite in a circular orbit above the Earth.
     
  9. Aug 19, 2011 #8

    PeterO

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    ??? That speed is appropriate for a satellite somewhere between the ISS and Hubble ??

    Escape velocity is closer to 40,000 km/hr as I recall. You didn't think the value was given in miles per hour did you?
     
  10. Aug 19, 2011 #9

    gneill

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    One could use this value of g and the given radius of the Earth to determine g(r) via the inverse square law.
     
  11. Aug 19, 2011 #10

    Doc Al

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    That's true. There are many ways to skin the cat.
     
  12. Aug 19, 2011 #11

    D H

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    I second that notion. Per my calculations, the satellite is orbiting at an altitude of -71 kilometers.
     
  13. Aug 19, 2011 #12

    Ray Vickson

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    You say the earth's radius is 6370000. What are the units? If they are km, that is way too large. Always, always state your units.

    RGV
     
  14. Aug 20, 2011 #13
    Hi Bobby. I'm new to the forums so I hope this response is Kosher. Ok so you have been given a speed for the orbiting satellite, lets call it Vsat. Ask your self the following two questions:

    1. Is Vsat a constant value? (does the speed ever change?)
    2. Is the path of the orbit circular in geometry?

    If you answer yes to both of the questions, start thinking about uniform circular motion (UCM) and the acceleration associated with that type of motion. Where is the acceleration usually directed for an object subjected to UCM.

    Now someone earlier brilliantly mentioned using netwon's 2nd law of motion; F = ma
    Ponder this in regard to the acceleration associated with UCM.

    Also think about the type of force that results in the above mentioned acceleration.

    Remember we are dealing with a satellite-earth system (one really big particle AND one comparatively smaller particle). These particles are seperated by a distance r, where r = radius of earth + height of satellite above earth surface.

    r = (Re + h)

    One last thing. Consider this inverse-square law:

    F= G(Mm)/r^2

    **where G is a constant
    **M is mass 1
    **m is mass 2
    **r is distance

    PM me if you need anymore help.
     
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