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Satellite and Earth

  1. Feb 19, 2007 #1
    How much work is required to lift a 1000-kg satelitte to an altitude of 2*10^6 m above the surface of the earth? The gravitational force is F = GMm/r^2, where M is the mass of the earth, m is the mass of the satellite, and r is the distance between them. The radius of the earth is 6.4*10^6m, its mass is 6*10^24kg, and in these units the gravitational constant, G, is 6.67*10^-11.

    I know that the formula of work is W = F * d

    d is distance. What is r here? The distance between the satellite and earth? Is it the 2*10^6 and what is d here? Thanks
     
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  3. Feb 19, 2007 #2

    HallsofIvy

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    Work= F*d if F is a constant. When F is a variable, you need
    [tex]W= \int f(x)dx[/tex]

    Here,
    [tex]W= 6.67*10^{11} \int_{6.4*10^5}^{8.4*10^6}\frac{Mm}{r^2}dr[/tex]
     
  4. Feb 19, 2007 #3
    M and m are also constant right? so we can pull that both outside from the integral??
     
    Last edited: Feb 19, 2007
  5. Feb 19, 2007 #4

    hage567

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    Yes, the masses are constant.
     
  6. Feb 19, 2007 #5
    why did I get the answer key from the text book:

    I got GMm * ((-1/8*10^6) + (1/6.4*10^6))

    is that right??
     
  7. Feb 19, 2007 #6

    hage567

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    Do you mean this is the answer in the book? Or this is the answer you got and it doesn't match the book?
     
  8. Feb 19, 2007 #7
    this is the answer I got and it didn't matches the book
     
  9. Feb 19, 2007 #8

    hage567

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    What's the answer in the book? How much are you off by?
     
  10. Feb 19, 2007 #9
    the answer in the book is 1.489*10^10 and I got 1.042*10^10
     
  11. Feb 19, 2007 #10

    hage567

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    I think you might be making a math error. I got the same answer as the book.
     
  12. Feb 19, 2007 #11

    hage567

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    One thing I see is that in your answer you have the height of the satellite as 8x10^6, when it should be 8.4x10^6. I don't think it is really going to change it too much though.
     
  13. Feb 19, 2007 #12
    where did you get 8.4 from?? there isn't 8.4 in the question
     
  14. Feb 19, 2007 #13

    hage567

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    In your question, you have 6.4x10^6 m as the radius of the Earth. Add the 2x10^6 m altitude of the satellite to that, and you get 8.4x10^6 m as your upper limit.

    You have to take all distances from the centre of the Earth.
     
    Last edited: Feb 19, 2007
  15. Feb 19, 2007 #14
    okay the problem is solved if I change it to 8.4*10^6
     
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