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Satellite Motion - This doesnt seem right

  1. Apr 15, 2008 #1
    Satellite Motion / Universal Gravitation - This doesnt seem right...

    1. The problem statement, all variables and given/known data
    How far is a 1Kg / 10N wieght, from the earths surface, when it is 5N?

    2. Relevant equations
    g= GM/r^2

    3. The attempt at a solution
    => W=mg
    => 5=1g
    g = 5

    g= GM/r^2
    =>5 = (6.67*10^-11)(1)/r^2

    => r= 3.56 x 10^-6 ????

    Could someone please assist me on where did I go wrong, because I dont think I missed any important details.
    Last edited: Apr 15, 2008
  2. jcsd
  3. Apr 15, 2008 #2
    As far as I can tell, I think the problem is that you used the wrong mass in the formula
    [tex] \mbox{g}=\frac{GM}{r^2}[/tex]
    You used the mass of the weight, but it should be the mass of the earth, which is approximately [tex]5.977\times10^{24} \mbox{ kg}[/tex]. The equation should be as follows:
    [tex] 5 = \frac{(6.67\times10^{-11})(5.977\times10^{24})}{r^2}[/tex]

    [tex] r = \sqrt{\frac{(6.67\times10^{-11})(5.977\times10^{24})}{5}} [/tex]

    [tex] r = 8.93\times10^6 m [/tex]

    Does that agree with the answer you have been given?
  4. Apr 15, 2008 #3

    You helped me solve this issue. Can you believe I made such a small mistake - Ridiculous!

    I really do appreciate your help.

    I think I need to think and take more care with my work! :)

    Kindest Regards,
  5. Apr 15, 2008 #4


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    Science Advisor

    You also need to be careful about what r represents. In your formula, r is the distance from the center of the earth. The problem asks for the distance from the earth's surface.

    You don't actually need to do all that calculation. You know that the object weighs 10 N at the earths surface: GM/R^2= 10. You are looking for r such that GM/r^2= 5. Dividing the first equation by the second, r^2/R^2= 2 so r= [itex]\sqrt{2}[/itex] R.

    Again, that r is distance from the center of earth. Since R is the radius of the earth, the distance from the surface of the earth is [itex]\sqrt{2}[/itex]R- R= ([itex]\sqrt{2}[/itex]- 1)R.
  6. Apr 15, 2008 #5
    My apologies, HallsofIvy is absolutely right. The answer I gave you was incomplete.

    The formula calculated the distance from the centre of the earth to the object, however the question asks for the distance from the surface of the earth to the object. This means we must subtract the earth's radius from the answer we calculated, as this will give the distance from the centre of the earth.

    [tex] r_{surface} = r - r_{earth} [/tex]
    [tex] r_{surface} = (8.93\times10^6)-(6.38\times10^6)[/tex]

    Object is [tex] 2.55\times10^6[/tex] metres away from the earth's surface. Does that make sense?
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