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Orbiting satellite

  1. Jun 28, 2005 #1
    When in orbit, a communication satellite attracts the earth with a force of 16.7 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.43*10^11 J. Find the satellite's altitude above the earth's surface. The radius of the earth is 6.38*10^6.

    OK, I must be making this harder than it needs to be. What I've been trying to do is to use the formulas for gravitational force to get an equation with two unknow variables (Mass of the satellite and height above earth's surface) And I do the same for gravitational potential energy. Then, since both equations have the same two unknown variables, I solve for one of them and substitute. Is there another way of doing this? Am I doing it copmletely wrong? Please help me!!! :confused:
  2. jcsd
  3. Jun 28, 2005 #2
    You sound like you have the right idea. Give it a shot.
  4. Jun 29, 2005 #3
    Total Energy of a satellite revolving around earth is given by:

    [itex]- \frac{GMm}{2r}[/itex]


    Note:This post has been edited after Older Dan's remarks.
    Last edited: Jun 29, 2005
  5. Jun 29, 2005 #4


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    There is no 2 in the potential energy. Perhaps you meant the total energy

    [itex] U = - \frac{GMm}{r} [/itex]

    [itex] T = \frac{1}{2}mv^2 = \frac{r}{2} \left( \frac{mv^2}{r} \right) = \frac{r}{2} \left| F_c \right| = \frac{r}{2} \left( \frac{GMm}{r^2} \right) = \frac{GMm}{2r} = -\frac{1}{2} U [/itex]

    [itex] E\ \ =\ \ T\ \ +\ \ U = \frac{GMm}{2r}\ \ -\ \ \frac{GMm}{r}\ \ =\ \ - \frac{GMm}{2r} [/itex]
  6. Jun 29, 2005 #5
    You are doing the right thing ninjagowoowoo, you should be able to find both the mass and height of the satellite with this method. In fact, this question was answered on this forum a long time ago (well it's in the archive)


    Hope that helps!
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