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Total energy of a geosynchronous satellite

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the total energy of a geosynchronous satellite (one that orbits over a fixed spot)
    with a mass of 1.5 x 10^3 kg, orbiting Earth st a height of 325 km with an orbital speed of 5.0 x 10^3 m/s.


    2. Relevant equations
    Ek = 1/2 mv^2
    Ep = mgh
    Fg = Gm1m2/r^2
    or
    PE = -Gm1m2/r

    3. The attempt at a solution
    Im wondering if a should find the force of gravity using Fg = Gm1m2/r^2 of the satellite and then plug that into Ep = mgh and add Ek = 1/2mv^2 (Ep + Ek = Et). OR if i need to use PE = -Gm1m2/r
     
  2. jcsd
  3. Jan 24, 2012 #2
    im wondering because i get different answers for potential energy

    Fg=Gm1m2/r^2
    =((6.67e-11 N(m^2)/kg)(5.98e24 kg)(1.5e3 kg))/(6.705e6)^2
    = 13308.24 N
    Ep= mgh
    = (13308.24 N)(325000 m)
    =4.325e9 J
    Ek=1/2mv^2
    =1/2(1.5e3kg)(5.0e3 m/s)^2
    = 1.875e10 J <<<<<<<<<<<<<<< wondering about these units here
    Et= 4.325e9 J + 1.875e10 J
    = 2.3075e10 J

    This is how i answered. Is this correct?
     
  4. Jan 24, 2012 #3

    gneill

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    Staff: Mentor

    mgh for potential energy only applies for h << rearth, that is, for objects close to the surface of the Earth. You should use the Newton's Law form for potential energy for larger separations.
     
  5. Jan 24, 2012 #4
    using the Fg = Gm1m2/r^2 i have basically found gravity at around 8.87 N at 325000 m and then used this for the Ep=mgh calculation. This will not work, I could understand this. When i use the PE = -Gm1m2/r i get such a large negative number the sum of PE & Ek is negative i dont understand this ?????
     
    Last edited: Jan 24, 2012
  6. Jan 25, 2012 #5

    gneill

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    Staff: Mentor

    Potential energy represents work required to bring an object from some arbitrary reference point to a given location. The formula [itex] -G\frac{M}{r}[/itex] carries the assumption that the arbitrary reference point is at infinity. When we do mechanics near the surface of the Earth we tend to choose a reference point for convenience, often the surface of the Earth itself. In this case the work required to bring an object from the reference point to a higher elevation than the reference height is positive, and thus we say that the potential energy is positive.

    Just keep in mind that potential energy is always measured with respect to some arbitrary reference point or height.

    For this problem, if you wish to make your potential energy reference point the surface of the Earth when using the Newton formula, simply take the difference between the potential at the surface of the Earth (at radius r0) and the desired location:

    [itex] PE = GMm \left[ \frac{1}{r_0} - \frac{1}{r}\right] [/itex]
     
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