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Gravitation and Orbiting Satellites

  1. Dec 9, 2007 #1
    1. Two satellites are orbiting around Earth. One satellite has a period of 1.4hours and
    is 200km above Earths surface. The other satellite has a period of 6.0h. Use Kepler's laws
    and the fact the radius of Earth is 6.37 x 10^6meters to determine the height of the second satellite above Earths surface.

    2. Keplers Third Law: (Ta/Tb)^2 = (Ra/Rb)^3
    G=6.67x10^-11 Nm^2/kg^2
    Msv2/r = GMeMs/r^2

    3. Ok first I know that Fg is equal to Fc, but I dont know the equation to find out Fc.
    So I just decided to attempt to find out the period but dividing 6hours by 1.4hours and using that number (4.29 rounded) multipled it with 200km and got 858km. I then used
    858km as the height of the 2nd satellite above the Earths surface. I also calculated the velocity of satellite one, which is 7.79 x 10^3 m/s (or just 7.8m/s) Now, since I am doing a long distance course and not good at all at algrebra, and the lack of good example question within the textbook and coursebook to go on, I stumped. The answer I have cannot be worth 5 marks from the work Ive shown, but I just cant find any way to prove how my answer is correct with so little to go on.
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2
    You will only need to use Kepler's Third Law.
  4. Dec 9, 2007 #3

    D H

    Staff: Mentor

    You were given the orbital periods for both satellites, the altitude of one of the satellites, and the radius of the Earth. That information plus Kepler's third law are all you need to solve the problem.
  5. Dec 9, 2007 #4
    So Ta=1.4hours and Tb=6hours
    What I got after doing the first part of the equation is 0.054

    Now Ra and Rb are pretty much the radius of the Earth, and when I finished
    that equation all I got on my calculator was 1E36.
    ...These numbers dont really seem like the height of the second satellite no matter how
    I look at it and add/subtract/multiple/divide them together.
  6. Dec 9, 2007 #5
    Are you using r in meters and T in seconds?
  7. Dec 9, 2007 #6

    D H

    Staff: Mentor


    Not so good.

    Solve for Ra (you don't use Kepler's laws here) and then solve for Rb (here you do use Kepler's law).

    That's not necessary here. The left-hand and right-hand side of the expression [tex](T_a/T_b)^2 = (R_a/R_b)^3[/tex] are unitless. If you so wished, you could even express length in furlongs and time in fortnights with this question, so long as all lengths are expressed in furlongs and all times are expressed in fortnights.

    What is important in this question is that all similar items (i.e., all lengths) be expressed in the same units. Here we have the radius of the Earth in meters and height in kilometers. Those units must be made commenserate with one another.
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