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Homework Help: Geosynchrously orbiting satellites

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data

    [PLAIN][PLAIN]http://i77.photobucket.com/albums/j64/mrbebu/Physics_orbit_problem.jpg [Broken]

    2. Relevant equations


    v = [tex]\sqrt{\frac{GM}{R}}[/tex]

    v = wR

    3. The attempt at a solution

    Am I correct in my methods and thinking?

    R_g = Geosynchrous orbit from earth's center = 4.22 x 10[tex]^{7}[/tex]m
    R_e = Radius of earth = 6.37 x 10[tex]^{6}[/tex]

    velocity of geosynchrous orbits (same for both satelites) --->
    v = [tex]\sqrt{\frac{GM}{R_g}}[/tex] = 3072 m/s

    Then I decided to find the angular velocity to relate it with angular displacement--->
    w = [tex]\frac{v}{R_g}[/tex] = 7.28 x 10[tex]^{-5}[/tex] rad/s

    so 10 orbits in a geosynchrous orbit is 240 hours = 864000 seconds

    The satellite that needs to catch up needs to complete 10.5 orbits in the same time.

    10.5 orbits = 21[tex]\pi[/tex] radians ---> [tex]\bar{w}[/tex] = [tex]\frac{d\theta}{dt}[/tex] = 7.64 x 10[tex]^{-5}[/tex] rad/s

    Ratio of the w's =
    [tex]\frac{7.28 x 10^{-5}}{7.64 x 10^{-5}}[/tex] = 0.95

    since w is inversely proportional to the radius, the satellite that needs to catch up will need to have a radius 0.95 times a geosynchrous one which --->

    0.95 x 4.22 x 10[tex]^{7}[/tex] = 4.01 x 10[tex]^{7}[/tex] m from earth's center

    But, i have a feeling that this is an incorrect assumption.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 3, 2007 #2
    I've done another way that is close to my previous answer, but i think is more accurate.

    using v = [tex]\frac{2\pi r}{T}[/tex]

    and substituting it in for v in--->

    v[tex]^{2}[/tex] = [tex]\frac{GM}{r}[/tex] ---> and that gives us --->

    T[tex]^{2}[/tex] = [tex]\frac{4\pi^{2}r^{3}}{GM}[/tex]

    so using the geosync satellite, we know that one period, T, takes 24 hours = 86400 seconds. And one T = 2[tex]\pi[/tex]r traveled

    But for the chasing satellite to catch the geo satellite in 10 orbits, it needs to travel and extra [tex]\frac{1}{2}[/tex] orbit during the full 10 orbits which = [tex]\pi[/tex] radians.

    so the chasing sat needs to cover an extra [tex]\frac{\pi}{10}[/tex] radians during the same 86400 second T. Therefore --->

    2[tex]\pi[/tex] + [tex]\frac{\pi}{10}[/tex] = [tex]\frac{21\pi}{10}[/tex]

    instead of [tex]\frac{2\pi r}{T}[/tex] , we use:

    ([tex]\frac{21\pi r}{10T}[/tex][tex])^{2}[/tex] (squaring the whole expression) = [tex]\frac{GM}{r}[/tex]

    Then solving for r = [tex]\sqrt[3]{\frac{100T^{2}GM}{441\pi^{2}}}[/tex]

    which = 4.09 * 10[tex]^{7}[/tex] meters from center of earth.

    I'm thinking this is the right answer
    Last edited: Nov 3, 2007
  4. Nov 3, 2007 #3

    D H

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    Staff Emeritus
    Science Advisor

    Good so far, but you still haven't answered the problem. How much lower is this orbit than a geosynchronous orbit?
  5. Nov 3, 2007 #4
    I'm not worried about that part. That's the easy part:wink: I'm worried about the actual radius i got for the new orbit. Good so far i guess means that it is correct. Which are you refering to, first solution or second?
  6. Nov 4, 2007 #5

    Doc Al

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    Staff: Mentor

    OK, but now express R in terms of [itex]\omega[/itex]:

    [tex]R^3 = \frac{G M}{\omega^2}[/tex]

    OK. Interpreting the problem as 10 geosynchronous orbits (it could also be 10 orbits of the faster satellite), I'd write it as:
    [tex]\omega_g T = 10(2\pi) = 20 \pi[/tex]

    I'd write that as:
    [tex](\omega_1 - \omega_g)T = \pi[/tex]

    By combining the two previous equations, I'd view that as:
    [tex]20\omega_1 = 21\omega_g[/tex]

    There's the problem with this solution: [itex]\omega[/itex] is not inversely proportional to the radius.
  7. Nov 4, 2007 #6

    D H

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    Staff Emeritus
    Science Advisor

    The second answer is correct. As Doc Al noted in the previous post, the first solution erroneously assumes "w is inversely proportional to the radius". The correct relationship is Kepler's third law,

    [tex]r \sim T^{2/3}[/tex]

    You can linearize this relationship as

    [tex]\frac {\Delta r}{r} \approx \frac 2 3 \frac {\Delta T}{T}[/tex]

    Thus decreasing the period by 1/21st (note well: not by 1/20th) of the original period corresponds to roughly a 1340 kilometer decrease in orbital radius at geosynchronous altitude. This is a lot closer to the correct value than the 2100 kilometers you obtained for your first solution. It's still not correct, as the approximation is just that -- an approximation.
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