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Calculating satellite's angular speed at perihelion

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A satellite is in elliptical orbit with a period of 8.00 x 10^4 s about a planet of mass 7.00 x 10^24 kg. At aphelion, at radius 4.5 x 10^7 m, the satellite's angular speed is 7.158 x 10^-5 rad/s. What is its angular speed at perihelion?


    2. Relevant equations

    V(orbital) = sqrt(GM/R)

    T=2∏sqrt(R^3/GM)

    V=ωr

    3. The attempt at a solution

    I used Kepler's 3rd law to find the semimajor axis of the satellite:

    T=2∏sqrt(R^3/GM)
    8.00 x 10^4 s = 2∏sqrt(R^3/(6.67 x 10^.11)(7 x 10^24 kg))

    and found that the semimajor axis (R) is 4.23 x 10^7 m

    I then reasoned that twice the semimajor axis minus the aphelion distance = the perihelion distance, so

    2(4.23 x 10^7 m) - 4.5 x 10^7 m = perihelion distance

    Perihelion distance (r) = 3.96 x 10^7 m

    I then used the value of the radius at perihelion (r) in the V(orbital) = sqrt(GM/r) expression and found:

    V(orbital) = sqrt[((6.67 x 10^-11)(7 x 10^24 kg))/(3.96 x 10^7 m)]
    V(orbital) = 3433.65 m/s

    I then plugged this into V=ωr:

    3433.65 m/s = ω x 3.96 x 10^7 m
    ω = 8.67 x 10^-5 rad/s

    However, the answer should be 9.29 x 10^-5 rad/s. My answer key shows that the conservation of angular momentum was used to find the answer. However, I don't see why my approach doesn't also work. It seems to yield a similar answer, but just far off enough to make me think that the approach is flawed. Any help would be greatly appreciated!
     
  2. jcsd
  3. Nov 13, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    The equation you used V(orbital) = sqrt(GM/R) is valid for R as the radius of curvature, which is equal to the radius of orbit only for circular motion.

    Use conservation of momentum.

    ehild
     
  4. Nov 13, 2011 #3
    Ok, will do. Thanks!
     
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