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Finding the Satellite Orbital Radius

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    If NASA wants to put a satellite in a circular orbit around the sun so it will make 2.0 orbits per year, at what distance (in astronomical units, AU) from the sun should that satellite orbit? The earth's orbit is 1.0 AU from the sun.

    [itex]T_E = \frac {1.0orbit}{yr}[/itex]
    [itex]T_s = \frac {2.0orbit}{yr}[/itex]
    [itex]r_E = 1.0AU[/itex]

    2. Relevant equations
    [itex]\frac {T_E^2} {r_E^3} = \frac {T_s^2} {r_s^3} => r_s = \sqrt[3]{\frac {T_s^2 r_E^3}{T_E^2}}[/itex]

    3. The attempt at a solution
    Using Kepler’s 3rd law of planetary motion, we know that the orbital period squared is proportional to the radius of the orbit cubed. Since the proportion is approximately the same for all objects orbiting relative to the Sun, then: [itex]\frac {T_E^2} {r_E^3} = \frac {T_s^2} {r_s^3}[/itex]

    Solving for the radius of the satellite and plugging in the values, we get:
    [itex]r_s = \sqrt[3]{\frac {T_s^2 r_E^3}{T_E^2}} = \sqrt[3]{\frac {(\frac {2.0orbit}{yr})^2 (1.0AU)^3}{(\frac {1.0orbit}{yr})^2}} = \sqrt[3]{4.0AU^3} = 1.5874AU \sim 1.6AU[/itex]

    Thank-you
     
    Last edited: Oct 25, 2014
  2. jcsd
  3. Oct 25, 2014 #2

    ehild

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    The satellite makes 2 orbits per year, what is its orbital period?
     
  4. Oct 25, 2014 #3

    SteamKing

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    The further away from the sun, the longer the orbital period. That's Kepler's (unpublished) Fourth Law.
     
  5. Oct 25, 2014 #4
    [itex]T_s = \frac {2 orbits} {1yr} x \frac {2\pi} {1 orbit} x \frac {1yr} {365days} x \frac {1day} {24hrs} x \frac {1hr} {3600s} = \frac {\pi} {7.884x10^6s}[/itex]
    [itex]G = 6.67x10^{-11} Nm^2/kg^2[/itex]
    [itex]M_\odot = 1.9885x10^{30} kg[/itex]
    [itex]1 AU = 1.4959x10^{11} m[/itex]

    [itex]T_s^2 = \frac {4\pi^2 r^3}{GM_\odot} \Rightarrow r = \sqrt[3]{\frac {T_s^2 GM_\odot}{4\pi^2}} = 81.102m x \frac {1 AU}{1.4959x10^{11} m} = 5.4216x10^{-10}AU[/itex]
     
  6. Oct 25, 2014 #5

    mfb

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    You don't need meters. The approach in post 1 was fine, you just plugged in the wrong orbital period.
    And no, an orbital radius of a few meters is certainly wrong (this would be inside the sun!!).

    T is the duration of an orbit, not the number of orbits per time (that is the inverse value).
     
  7. Oct 25, 2014 #6

    ehild

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    2 total orbits last one year, how long time is one orbit? (in years)
     
  8. Oct 25, 2014 #7

    SteamKing

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    5.4216*10-10AU works out to be about 81 meters. Does this sound like a reasonable answer?

    If you make 2 orbits per year, the orbital period is going to be a half year for each orbit.

    You know that the earth's orbital period = 1 year (it's kind of a definition) and you have Kepler's third law which relates orbital distance to orbital period.

    Try again, and put a little thought into your calculations.
     
  9. Oct 25, 2014 #8
    For the satellite, one orbit (in years) is half a year. Thus,

    [itex]T_E = 1yr[/itex]
    [itex]T_s = 0.5yr[/itex]
    [itex]r_E = 1.0AU[/itex]

    [itex]r_s = \sqrt[3]{\frac {T_s^2 r_E^3}{T_E^2}} = \sqrt[3]{\frac {(0.5 yr)^2 (1.0 AU)^3}{(1 yr)^2}} = \sqrt[3]{0.25 AU^3} = 0.62996 AU \sim 0.63 AU[/itex]

    In order for the orbital period to decrease, the object must be closer to the Sun. 0.63AU is closer than the Earth and has a shorter orbital period (0.5 yr), therefore, numerically the result corresponds to the aforementioned reasoning.

    Using the second method:
    [itex]T_s = 0.5 yr x \frac {365 days} {1 yr} x \frac {24 hrs} {1day} x \frac {3600 s} {1 hr} = 1.5768x10^7 s[/itex]
    [itex]G = 6.67x10^{-11} Nm^2/kg^2[/itex]
    [itex]M_\odot = 1.9885x10^{30} kg[/itex]
    [itex]1 AU = 1.4959x10^{11} m[/itex]

    [itex]T_s^2 = \frac {4\pi^2 r^3}{GM_\odot} \Rightarrow r = \sqrt[3]{\frac {T_s^2 GM_\odot}{4\pi^2}} = 9.4177x10^{11} m x \frac {1 AU}{1.4959x10^{11} m} = 0.62957 AU \sim 0.63 AU[/itex]

    The answers from both methods match.

    Thank-you
     
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