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Centripetal motion and universal gravitation question: Mars and Sun question

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Mars travels around the Sun in 1.88 (Earth) years, in an approximately circular orbit with a radius of 2.28 x 10^8 km
    Determine
    a) the orbital speed of Mars (relative to the Sun)
    b) the mass of the Sun


    2. Relevant equations

    acceleration centripetal = 4(pi^2)(r) / (T^2)
    universal attraction = (G)(m1)(m2) / (radius)^2


    3. The attempt at a solution

    Given:
    T = 1.88 years = 59287680s = 5.9x10^7s
    r = 2.28 x 10^8km = 2.28 x 10^11m


    Orbital speed means centripetal acceleration yes?
    Then,

    acceleration centripetal = 4(pi^2)(r) / (T^2)
    acceleration centripetal = 4(9.869604401)(2.28x10^11m) / (34.81x10^14s)
    acceleration centripetal = 90.01x10^11m / 34.81x10^14s
    acceleration centripetal = 2.59x10^-3m/s = 0.00259m/s

    How come its so slow?
     
  2. jcsd
  3. Mar 1, 2009 #2

    Doc Al

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    Staff: Mentor

    No. Speed is distance divided by time, not acceleration.
     
  4. Mar 1, 2009 #3
    Ok so,
    Mars travels around the Sun once every 1.88 years. So I have to find the distance of that circle and divide it by the time to find the speed. I have the radius, so I have to find the perimeter.

    Perimeter of circle is 2(pi)(r)
    So the distance traveled is 14.33x10^11m

    In 1.88 years = 5.9x10^7s
    So the speed is 14.33x10^11m / 5.9x10^7s = 2.43x10^4m/s yes?
     
  5. Mar 1, 2009 #4

    Doc Al

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    Staff: Mentor

    Good.
     
  6. Mar 1, 2009 #5
    Now can I use this formula to solve for mass of Sun?
    V^2 = (G)[m1(sun)] / r
    [m1(sun)] = (V^2)(r) / (G)
    [m1(sun)] = (2.43x10^4m/s)^2(2.28 x 10^11m) / [6.67x10^-11(N)(m^2)/kg^2]
    [m1(sun)] = (5.9x10^8m^2/s^2)(2.28 x 10^11m) / [6.67x10^-11(N)(m^2)/kg^2]
    [m1(sun)] = (13.45x19^19m^3/s^2) / [6.67x10^-11(N)(m^2)/kg^2]
    [m1(sun)] = 2.02x10^8(N)m/kg^2

    Why are the units all weird?
     
  7. Mar 2, 2009 #6

    Doc Al

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    Staff: Mentor

    Good.
    In your last step you didn't divide the units properly. You should have gotten:
    [m^3/s^2]/[(N)(m^2)/kg^2] = [m^3/s^2]*[kg^2/(N)(m^2)] = [m kg^2]/[N s^2]

    To simplify further, express Newtons in terms of more fundamental units:
    N = kg-m/s^2
     
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