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Calculating the Orbital Period

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the Orbital Period for the following: Earth's orbit to the Sun. I get stuck towards the end. I must use Newton's second law and have the Universal Law of Gravitation equal the Centrifugal Force.
    Newton's gravitational constant: G= 6.67*10^-11 Nm^2/kg^2
    Mass of Sun = 1.98*10^30 kg
    Mass of the Earth 5.97*10^24 kg
    Distance of the Earth from the Sun: 149.6*10^6
    T = time


    2. Relevant equations
    Centrifugal Force = (mv^2)/r
    v=(2(pi)(r))t
    Force of Gravity = (GMm)/d^2
    Mm being mass, and G/d^2 being acceleration.


    3. The attempt at a solution
    I multiplied Newton's gravitational constant by the mass of the Earth by the mass of the Sun, and then divided it all by the distance of the Earth from the Sun. I got 5.270280882*10^35 km. This was for the gravitational pull.

    For the Centrifugal Force, I multiplied the mass of the Earth by v=((2(pi)(r))^2)/t^2. And put it over 149.6*10^6 km. I got (3.52229083*10^34 kg*km)/t^2.

    I then set my two results equal to each other. Next, I multiplied by t^2 as that is the variable I am trying to find. This gave me 5.270280882*10^35 km*t^2 = 3.5229083*10^34 kg*km. Then, I divided by km on both sides to cancel it. I now have 5.270280882*10^35 t^2 = 3.5229083*10^34 kg. If I divide by the number on the left side of the equation and then square root both sides, I get a number than can't possibly be the Orbital Period of the Earth.


    Can you tell where I am going wrong? Any advice is appreciated.
    Thanks.
     
  2. jcsd
  3. Mar 8, 2010 #2

    mgb_phys

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    Centrifugal Force = (mv^2)/r
    Force of Gravity = (GMm)/d^2

    For an orbit these two are equal so you have
    mv^2/r = GMm/r^2

    The mass of the Earth cancels, orbit's don't depend on the mass of the small object - that's why a spaceman and a space shuttle can float along in the same orbit.

    v^2/r = GM/r^2

    Work out v in terms of the circumference and the period ( v= 2 PI r/t) , do a bit of rearranging and you're there.
     
  4. Mar 8, 2010 #3
    OK, thanks. I will try this and let you know how it goes.
     
  5. Mar 8, 2010 #4
    ((4(pi^2)(149.6 km^2))/t^2)/149.6 km=(6.67*10^-11(1.98*10^30 kg))/149.6 km^2

    I do not understand what to do. If I cancel the 149.6 km from the denom. on the left side, I can't go much further.
     
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