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Velocity of earth's orbit?

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Given that the earth's distance from the sun varies from 1.47 to 1.52x10^11m, determine the minimum and maximum velocities of the Earth in it's orbit around the sun.


    2. Relevant equations

    [itex] F=G\frac{m1m2}{r^2} [/itex]

    [itex] E=K+U [/itex] ???

    3. The attempt at a solution

    I think the way to do this is with K1+U1 = K2+U2 , where one side of the equation is the earth at its closest point to the sun and the other side is the earth at its farthest point. Let Me = mass of earth, Ms = mass of sun, Rn = distance at nearest point, Rf= distance at farthest point, Vn = velocity at nearest point, Vf = velocity at farthest point.

    [itex] K1+U1 = K2 + U2 [/itex]

    [itex] \frac{MeVn^2}{2} + G\frac{MsMe}{Rn} =\frac{MeVf^2}{2} + G\frac{MsMe}{Rf} [/itex]

    the Me's cancel. to solve for Vn replace Vf with [itex] \frac{2piRf}{T} [/itex]

    [itex] \frac{Vn^2}{2} + G\frac{Ms}{Rn} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} [/itex]

    [itex] \frac{Vn^2}{2} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn} [/itex]


    [itex] Vn^2 = 2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}) [/itex]

    [itex] Vn = (2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}))^\frac{1}{2} [/itex]

    then after plugging in I would go back and solve for Vf. Would this give me the correct answer?
     
  2. jcsd
  3. Apr 10, 2014 #2

    BiGyElLoWhAt

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    well, one thing I want to say is that the earth is revolving, not translating. You don't want 1/2 mv^2, you want 1/2 Iw^2 for your KE, and I'm sure Mr^2 is good enough for I. other than that I think you're on the right track.
     
  4. Apr 10, 2014 #3

    BiGyElLoWhAt

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    ... I was just looking at your eq. again, and I'm not sure what you're doing with the 2pi*r/T, that is w, but I don't think you're looking at the KE in this situation correctly.
     
  5. Apr 10, 2014 #4
    Ok cool thanks. What does Iw represent?
     
  6. Apr 10, 2014 #5
    I was using the equations for circular motion where v = 2(pi)r/T . I am only vaguely sure of what I am doing though.
     
  7. Apr 10, 2014 #6

    BiGyElLoWhAt

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    ok, looking again (again)... that'll get you the tangential velocity, but not angular, I'm not sure what you're after exactly.

    Also, I think it's good practice to start with the angular and convert over, seeing as you have KE as a result of rotation, not translation. I think that will get you the correct result, though (this time).
     
  8. Apr 10, 2014 #7

    BiGyElLoWhAt

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    Iw is moment of inertia * angular velocity
     
  9. Apr 10, 2014 #8
    what exactly is the angular velocity? Im not sure If I am up to that yet in my physics class. The section this problem was from was covering universal gravitation as well as kepler's laws. The question just says find the maximum and minimum velocities it doesn't specify between angular or tangential. The KE I am using wont work?
     
  10. Apr 10, 2014 #9

    BiGyElLoWhAt

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    well angular velocity is the rate of revolution measured in radians/seconds, or also commonly 1/seconds (radians are funny that way). w = 2pi/T =v/r, so what you're doing will work I think.
     
  11. Apr 10, 2014 #10

    BiGyElLoWhAt

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    just out of curiosity is it a general physic course?
     
  12. Apr 10, 2014 #11
    Suppose it was a perfectly circular orbit around the sun, at the average distance of 1.495x10^11 m. Would you know how to determine the tangential velocity then (say using F = ma)? What value would you get for the tangential velocity?

    Chet
     
  13. Apr 10, 2014 #12

    vela

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    Using ##V_f = \frac{2\pi R_f}{T}## isn't correct. This assumes that the Earth covers a distance of ##2\pi R_f## in one year, which would be true if it followed a circular path of radius ##R_f##, but the Earth covers less distance because it moves toward the Sun.

    You have two unknowns, so you're going to need another equation. You used conservation of energy already (though the sign of the potential energy is wrong). What's another conserved quantity you can use?
     
  14. Apr 10, 2014 #13

    vela

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    This is completely wrong. You've misread the problem.
     
  15. Apr 10, 2014 #14
    I think energy is the only conserved quantity I know so far. The only one that definitely stays the same over time. Also, do you mean another equation involving velocity?
     
  16. Apr 10, 2014 #15

    ehild

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    What about angular momentum?

    ehild
     
  17. Apr 10, 2014 #16

    vela

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    It's a central force, so angular momentum is conserved.
     
  18. Apr 11, 2014 #17
    I havent learned about momentum yet. There is no other way to solve this?
     
  19. Apr 11, 2014 #18

    vela

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    Use F=ma to get another relationship between V and R.
     
  20. Apr 11, 2014 #19

    ehild

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    You certainly know Kepler's Laws. What do the first two say?

    ehild
     
  21. Apr 11, 2014 #20
    do i have to use T^2 = r^3
     
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