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Homework Help: Orbital Period

  1. Feb 25, 2015 #1
    In the far future(10^85 years) an “element” called positronium will develop with a diameter of
    the current observable universe of 93 billion light years. (Remember that light travels at 3 × 10^8 m/s). This element consists of an electron and a positron, both of which have a mass 9.11 × 10^−31 kg orbiting a common center of mass. The force between them is given by Coulombs law F = Ke^2/r^2 where K = 8.99 X 10^9Nmi^2/C^2, e = 1.602X10^-19C, and r is the distance between particles. How long does it take for these particles to orbit each other? I.e. what is their orbital period?

    Edit: For the sake of physics homework, we can suspend our knowledge of atomic structure and just treat this as a question about orbital period.

    Have I posted in the right section?
    Last edited: Feb 26, 2015
  2. jcsd
  3. Feb 26, 2015 #2


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    Gold Member

    No, I think this belongs in the science fiction section since it isn't physics.

    Also, in the far future, the size of the observable universe will not be the same as it is now (but that's just a quibble and can be fixed by adding the word "current" to your "diameter of ... "

    An "element" with the diameter of current observable universe isn't even decent science fiction, it's just silly.

    [strike]An electron and a positron will not make up an element.[/strike]EDIT: well, apparently I'm wrong on that. Your "positronium" is said to be an atom.

    An element would be an atom and the electron(s) in an atom does not have a circular orbit the way you are thinking of.
    Last edited: Feb 26, 2015
  4. Feb 26, 2015 #3


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    Staff: Mentor

    If the fantastic back story is ignored and a flat, Newtonian universe is assumed, how would you go about solving the problem? What are the basic elements of the problem?
  5. Feb 26, 2015 #4
    Ok, here is a shot.

    93bill/2= 46.5 billion x 3x10^8 (speed of light) x 365 days x 24 hours x 3600 seconds = r = 4.399x10^26

    F = Ke^2/r^2 = 8.99x10^9 x (1.602x10^-19)^2 / 4.399x10^26 = .00327N

    F-MxV^2/r = .00327 = 9.11x10^-31 x V^2 / r

    v^2 = (square root of) ((.00327 x 4.399x10^26) / 9.11x10^-31) = 1.25x10^/4 m/s

    Circumference = pi r^2 = 3.14 x (4.399 x 10^26)^2 = 1.381 x 10 ^ 53 m

    P = v/c = 1.25 x 10 ^-4 / 1.381 x 10^53 = 9.049 x 10^-58 s

    How am I looking with all that?
  6. Feb 26, 2015 #5


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    Staff: Mentor

    Your value for the force seems highly unlikely: very much too large. You failed to square the radius in the denominator.
    Your formula for the circumference is not correct, which should be obvious because it yields units of square meters.
  7. Feb 27, 2015 #6
    Ok, another shot.

    F = Ke^2/r^2 = 8.99x10^9 x (1.602x10^-19)^2 / (4.399x10^26)^2 = 3.27 x 10^-81

    C = PI X 2R = 3.14 x (2X (4.399 x 10^26)) = 2.76 X 10^ 27M

    v^2 = (square root of) ((3.27 X 10^-81 x 4.399x10^26) / 9.11x10^-31) = 1.256 X 10^-43

    P = V/C = 1.256 X 10^27 / 2.76 X 10^27 = 4.55 X 10 ^-17 s

    Getting warmer?
  8. Feb 27, 2015 #7


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    Don't you expect a P of several (tralatrillion)4 years instead of 46 attoseconds ?

    The distance between the particles isn't 93bill/2 but 93bill

    The force is still wrong if I correct for that. re-calculate ?
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