1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orbital Frequency problem

  1. Feb 25, 2008 #1
    1. A positron is an elementary particle identical to an electron except that its charge is
    + e. An electron and a positron can rotate about their center of mass as if they were a dumbbell connected by a massless rod.

    What is the orbital frequency for an electron and a positron 1.70 nm apart? (Answer in Hz)

    2. If someone could just define orbital frequency for me or point me in the direction of the right type of equation to use... I'm lost on this one and it's not in our textbook.

    Possible relevant equation - electric field of a sphere of charge

    E = Q/4[tex]\pi[/tex][tex]\epsilon\underline{}0[/tex]r[tex]\overline{}2[/tex]

    that should read Q divided by (4 times pi times epsilon times r squared)

    3. The fact that the positron has a +e charge but is identical tells me that the mass is still the same.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 26, 2008 #2
    Think about a reduced mass system for say, planetary motion. Orbital frequency is 1/orbital period, which is how long it takes to make a full revolution. A more generael Kepler's law could help you out.
  4. Feb 26, 2008 #3
    I also am stuck on this exact problem

    Although the distance between the two particles is different in my question, we have the exact same problem.

    I was thinking of approaching this problem as a dipole since the problem states the particles "can rotate about their center of mass as if they were a dumbbell connected by a massless rod."

    Also, I know that dipoles rotate when in electric fields, where the torque [tex]\tau[/tex] = pEsin([tex]\theta[/tex]) where E is the field strength, [tex]\theta[/tex] is the angle made by the dipole and field and p = qs where q is the charge and s is the distance between the two charges on the dipole. But you need an angle [tex]\theta[/tex] for this and we aren't given one. Additionally, there is no mention of the field so I think this idea is out.

    Could you say that the positron is fixed at the origin and the electron moves in uniform circular motion around it? That way, you know that E = mv[tex]^{2}[/tex]/(qr). And, you also know that E = kq/r[tex]^{2}[/tex]. Solving for v, you could then find frequency because v = 2 [tex]\pi[/tex]rf.

    Do I sound like I'm on the right track?
    Last edited: Feb 26, 2008
  5. Feb 26, 2008 #4
    That sounds like a good way to solve the problem. I am not sure if you can just assume the positron is fixed at the origin. It sort of sounded like they were both rotating around the same center of mass, but I wasn't sure. Let me know if you come up with anything.
  6. Feb 27, 2008 #5
    Exactly, they are both rotating around the same center of mass, so just change your frame to be in the center of mass frame. Using CM and reduced mass will get you there. I'm sure there are others, but this is the most direct and easy way I can think of.
  7. Feb 27, 2008 #6

    thanks. I got it just using the F=qE and just solving for one of them
  8. Feb 27, 2008 #7
    Alright, well I said that

    F=qE = m v[tex]^{2}[/tex] / r = 4 [tex]\pi[/tex] [tex]^{2}[/tex] r f[tex]^{2}[/tex].

    And you know

    E = k q / r[tex]^{2}[/tex].

    Did you just use r to be 1/2 of the distance between them and solve for f?

    I'm still stuck
  9. Feb 27, 2008 #8
    I used the whole distance in the kq/r squared equation and just solved for the electron.

    Then I used half of r as the "r" in mv squared over r and also in the v = 2 x pi x r x f equation.
  10. Feb 27, 2008 #9
    Got it too

    Hey thanks. I figured out that I had been doing it correctly all along; I just was using millimeters instead of nanometers. The nm looked like mm despite the absurd notion that these particles were that far apart. I guess I'm just a steep noob. Thanks tho.

    And just for anyone else who might look at this thread, I did the following:

    Used F = qE = k*q^2*r^-2 to find E using the total distance between them for r.

    Used m*v^2/r = F = 4*pi^2*m*r*f^2 to find frequency, using half the total distance for r.

    Thanks again.
  11. Feb 28, 2008 #10
    Harshman Is Watching You.
  12. Jul 9, 2008 #11
    This may be a silly question, but this stuff really confuses me and I was wondering what the q would be in this problem? I know what to use for r and k, but not for q to solve for F=qE=kq^2/r^2. Could someone help me? And as for solving for mv^2/r, what's v?
  13. Jul 9, 2008 #12
    Okay so q in this problem is the charge of the positron, which is +e where e = 1.6[tex]\times[/tex]10[tex]^{-19}[/tex].

    v is the tangential velocity of the two particles. Be careful, because this is different than the angular velocity [tex]\omega[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Orbital Frequency problem
  1. Orbital Frequency (Replies: 1)

  2. Orbits problem (Replies: 9)

  3. Orbit Problem (Replies: 2)