What is the Orbital Period of Positronium in a Circular Orbit?

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SUMMARY

The discussion centers on calculating the orbital period of positronium, a bound state of an electron and a positron. The relationship derived is T^2 = (16)(pi^3)(E0)(me)(mp)(r^3) / (e^2)((me) + (mp)), which is a direct application of Kepler's third law for electrical orbits. The participants also explore the implications of this formula when substituting a proton for the positron, demonstrating the consistency of the derived expression. The discussion emphasizes the importance of equating centripetal and electrostatic forces in circular orbits.

PREREQUISITES
  • Understanding of Kepler's laws of planetary motion
  • Familiarity with electrostatics and Coulomb's law
  • Knowledge of centripetal force and its application in circular motion
  • Basic concepts of atomic structure, specifically positronium
NEXT STEPS
  • Study the derivation of Kepler's third law in the context of electrical orbits
  • Explore the properties and behavior of positronium in quantum mechanics
  • Learn about the electrostatic force and its role in particle interactions
  • Investigate the differences between positronium and hydrogen-like atoms
USEFUL FOR

Students and researchers in physics, particularly those focusing on atomic physics, quantum mechanics, and electrostatics, will benefit from this discussion.

PinkFlamingo
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Hi there! I hope someone can help me with this problem. I've been working on this for over 5 hours and I've gotten nowhere!

A positron is a particle with the same mass as an electron but with a positive charge. A positron and an electron can briefly form an unusual atom known as positronium. Imagine a situation where the two particles are in a circular orbit about their center of mass. Since the particles have equal mass, the center of mass is midway between them. Let r be the separation of the particles (so that the orbits are each of radius r/2).

(a) Show that the orbital period T is related to the separation distance r by:

T^2 = (16)(pi^3)(E0)(me)(mp) (r^3)
---------------------
(e^2)[(me) + (mp)]

This is a consequence of Kepler's third law for electrical orbits.

(b) Show that if an electron and a proton are in circular orbits about their center of mass (which is not at the midway point between them but much closer to the proton), then the same expression results.

* * * * *

OK, so so far, I'm guessing that I somehow use the formulae:

q = ne

F = 1 |Q||q|
-------- x ---------
4(pi)(E0) (r^2)

But I'm not really sure where the rest of it comes from

If someone could help me out, I would really appreciate it!

Thanks!

Mandy
 
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The bad idea of Keplerian orbits being used aside, if the orbit is indeed a circle, then the force each particle feels must obey the centripetal force equation, F=mv2/r. In this case the centripetal force is the electrostatic force, so they two equations can be set equal to each other.
 

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