How Do You Calculate the Orbital Frequency of an Electron and Positron System?

• nerdgirl909
In summary: In this case, v = \omega r where r is the distance between the two particles. In summary, the conversation involves a discussion of the orbital frequency for an electron and positron rotating around their center of mass. The equation used is F=qE=kq^2/r^2, where q is the charge of the positron, and E is the electric field strength. The frequency is then found using the equation mv^2/r = F = 4*pi^2*m*r*f^2, where v is the tangential velocity of the particles. The solution involves changing the frame of reference to the center of mass and using the reduced mass system. The conversation also includes a discussion on using a dipole approach to solving
nerdgirl909
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$$\pi$$$$\epsilon\underline{}0$$r$$\overline{}2$$

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.

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.

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 $$\tau$$ = pEsin($$\theta$$) where E is the field strength, $$\theta$$ 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 $$\theta$$ 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$$^{2}$$/(qr). And, you also know that E = kq/r$$^{2}$$. Solving for v, you could then find frequency because v = 2 $$\pi$$rf.

Do I sound like I'm on the right track?

Last edited:
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.

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.

solved

thanks. I got it just using the F=qE and just solving for one of them

Alright, well I said that

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

And you know

E = k q / r$$^{2}$$.

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

I'm still stuck

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.

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.

Harshman Is Watching You.

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?

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?

Okay so q in this problem is the charge of the positron, which is +e where e = 1.6$$\times$$10$$^{-19}$$.

v is the tangential velocity of the two particles. Be careful, because this is different than the angular velocity $$\omega$$

1. What is the Orbital Frequency problem?

The Orbital Frequency problem is a mathematical problem that involves calculating the frequency at which an object orbits around a central body. This is important in understanding the motion and behavior of objects in space, such as planets, moons, and artificial satellites.

2. How is Orbital Frequency calculated?

Orbital Frequency (f) is calculated using the formula f = 1 / T, where T is the orbital period of the object. The orbital period is the time it takes for an object to complete one full orbit around the central body. The units for orbital frequency are usually in Hertz (Hz) or revolutions per minute (RPM).

3. What factors affect Orbital Frequency?

The main factor that affects Orbital Frequency is the distance between the orbiting object and the central body. The closer the object is to the central body, the higher the orbital frequency will be. Other factors that can affect orbital frequency include the mass of the central body, the shape of the orbit, and the presence of other objects in the vicinity.

4. Why is Orbital Frequency important in space exploration?

Orbital Frequency is important in space exploration because it helps us understand the motion and behavior of objects in space. By knowing the orbital frequency, scientists and engineers can determine the optimal orbit for satellites and spacecraft, and also predict the movement of planets and other celestial bodies.

5. Can Orbital Frequency change over time?

Yes, Orbital Frequency can change over time. This is due to various factors such as gravitational forces from other objects, atmospheric drag, and changes in the central body's mass. These changes can cause the orbiting object to speed up or slow down, resulting in a change in orbital frequency.

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