Could use some help on this e&m question

[blur13]

Hi, I've been trying to figure out this problem for a while now and I could use some help. The problem reads "An electron and proton attract each other with a $$1/r^2$$ electric force, just like the gravitational force. Suppose that an electron moves in a circular orbit about a proton. If the period of motion is 24 hours, what is the radius of the orbit?"

My train of thought so far has been that I would need to equate the $$1/r^2$$ to the equation for centripetal force $$F=mv^2/r$$, and solve that for r. I also know that the period comes into play, so I'm guessing that $$T=2\Pi r/v$$ would also be used somehow. However, I can't figure out how I would solve for the velocity just by plugging in the period to that equation, since they don't give us the value for the radius, and I need the value of the velocity to solve for the radius in the force equation. Am I on the right track or totally off?

 

[siddharth]

The electron and proton attract each other with a force that is propotional to $$1/r^2$$. The actual force of attraction is given by coloumb's law.

You are on the right track by equating $$mv^2/r$$ and the force of attraction between the proton and electron.

Also, you do use $$T=2 \pi/w$$. Use the fact that for a revolution of $$2 \pi$$ radians, it takes 24*3600 seconds. So what is the angular velocity in radians per second? You should get the answer from this.

 

[quicknote]

Hello there! It seems like you are on the right track with your thinking. Let's break down the problem and see if we can come up with a solution together.

First, we know that the electron and proton are attracted to each other with a 1/r^2 electric force, just like the gravitational force. This means that the force between them is directly proportional to the inverse of the square of the distance between them.

Next, we are given that the electron is moving in a circular orbit around the proton, and we are asked to find the radius of this orbit. We can use the equation for centripetal force, F=mv^2/r, to help us solve this problem.

To start, let's set up the equation for the electric force between the two particles: F = kQq/r^2, where k is the Coulomb's constant, Q is the charge of the proton, and q is the charge of the electron.

Now, we can equate this to the equation for centripetal force and solve for r:

kQq/r^2 = mv^2/r

r^3 = kQq/mv^2

r = (kQq/mv^2)^1/3

Next, we need to bring in the period of motion, which is given to us as 24 hours. We can use the equation T=2πr/v to find the velocity of the electron:

24 hours = 2πr/v

v = 2πr/24 hours

Now, we can substitute this value for v into our equation for r:

r = (kQq/m(2πr/24 hours)^2)^1/3

Solving for r, we get:

r = (kQq/4π^2m)^1/3 * (24 hours)^2/3

Finally, all we need to do is plug in the given values for k, Q, q, and m to find the radius of the orbit.

I hope this helps you with your problem! Remember to always carefully analyze the given information and use the appropriate equations to solve the problem. Good luck!