Electron orbiting a proton with 24-hour period

[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.