Show that orbital period squared is proportional to radius cubed

[Nusc]

The position of a particle of mass m moving in a circular orbit with radius r = ro is given by: r(t) = ro <er>

Show using Newton's second law of motion, that if the force responsible for this motion is given by:

F = -k/r^2 <er>

a) the angular velocity is constant in time, and
b) the square of the orbital period is proportional to the cube of the orbital radius. Compute the proportionality constant.

for far part a) L = |r x mv| = |mr^2(theta dot) <ez>|
therefore theta dot = L/mr^2

how do I start part b)?

 

[mezarashi]

With the given proportionality, simply plug in any two points (ro and its corresponding To) and solve for the missing constant.

 

[Nusc]

I forgot some parts of the question.

Now first off all, what is the proportionality constant?

And what is To?

What formula do I begin with?

Thanks

 

[mezarashi]

A proportionality constant is a constant such as k in Coloumb's law or G in Newton's law of gravitation. I guess its obvious now you didn't know, but the question is saying that the motion obeys Kepler's law (although you could've done it without knowing this), which takes the form:

$$T^2 = KR^3$$

If you know the period and the radius of orbit, you can find K right?

 

[Nusc]

K = (To^2)/(Ro^3)?

There is absolutely no way this question is that trivial. Remember I did forget to mention something in the question which had has now been in included, if it makes a difference.

 

[Nusc]

How do I take into account Newton's second law F = -k/r^2 <er>?

 

[mezarashi]

To find part (a), the question probably wanted you to do:

$$F= \frac{k}{r^2} = m \omega^2 r$$

In fact, by arranging this equation, you will be able to get the constant K.