How Do You Derive Kepler's Third Law Using Newton's Law of Gravitation?

[petern]

Derive Kepler's Third Law of Planetary Motion from Newton's Law of Universal Gravitation.

I know the Law of Universal Gravitation is Fg = (Gm1m2)/(r^2) and the Third Law of Planetary Motion is T^2 = kr^3


What should I do next?

 

[rock.freak667]

Start from here:
The gravitational force of attraction between two masses provides the centripetal force required to keep the mass in a circular orbit.

so that (Gm1m2)/(r^2)=m1(w^2)r

 

[petern]

Which equation does the m1(w^2)r come from?

 

[rock.freak667]

the centripetal force is given by these equations:
F_C=m\omega^2r = \frac{mv^2}{r}=mv\omega

I just used the first equality

 

[petern]

We haven't learned about m(w^2)r yet so I don't think that's what we're suppose to use. What is the w? However, we've learned about (mv^2)/r.

 

[petern]

I think I set (Gm1m2)/(r^2) = (mv^2)/(r). I then plug the equation v = (2*pi*r)/T into the one I previously listed. After that, I cancel m1 out and get r1^3 = (Gm2T^2)/(4*pi). After that I don't know how to get rid of the G, m1, 4, and pi so that I'll end up with the equation T^2 = kr^3.

 

[petern]

Please help. I don't know what to do.

 

[Galileo's Ghost]

I think I set (Gm1m2)/(r^2) = (mv^2)/(r). I then plug the equation v = (2*pi*r)/T into the one I previously listed. After that, I cancel m1 out and get r1^3 = (Gm2T^2)/(4*pi). After that I don't know how to get rid of the G, m1, 4, and pi so that I'll end up with the equation T^2 = kr^3.

Then you have done it! G, m, 4, and pi are all constants. You don't need to get rid of them. They can be rolled up into one constant value that you are calling "k".