# Keplers Third Law

1. Feb 23, 2009

### skiing4free

1. The problem statement, all variables and given/known data
By considering the centripetal force which acts on a planet in a cirlar orbit, show that T^2$$\propto$$R^3, where T is the time taken for one orbitaround the Sun and R is the radius of the orbit.

2. Relevant equations
Fc=GMm/r^2
1/2Mv^2=GMm/r^2

3. The attempt at a solution
I showed that F$$\propto$$1/d^2 but then could not incorporate R^3 into the equation.......

2. Feb 23, 2009

### Gib Z

Well, you made it clear that you know that the gravitational force IS the centripetal force on the planet, and you have the expression for the gravitational force. What is the formula for the centripetal force on an object? Equate these two formulas.

3. Feb 23, 2009

### davieddy

This equation isn't even dimensionally correct,
and KE <> PE.

4. Feb 24, 2009

### skiing4free

Thats exactly what i showed, Ek=Epg. (<>, that doesn't make sense???)

5. Feb 24, 2009

### davieddy

<> means NOT equal to.
Anyway potential energy is -GMm/r.

Even if you had erroneously said KE= PE you could have deduced
T^2 proportional to r^3.

6. Feb 24, 2009

### alseth

as was previously said just equate the expressions for the gravitational force and the centripetal force GMm/r^2=m(v^2)/r and play with it a bit then use some basic formulas for circular motion for angular velocity and time period and you will get the expression for T and R with some constants in it. Where did you find this : 1/2Mv^2=GMm/r^2 ???

7. Feb 24, 2009

### skiing4free

Sorry i got really confused with the energy equations this is what i meant to write;

F=(m4$$\pi$$^2 r)/T^2

F=GMm/r^2

so

(m4$$\pi$$^2 r)/T^2=GMm/r^2

m cancels, left with 4$$\pi$$^2 r^3=GMT^2

$$\Rightarrow$$ r^3$$\propto$$T^2