# A quick question about Elliptical/Circular Orbit!

1. Sep 12, 2014

### Parto0o

Hi everyone.

So I got this question in a test, but I'm not exactly sure about the answer.

Q: For a planet moving in an elliptical orbit, why is 1/2Iω^2 not the correct expression for its kinetic energy?

2. Sep 12, 2014

### Staff: Mentor

How did you answer that question? Or were you so not sure that you had to leave it blank? You'll get much more helpful answers if we know how much you already know.

3. Sep 13, 2014

### Parto0o

I think I have no idea about it. I already know the basic formulas, but cannot relate them to this question.

4. Sep 13, 2014

### Staff: Mentor

OK, then here's one way of thinking about it. At any given moment, the velocity vector of the orbiting object can be written as the sum of radial and tangential components. The kinetic energy depends on both (to be precise, $E_k=mv^2/2=\frac{m}{2}\sqrt{v_r^2+v_\theta^2}$). The angular velocity depends only on the tangential component ($\omega=v_\theta/r$).

Just by looking those relationships, you can see that if two different objects have the same $r$ and $v_\theta$ they will have the same $\omega$ but can still have different $v_r$ values and hence different kinetic energies. Therefore, you will get different relationships between $\omega$ and kinetic energy for different values of $v_r$. The formula you quote doesn't have that property so it can only be correct for a particular value of $v_r$; that value is zero, which describes uniform circular motion.

5. Sep 13, 2014

### Creator

Good explanation, Nugatory; but the final answer for instantaneous Kinetic energy in elliptical orbit should be...

$E_k=mv^2/2=\frac{m}{2}[\sqrt{v_r^2+v_\theta^2}]^2=\frac{m}{2}({v_r^2+v_\theta^2})$

Because the resultant velocity vector itself, $\sqrt{v_r^2+v_\theta^2}$,must be squared...
Thus the final answer, $\frac{m}{2}({v_r^2+v_\theta^2})$, reveals that a zero radial velocity gives the correct kinetic energy for circular orbit.