# Orbital Period

1. Nov 15, 2013

### Psyguy22

1. The problem statement, all variables and given/known data
A planet with a mass of 8.99·1021 kg is in a circular orbit around a star with a mass of 1.33·1030 kg. The planet has an orbital radius of 1.21·1010 m.

a) What is the linear orbital velocity of the planet?

b) What is the period of the planets orbit?

c) What is the total mechanical energy?

2. Relevant equations
Keplers 3rd law
$(T^2/R^3)=(4π^2/G M)$

3. The attempt at a solution
I figured it would be easier to solve b first, So I solved for T
$T=sqrt((4π^2*R^3)/(G M))$

and came up with 6.2784...*10^5. But I am confused on what the units would be. Both seconds and years are wrong.

2. Nov 15, 2013

### cepheid

Staff Emeritus
When you use this version of Kepler's third law, everything is in SI units. So, your value of T would be in seconds.

It's only the case when T is in years and R is in AU that the simpler version of Kepler's 3rd Law (T^2 = R^3) works. This is because you happen to have chosen a unit system in which the constants (4pi^2/GM) are equal to 1.

If you are getting the wrong answer in seconds, you need to double check your arithmetic.

3. Nov 15, 2013

### hilbert2

You don't need Kepler's 3rd law here. You know the centripetal acceleration of the planet from equation $a=G\frac{M}{r^{2}}$, where $M$ is the mass of the star and $r$ is the orbital radius. You also know that for motion in a circular orbit, $a=\frac{v^{2}}{r}$. From these eqs you can solve the orbital velocity $v$, and knowing the velocity the calculation of orbital period and mechanical energy are trivial.

4. Nov 15, 2013

### cepheid

Staff Emeritus
Kepler's 3rd Law in the form above can be derived easily by equating the centripetal and gravitational accelerations exactly as you have done above. So I don't see what is the harm in using a mathematically-equivalent thing. (You'll get the period first, and knowing the period, the calculation of orbital velocity and mechanical energy are trivial).

5. Nov 15, 2013

### hilbert2

^ Actually, Keplers 3rd law holds for more general elliptical orbits, not only circular, and it can't be derived simply from the formulae for circular orbits. Of course it doesn't matter whether one uses Kepler's laws or more detailed mechanical considerations in solving the problem.

6. Nov 15, 2013

### Psyguy22

Thanks both of you. @ Cephid, you were right. I must of typed numbers wrong the first time.

7. Nov 15, 2013

### cepheid

Staff Emeritus
I know that. The point I was making was that if you assume a circular orbit, and equate centripetal and gravitational accelerations, you will obtain the *exact* equation for Kepler's 3rd Law that is given in the OP, which is a special case of Kepler's 3rd for circular orbits. I know that if you derive it for the most general case, you will find that r is replaced by the semi-major axis of the ellipse, but that's not relevant to this problem.

In any case, you seem to have missed the actual point I was making, which is that I see no utility in advising the OP to use a second method that is mathematically-equivalent to the one he/she is already using.