# Kepler's Law Program With Planets

1. Apr 16, 2012

### hardygirl989

1. The problem statement, all variables and given/known data

Two planets X and Y travel counterclockwise in circular orbits about a star as shown in the figure below. The radii of their orbits are in the ratio 4:3. At one moment, they are aligned as shown in figure (a), making a straight line with the star. During the next five years, the angular displacement of planet X is 90.0° as shown in figure (b). What is the angular displacement of planet Y at this moment?

2. Relevant equations

ƩF=ma

3. The attempt at a solution

ƩF=ma
(GMplanetMstar)/r^2 = ma = Mplanet V^2 / r
GMstar = r^3ω^3=Rx^3ωx^3 = Ry^3ωy^3
ωy = (90 degrees/5 years)(4/3)^(3/2) = 16√3 degrees/year ≈ 27.712812921102 degrees/year

I think I have figured out most of this problem, but I can not figure out how to convert into revolutions. Can anyone help? Thanks.

2. Apr 16, 2012

### Steely Dan

One revolution of a planet corresponds to completing a full circular orbit. The number of degrees in a circle is 360, so there are that many degrees in a full revolution. Then the fraction of a revolution you've traveled is just the fraction of 360 degrees that you've traveled.

3. Apr 16, 2012

### hardygirl989

If I divide by 360, I get 2√3/45, but this would make the number in revolutions per year then. How do I get the number into just revolutions?

4. Apr 16, 2012

### Steely Dan

Well, if you have an angular speed, what you've got is essentially a number of revolutions per year that Planet Y traverses. You are told that the situation runs its course for five years. So how many revolutions does Planet Y traverse in that time?

While I'm not exactly sure what your work is saying, it does seem that there's an algebraic error in the second line going to the third line, since it would seem there should only be $\omega^2$, not cubed.

5. Apr 16, 2012

### hardygirl989

I noticed that I forgot to post the image
http://www.webassign.net/serpse8/13-p-020.gif

You are right about the W^3 thing. It should be W^2, but I have it written that way on my paper...sorry. The R should be cubed tho because you multiply by the other R.

So I think in 5 yrs it goes 1/4 of a revolution? So, I do... (1/4)*360*27.7128 = 2494.15? That number seems too big...

6. Apr 16, 2012

### Steely Dan

You have already accounted for the quarter of a revolution when solving for the angular speed of planet Y to begin with. That is, the correct form of your last line is

$$\omega_y^2 = \omega_x^2 \frac{R_x^3}{R_y^3},$$

and to calculate $\omega_x$ you would say that it completes 1/4 revolutions per 5 years, instead of 90 degrees per 5 years. Or you could do what you did, and solve for a number of degrees per year, and convert it to revolutions per year. The point is, once you have $\omega_y$, you've found the number of revolutions Planet Y will complete in one year. It is simple from there to figure out how many it will complete in five years.

7. Apr 16, 2012

### hardygirl989

Oh! I see now.

ωy = (1/4 revolutions in 5 years)(4/3)^(3/2)
ωy = (1/4)(4/3)^(3/2) = 0.385 rev

This is the correct answer! Thank you! :)

8. Apr 17, 2012

### hardygirl989

How do I mark this thread as solved?