# Sun and Earth Problem

1. Nov 28, 2007

### quickclick330

More sample exam problems...

2) The mass of our sun is ~ 2 x 10^30 kg. The mass of a planet is 6 x 10^24 kg. If the
distance between the center-of-mass of each is 100 x 10^11 m, what is the distance
in meters from the center-of-mass of the sun to the center-of-mass of the Sun-
Earth system?

2. Nov 28, 2007

### temaire

Well, you could start by writing down some relevant formulas, such as:

$$F_{g} = \frac{Gm_{1}m_{2}}{r^{2}}$$

$$\frac{T_{A}^{2}}{r_{A}^{3}} = \frac{T_{B}^{2}}{r_{B}^{3}}$$

3. Nov 28, 2007

### quickclick330

What is the second equation? I'm not familiar with that one.

4. Nov 28, 2007

### Telmerk

That is the third law of Johannes Kepler. ;-)
r: sun-planet distance
T: period

5. Nov 28, 2007

### quickclick330

That one would help. So how would you relate them together? Energy Principle?

6. Nov 28, 2007

### rock.freak667

The second equation is what happens as a result of some fancy manipulation of centripetal force and gravitational force of attraction.

7. Nov 28, 2007

### quickclick330

ooh okay. I went to a help center and we had the centripetal force and gravitational force set equal to each other but the grad TA couldn't figure out how to get rid of velocity. So that second equation is really all i need right?

8. Nov 28, 2007

### rock.freak667

Use the centripetal force as $m\omega^2r$ where $\omega=\frac{2\pi}{T}$ to not have v and have the period of revolution

9. Nov 28, 2007

### quickclick330

will r be the distance from planet to center of mass?

10. Dec 4, 2007

### Telmerk

Yes.

11. Dec 4, 2007

### D H

Staff Emeritus
I realize this is too late, but the help here has been quite bad. There is no need to invoke any of Kepler's laws to solve this problem. This is a simple problem. Quickclick, what is the equation of the center of mass for a group of particles?