1. Jan 24, 2004

### ACLerok

With only two lectures in my back pocket, I still can't find a reasonable solution for this problem.

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. The orbital period of Mercury is 88.0 days.

What would be the orbital period of such a planet?

and also, in general:

Four identical masses of mass 500 kg each are placed at the corners of a square whose side lengths are 15. cm.

What is the magnitude of the net gravitational force on one of the masses, due to the other three?

2. Jan 24, 2004

### jamesrc

First one:
You can use Kepler's Third Law (or equate the centripetal force with the gravitational attractio if you'd rather):

$$\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}$$

With T1 = 88 days and R2 = 2*R1/3 to find the period of the mysterious new planet.

Second one:

Just do a little vector addition here. You know that the gravitational attraction between two masses separated by a distance r is given by

$$\vec{F_{12}} = -\frac{{\rm G}m_1 m_2}{{\vec{|r_{12}|}}^3}\vec{r_{12}}$$

Just so we have our notation straight, F12 indicates the force on 1 due to 2, while r12 indicates the position vector from 1 to 2.

Just sum the forces (good old law of superposition). Take the top left mass as an example: you'll have a force pointing down due to the bottom left mass, a force pointing to the right due to the top right mass, and a force pointing diagonally toward the bottom right mass due to that mass. The first two forces will have the same magnitude, while the third force will have a magnitude of half of either of the first two forces.

Hint: the net force on each mass in this configuration should be directed radially inward to the center of the square the 4 masses make.