# Gravitational field for 2 particles

1. Oct 9, 2008

### Symstar

1. The problem statement, all variables and given/known data
Two identical particles, each of mass m, are located on the x axis at x = +x0 and x = -x0.

Determine a formula for the gravitational field due to these two particles for points on the y axis; that is, write $$\vec{g}$$ as a function of y, m, x0, and so on.
Express your answers in terms of the variables y, m, x0, and appropriate constants. Answer in the form gx,gy.

2. Relevant equations
$$\vec{g}=\frac{\vec{F}}{m}$$
$$\vec{g}=\frac{GM}{r^2}\hat{r}$$

3. The attempt at a solution
I'm really at a loss as to how to approach this problem. I would assume that, graphically, the y axis will be between the two particles and that we are trying to find the components of g in terms of the point on the y axis.

I suppose that in this frame we have a triangle with our particles on two corners and our point on the y axis at the third corner.

I'm still very unsure as to how gravitational fields work in cases like this, but here's what I attempted:

$$\vec{g}=\frac{\vec{F_1}}{m} + \frac{\vec{F_2}}{m}$$
$$\vec{g}=\frac{mG}{r^2_1}\hat{r_1} + \frac{mG}{r^2_2}\hat{r_2}$$
$$r^2_1=-x^2_0+y^2$$
$$r^2_2=x^2_0+y^2$$
$$\vec{g}=\frac{mG}{-x^2_0+y^2}\hat{r_1} + \frac{mG}{x^2_0+y^2}\hat{r_2}$$

At this point, assuming I haven't made a mistake (I doubt that I haven't) I don't know what to do. Could someone give me a hand and some explaining, please?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 9, 2008

### nasu

You are on the right track, kind of.

Correction:
r1^2= xo^2+y^2=r2^2

Then find the x and y components for each one of the two forces (or gravitational fields).
What can you say about the x component, just from the symmetry of the problem?

3. Feb 17, 2011

### interyeti

I'm really stuck on how to resolve the forces into x and y components without an angle. If we could use trig functions, it would be no problem, but the formula does not depend on theta, according to the answer. Is there some geometric trick I'm missing or what?

The x components (whatever they are) should just cancel to 0 right?

Last edited: Feb 17, 2011