Gravitational field for 2 particles

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SUMMARY

The discussion focuses on calculating the gravitational field generated by two identical particles, each of mass m, positioned at x = +x0 and x = -x0 along the x-axis, for points along the y-axis. The gravitational field \(\vec{g}\) is expressed as a function of y, m, and x0, utilizing the equations \(\vec{g} = \frac{\vec{F}}{m}\) and \(\vec{g} = \frac{GM}{r^2}\hat{r}\). The participants clarify that the x-components of the gravitational fields from both particles cancel each other due to symmetry, leading to a net gravitational field that only has a y-component.

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Homework Statement


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 [tex]\vec{g}[/tex] 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.

Homework Equations


[tex]\vec{g}=\frac{\vec{F}}{m}[/tex]
[tex]\vec{g}=\frac{GM}{r^2}\hat{r}[/tex]


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:

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

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?
 
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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).
Add the components.
What can you say about the x component, just from the symmetry of the problem?
 
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:

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