Net gravitational force on an object

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SUMMARY

The net gravitational force on an object is calculated by summing the x and y components of the forces exerted by other masses. For Object 1 located at (d, 0), the gravitational force exerted on Object 0 is represented as \(\frac{Gm^2}{d^2}\left<-1, 0\right>\). Object 2, positioned at (-d, l), exerts a gravitational force of \(\frac{2Gm^2}{d^2 + l^2}\left\) on Object 0. The resultant gravitational force combines these vectors, yielding \(\left<\frac{-Gm^2}{d^2}, 0\right> + \left<\frac{2dGm^2}{(d^2 + l^2)^{3/2}}, \frac{-2lGm^2}{(d^2 + l^2)^{3/2}}\right>\).

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Jaccobtw
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Homework Statement
A test object of mass m is placed at the origin of a two dimensional coordinate system. An object 1, of the same mass, is at (d, 0), and an object 2, of mass 2m, is at (-d, l). What is the magnitude of the vector sum of the gravitational forces exerted on the test object by the other two objects?
Relevant Equations
F(gravity) = G * (M * m)/ (r^2)
G = 6.6738 * 10^-11 N * m^2
IMG_0083.jpg


I suppose we can just find the net x components and y components and then go from there.
Σ Fx = F(mass 1) - Fx(mass 2)
G* (m^2)./d^2) - something

I'm not sure how to express the component forces of the 2nd mass
 
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Object 1 is at (d, 0). The distance from object 0 to object 1 is r=d. The direction vector from object 1 to object 0 is <-d, 0> the unit vector in that direction is <-1, 0>. The magnitude of the gravitational force object 1 exert on object 0 is \frac{Gm^2}{d^2} and the vector force is \frac{Gm^2}{d^2}\left&lt;-1, 0\right&gt;. Since you are asking about object 2 I presume you got that.

Object 2 is at (-d, l). The distance from object 0 to object 2 is r=\sqrt{d^2+ l^2}. The direction vector from object 2 to object 0 is <d, -l>. The magnitude of the gravitational force object 1 exerts on object 0 is \frac{2Gm^2}{d^2+ l^2} and the vector force is \frac{2Gm^2}{d^2+ l^2}\left&lt;d, -l\right&gt;.

The resultant is the sum of those, \left&lt;\frac{-Gm^2}{d^2}, 0\right&gt;+ \left&lt;\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right&gt;.
 
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HallsofIvy said:
The resultant is the sum of those, \left&lt;\frac{-Gm^2}{d^2}, 0\right&gt;+ \left&lt;\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right&gt;.
Of course you meant to write
"The resultant is the sum of those, \left&lt;\frac{-Gm^2}{d^2}, 0\right&gt;+ \left&lt;\frac{2dGm^2}{(d^2+ l^2)^{3/2}}, \frac{-2lGm^2}{(d^2+ l^2)^{3/2}}\right&gt;."
 

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