Gravitational force of particles

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To solve the gravitational force problem involving three particles, the key is to apply Newton's law of gravitation, F=Gm1m2/r^2, while considering the forces acting on each particle. The third particle's position must be determined such that the gravitational forces on both Particle 1 and Particle 2 are doubled. By placing the third particle between the first two, the forces can be expressed as the sum of attractions, while positioning it outside requires subtracting forces. The mass of the third particle is irrelevant as it cancels out in the calculations. Ultimately, the solution involves setting up equations based on the distances and ensuring the forces are balanced to achieve the desired outcome.
sweetrose
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I was working on some gravitational force problems, and this one was particularly challenging to me:

"Two particles are located on the x axis. Particle 1 has a mass m and is at the origin. Particle 2 has a mass 2m and is at x=+L. Where on the x-axis should a third particle be located so that the magnitude of the gravitational force on BOTH particles 1 and particle 2 doubles? Express your answer in terms of L. Note that there are two answers."

I don't know where to get started on this problem, I know I will probably have to cancel out variables using the equation F=Gm1m2/r^2. I can picture the problem in my mind, but I don't know what I can solve for. Can anybody enlighten me by explaining to me where they would start?

*I started out simply setting up an equation for the force acting between particle 1 and particle 2. My diagram looks a little bit like this right now:


-----p1--------p2
...<--L------>

I know that if I place p3 in the somewhere between p1 and p2, the force on p1= F(between p1 and p3) + F(between p1 and p2), and the force on p2 would be F(between p2 and p3) + F(between p1 and p2).

When I try to picture the particle either on behind p1 or pass p2, it becomes a bit more difficult for me to picture, but I still get that you're going to have to subtract the forces, instead of adding them. The mass for p3 is not given, so I'm thinking that it cancels out somewhere in the problem. That's all I have for this problem. :shy:
 
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The forces on the two original masses are equal in magnitude. If that needs to be doubled by a third mass it implies that the additional force caused on the two masses by the new mass must also be equal for both. Use this and write the distances in the equation in terms of x and L - x.
 
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