Universal gravitation’s problem -- Balancing gravitational forces from 2 masses

[Riccardo Finessi]

Homework Statement

I have this problem that I’ll must have done for tomorrow but I can’t figure it out, I have done an attempt but dosean’t take anywhere, can anyone help me figure it out


Two particles are on the x-axis. Particle 1 has mass m and is at the origin of the axis, while particle 2 has mass 2m and its position is x = +L. Between these two particles there is a third particle.
At what position on the x-axis would the third particle have to be for the magnitude of the gravitational attractive force acting on both particles 1 and 2 to double? Express your answer in terms of L.


result:0,414L

Relevant Equations

F=G m1 m2 / d*2

 

[berkeman]

Welcome to PF.

What is "t" in your equations? Time?

 

[TSny]

Welcome to PF!

When posting your work, we ask that you type your work (using Latex if possible), rather than posting images of your work. More guidelines for posting questions about homework problems can be found here.

It's a little hard to follow your work in the image. There are no words to describe your train of thought. However, I think your work is ok up to the following

 

[BvU]

With a small sketch things become easier:

In number 1 you have $$
F={G\,m_1m_2\over L^2}
$$so in number 2 you want $$
{G\,m_1m_3\over x^2L^2}={G\,m_1m_2\over L^2}\quad\Rightarrow \quad
m_2 = {m_3\over x^2} \quad \Rightarrow \quad m_1 = {m_3\over 2x^2}
$$Then in number 3 you require $$
{G\,m_2m_3\over (1-x^2)L^2} = {G\,m_1m_2\over L^2}\quad\Rightarrow \quad m_1 = {m_3\over (1-x)^2}
$$this way it's easy to see your

was still correct, as @TSny replied.

$\$