please help with Physics Question

Question

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. A third particle is placed between particles 1 and 2. Where on the x axis should the third particle be located so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles? Express your answer in terms of L. 



D = ? L

Answer 1

First let assign the force between particle 1 and 2 is f₀. Next, let's assume the mass of the third particle is M, and the position is a, the gravitation force generated by the third particle to the first and second particles are f1 and f2, respectively. If "magnitude of the gravitational force on both particle 1 and particle 2 doubles", that means f₀+f₁ = 2f₀ and f₀+f₂ = 2f₀, or f₁=f₀ and f₂=f₀. Since we have:

,

,

,

so with f₁=f₂=f₀, we will get two equations:

,

.

Solve these two equations we can get:

and .

So the third particle must place at the location of x=a and its mass must be M as we calculated above.

Answer 2

Place the third particle at some unknown position r. The third particle also has unknown mass M.

Before the third particle was placed, the gravitational force between the first and second particles was G(2m²)/L².

F₁ (the gravitational force on particle one) = F₂ (the gravitational force on particle 2) = G(2m²)/L².

After placing the third particle down at position r, we get

F₁ = G(2m²)/L² + GmM/r²

F₂ = G(2m²)/L² + G(2m)M/(L-r)²

Because we're trying to double both gravitational forces,

G(2m²)/L² + GmM/r² = 2 * G(2m²)/L² and

G(2m²)/L² + G(2m)M/(L-r)² = 2 * G(2m²)/L²

Simplifying gives,

GmM/r² = G(2m²)/L² and

G(2m)M/(L-r)² = G(2m²)/L²

Divide by Gm in both equations

M/r² = 2m/L²

2M/(L-r)² = 2m/L²

Divide the two equations

(L-r)²/(2r²) = 1

Lastly, solve for r