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.

The solution would be like this for this specific problem:

The force on m is:

GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:

GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

2mx^2 / L^2 = m(L - x)^2 / L^2 

2x^2 = (L - x)^2 

xsqrt2 = L - x 

x(1 + sqrt2) = L 

x = L / (sqrt2 + 1) From here, we rationalize. 

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1) 

x = L(sqrt2 - 1) / (2 - 1) 

x = L(sqrt2 - 1)

= 0.414L

Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.

he kicked it 10 feet straight up into the air? i think this works.