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^2Â
Since the Ms are the same, thenÂ
2mx^2 / L^2 = m(L - x)^2 / L^2Â
2x^2 = (L - x)^2Â
xsqrt2 = L - xÂ
x(1 + sqrt2) = LÂ
x = L / (sqrt2 + 1) From here, we rationalize.Â
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)Â
x = L(sqrt2 - 1) / (2 - 1)Â
x = L(sqrt2 - 1)
Â
= 0.414L
Â
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.