Answer
The particle of mass $16m$ must be placed between the two particles but closer to the less massive particle.
The answer to part (a) does not change if the third particle has a mass of $16m$
Work Step by Step
We can find the magnitude of the gravitational force on the particle of mass $16m$ due to the particle of mass $m$:
$F = G~\frac{(m)(16m)}{r_1^2} = \frac{16Gm^2}{r_1^2}$
Note that this force is directed toward the particle of mass $m$.
We can find the magnitude of the gravitational force on the particle of mass $16m$ due to the particle of mass $2m$:
$F = G~\frac{(2m)(16m)}{r_2^2} = \frac{32Gm^2}{r_2^2}$
Note that this force is directed toward the particle of mass $2m$.
For the net gravitational force on the particle of mass $16m$ to be 0, the two gravitational forces must point in opposite directions. Therefore, the particle of mass $16m$ must be placed between the two other particles.
For the net gravitational force on the particle of mass $16m$ to be 0, the two opposing gravitational forces must have the same magnitude. Therefore, the particle of mass $16m$ must be placed closer to the particle of mass $m$ such that $r_1 \lt r_2$
The particle of mass $16m$ must be placed between the two particles but closer to the less massive particle.
The answer to part (a) does not change if the third particle has a mass of $16m$.
