Question

Two stationary particles of masses m1, and m2 are
placed a distance r apart. A third particle is placed
inbetween them on the line joining them in such a
way that it experiences no force. Calculate the
distance of third particle from 'm1'. ​

Answer 1

Given:

Distance between m1 and m2 = r

Let the distance from m1 be x

And from m2 be (r-x)

Let the third particle be m

Since net force on particle is 0, then Forces of both both masses on particle is equal.

We know

$\boxed{Gravitational \: Force = \dfrac{GMm}{R^{2}}}$

According to question,

$\dfrac{G \times m_{1} \times m}{x^{2}} = \dfrac{G \times m_{2} \times m}{{(r-x)}^{2}}$

m1/x² = m2/(r-x)²

(r-x)²/x² = m2/m1

$\boxed{x = r\dfrac{\sqrt{m_{1}}}{\sqrt{m_{1}}+ \sqrt{m_{2}}}}$

Answer 2

Given:

Distance between m1 and m2 = r

Let the distance from m1 be x

And from m2 be (r-x)

Let the third particle be m

Since net force on particle is 0, then Forces of both both masses on particle is equal.

We know

According to question,

m1/x² = m2/(r-x)²

(r-x)²/x² = m2/m1