Question

A particle of mass 3m is located 1.00 m from a particle of mass m. (a) Where should you put a third mass M so that the net gravitational force on M due to the two masses is exactly zero?

Answer 1

Explanation:

It is given that net gravitational force on M is exactly equal to zero. Hence, distance to M from the bigger mass is 3m. Therefore, expression for net force will be as follows.

           $F_{net} = F_{1} + F_{2} = 0$  

So,  

          $\frac{-G(3m)(M)}{x^{2}} + \frac{G(m)(M)}{(1 - x)^{2}} = 0$

The first term is negative as the third mass is located between the other two masses. This means that 3 m will be pulling it leftwards (negative x direction) and m will be pulling it rightwards (positive x direction).

      $\frac{G(m)(M)}{(1 - x)^{2}} = \frac{G(3m)(M)}{(x)^{2}}$

On dividing both sides of the equation by G.m.M, we get the following.

      $\frac{1}{(1 - x)^{2}} = \frac{3}{x^{2}}$

               $x^{2} = 3 - 6x + 3x^{2}$

                    0 = $3 - 6x + 2x^{2}$

Using the formula, $\frac{-b \pm \sqrt{(b)^{2} - 4ac}}{2a}$   the value of x comes out to be equal to +2.37 (not usabale) and -0.634 (usable).

Hence, we can conclude that the third mass will be located 0.634 meters away from the 3 m mass.