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.

Answer 2

The point at which the third mass M should be placed so that the net gravitational force on M due to the two masses is exactly zero is;

0.634 m from the 3m particle

We are told that a particle of mass 3m is located 1 m from a particle of mass m.

Formula for gravitational force is;

F = GMm/r²

Now we want to find where to put a third mass with mass M so that the net gravitational force on M due to the two masses is exactly zero.

Let the particle to the left of the third mass be the one with the mass of 3m and let the particle with the mass of m be to the right of the third mass.

Since F_net = 0, then it means that;

F₁ = F₂

Force of gravity of left particle; F₁ = GM(3m)/r²

Force of gravity of right particle; F₂ = GMm/r²

Let the distance we want to place the third particle be x m from the left particle. Thus;

F₁ = GM(3m)/x²

F₂ = GMm/(1 - x)²

From the net force equation, we now have;

GM(3m)/x² = GMm/(1 - x)²

GM and m will cancel out to give;

3/x² = 1/(1 - x)²

x² = 3(1 - x)²

x² = 3(1 - 2x + x²)

x² = 3 - 6x + 3x²

3x² - x² - 6x + 3 = 0

2x² - 6x + 3 = 0

Using online quadratic equation solver gives us;

x = 0.634 m from the 3m particle

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