Question

The point masses m and 2m lie along the x-axis, with m at the origin 2m and at x = L. A third point mass M is moved along the x-axis. (a) At what point is the net gravitational force on M due to the other two masses equal to zero? (b) Sketch the x-component of the net force on M due to m and 2m, taking quantities to the right as positive. Include the regions x < 0, 0 < x < L , and x > L . Be especially careful to show the behavior of the graph on either side of x = 0 and x = L.

Short answer

(a) The point in which the net gravitational force is zero is at 0.414L.

(b)

Step-by-step solution

Identification of the given data

The given data is listed below as:

• The mass of the first point object is, m
• The mass of the first point object is, 2m
• The distance of the first point object from the origin is, x
• The distance of the second point object from the origin is, x = L

Significance of the gravitational force

The gravitational force is described as the attraction force that acts between one or more objects. The gravitational force is directly proportional to the masses of the objects and inversely proportional to the distance between the objects.

(a) Determination of the point at which the net gravitational force is zero

The equation of the net gravitational force due to both the masses is expressed as:

$\begin{array}{r}\frac{-GmM}{x^2}+\frac{G\left(2m\right)M}{(L-x{)}^2}=0\\ \frac{-GmM}{x^2}=\frac{G\left(2m\right)M}{(L-x{)}^2}\end{array}$

Here, G is the gravitational constant, m and 2m are the masses of the first and second point objects, L is the distance of the second point object from the origin and x is the distance of the first point object from the origin.

The above equation has been solved below.

$\begin{array}{r}\frac{-GmM}{x^2}=\frac{G\left(2m\right)M}{(L-x{)}^2}\\ \frac{2}{(L-x{)}^2}=\frac{1}{x^2}\\ 2x^2=(L-x{)}^2\\ \sqrt{2}x=(L-x)\end{array}$

Hence, further as:

$\begin{array}{r}\sqrt{2}x+x=L\\ x(1+\sqrt{2})=L\\ x=\frac{L}{(1+\sqrt{2})}\\ =0.414L\end{array}$

Thus, the point in which the net gravitational force is zero is at 0.414L.

(b) Sketching the x-component of the net force on M

The -component of the net force on has been provided below:

From the above graph, it can be identified that the objects in the reality are not the point masses. Hence, it is impossible to have both the point masses at x = 0 for the mass 2m and x = L for the mass M. Apart from that, the graph also shows that as the point masses approach towards each other, then the gravitational force amongst the two masses eventually approaches infinity.

Thus,