As seen in the figure, two spheres of massmand a third sphere of massMform an equilateral triangle, and a fourth sphere of massis at the center of the triangle. The net gravitational force on that central sphere from the three other spheres is zero. (a) What isMin terms ofm? (b) If we double the value of, what is the magnitude of the net gravitational force on the central sphere?

From figure, masses of three spheres at corner of the triangle =m

Mass of the central sphere =M

This problem is based on Newton’s gravitational law. We can find the net gravitational force by using vector addition of forces.

Formula:

Force of Gravitation,$F=\frac{G{m}_1{m}_2}{r^2}$ (i)

Using equation (i), Force due to mass m on particlecan be written by:

$F_m=\frac{Gm{m}_4}{r^{2^{}}}$

Along x axis, the form of the force can be written as:

$F_{mx}=\frac{Gm{m}_4}{r^{2^{}}}\cos \left(\theta \right)$ (ii)

Similarly, along y axis, the force can be written as:

$$\begin{gathered} F_{my}=\frac{Gm{m}_4}{r^{2^{}}}\sin \left(\theta \right) \\ =\frac{Gm{m}_4}{r^{2^{}}}\cos \left(\theta \right) \end{gathered}$$ (iii)

Using equation (i), Force due to masscan be written as:

$F_M=\frac{GM{m}_4}{r^{2^{}}}$

Using equation (ii), the force due to mass M along x-axis can be written as:

$F_{Mx}=0$

Similarly, the force due to mass M along y-axis can be written as:

$F_{My}=\frac{GM{m}_4}{r^{2^{}}}$ (iv)

As net force on the central sphere due to the three masses m, m, M is zero. (Given) $$\begin{gathered} \underset{}{\sum F_y=0} \\ {F}_{My}-2F_{my}=0 \end{gathered}$$

From equation (iii) and (iv), we get

$\frac{GM{m}_4}{r^{2^{}}}-2\frac{Gm{m}_4}{r^{2^{}}}\sin \left(\theta \right)=0$

Hence,

$M=2m\sin \left(30\right)$

Therefore, the mass M is

As we double mass, it will double all the three forces. Direction of forces will not change here. So, the net force will be zero.