A particle of mass 3m is located $\mathbf{1}.\mathbf{00}\mathbf{m}$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 precisely zero?

(b) Is the equilibrium of M at this point stable or unstable (i) for points along the line connecting m and 3m, and (ii) for points along the line passing through M and perpendicular to the line connecting m and 3m?

The given data can be listed below as follows,

• The mass of the first and second particles are 3mandm,respectively.
• The first particle is located$1.00\mathrm{m}$ from the second particle.
• The mass of the third particle is M.

The law illustrates that the force acting on a particle is inversely proportional to the difference between the particle and the masses and directly proportional to the product of the masses.

The net gravitational force on Mdue to the two masses is zero by equating the two different forces acting on M. The stability of the equilibrium of Mis also identified by equaling the troops.

The free-body diagram has been drawn below-

Figure 13.9

${\mathrm{F}}_1=\mathrm{G}\frac{\mathrm{mM}}{{\mathrm{x}}^2}$ ………………………………………………………..i)

Where G is the gravitational constant, mis the second particle, and Mis the third mass. Besides, xis the distance between the mass mand the mass M.

The force ${\mathrm{F}}_2$ acting on Mcan be expressed as:

${\mathrm{F}}_2=\mathrm{G}\frac{\mathrm{M}\left(3\mathrm{m}\right)}{{(1-\mathrm{x})}^2}$……………………………………………….ii)

Where 3mis the mass of the first particle and${(1-\mathrm{x})}^2$ is the distance between the group 3mand the mass M.

To make the gravitational force due to the first and the second mass on Mzero, both the equations i) and ii) must be equal.

${\mathrm{F}}_1={\mathrm{F}}_2$

Substituting the values in the above equation, we get,

$\begin{array}{c}\frac{\mathrm{GmM}}{{\mathrm{x}}_2}=\frac{\mathrm{GM}\left(3\mathrm{m}\right)}{{(1-\mathrm{x})}^2}\\ \frac{1}{\mathrm{x}}=\frac{\sqrt{3}}{1-\mathrm{x}}\\ \mathrm{x}=\frac{1}{\sqrt{3}+1}\\ =0.36\text{\hspace{0.17em}m}\end{array}$

Here, Mneeds to be located$0.36\mathrm{m}$from m.

Thus, M needs to be located $1-0.36\text{\hspace{0.17em}m}=0.64\text{\hspace{0.17em}m}$.from the 3m.

i) The equilibrium of Mat the point along the line connecting mand 3mis unstable due to having more distance.

ii) The equilibrium of Mfor points along the line passing through Mand perpendicular to the line connecting mand 3mis stable as the distance of the line is more diminutive.