In the figure, a square of edge length20.0 cmis formed by four spheres of masses${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{g}\mathbf{,}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{g}\mathbf{,}{\mathit{m}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{g}\mathbf{,}{\mathit{m}}_{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{g}\mathbf{.}$ In unit-vector notation, what is the net gravitational force from them on a central sphere with mass${\mathbf{m}}_{\mathbf{5}}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{g}$?

From the figure, edge length of the square = 0.2 m

Mass of four spheres at the edges

$$\begin{gathered} m_1=0.005kg \\ {m}_2=0.003kg \\ {m}_3=0.003kg \\ {m}_4=0.005kg \end{gathered}$$

Mass of the central sphere,$m_5=0.0025kg$

Distance between each edge sphere and the central sphere =0.14 m

Gravitational constant, $G=6.67\times {10}^{-11}N-m^2/k{g}^2$

Force between two masses can be calculated by using Newton’s law of gravitation. According to Newton’s law of gravitation, the force of two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By vector addition, we can find the net force.

Formula:

Gravitational force of attraction, $F=\frac{G{m}_1{m}_2}{r^2}$ (i)

Here, force due to masses $m_1$and $m_4$are same in magnitude because of their same masses and opposite in direction, so they cancel out each other.

Hence, net force is due to masses $m_2$and $m_3$only.

Using equation (i), Force due to mass $m_2$is given by:

$$\begin{gathered} F_{25}=\frac{G{m}_2{m}_5}{r_{15}^2} \\ =\frac{6.67\times {10}^{-11}\times 0.{003}^{*}0.0025}{{\left(0.14\right)}^2} \\ =2.5\times {10}^{-14}Nalongdigonal\left(45°withbase\right) \end{gathered}$$

Using equation (i), Force due to mass $m_3$is given by:

Net force:

$$\begin{gathered} \overrightarrow{F}={\overrightarrow{F}}_{25}-{\overrightarrow{F}}_{35} \\ =\left(2.5\times {10}^{-14}-8.35\times {10}^{-15}\right) \\ =1.67\times {10}^{-14}N\left(45°withbase\right) \end{gathered}$$

Hence,

$$\begin{gathered} \overrightarrow{F}=\left(1.67\times {10}^{-14}\cos \left(45\right)\hat{i}\right)+(1.67\times {10}^{-14}\sin \left(45\right)\hat{j} \\ =1.16\times {10}^{-14}\hat{i}+1.16\times {10}^{-14}\hat{j} \end{gathered}$$

The net gravitational force is$1.16\times {10}^{-14}\hat{i}+1.16\times {10}^{-14}\hat{j}$