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(B) Let the four particles be at the corners of a square of side $a$. The distance of each particle from the center of the square is $r = \frac{a}{\sq

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$1.16\sqrt{\frac{GM}{a}}$

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(B) Let the four particles be at the corners of a square of side $a$. The distance of each particle from the center of the square is $r = \frac{a}{\sqrt{2}}$.Consider one particle of mass $M$. The gravitational forces acting on it due to the other three particles are:$1$. Two forces of magnitude $F_1 = \frac{GM^2}{a^2}$ directed along the sides of the square.$2$. One force of magnitude $F_2 = \frac{GM^2}{(a\sqrt{2})^2} = \frac{GM^2}{2a^2}$ directed along the diagonal.The resultant force $F_c$ towards the center is the sum of the components of these forces along the diagonal:$F_c = F_1 \cos(45^\circ) + F_1 \cos(45^\circ) + F_2$$F_c = 2 \left( \frac{GM^2}{a^2} \right) \frac{1}{\sqrt{2}} + \frac{GM^2}{2a^2} = \frac{GM^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right)$.This force provides the necessary centripetal force for circular motion: $F_c = \frac{Mv^2}{r}$.Substituting $r = \frac{a}{\sqrt{2}}$:$\frac{Mv^2}{a/\sqrt{2}} = \frac{GM^2}{a^2} \left( \sqrt{2} + 0.5 \right)$$v^2 = \frac{GM}{a} \left( \frac{\sqrt{2}}{\sqrt{2}} + \frac{0.5}{\sqrt{2}} \right) = \frac{GM}{a} (1 + 0.3535) \approx 1.3535 \frac{GM}{a}$.$v \approx 1.16 \sqrt{\frac{GM}{a}}$.

Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed if each of them revolves under the influence of the others' gravitational field in a circular orbit circumscribing the square?

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