MHT CET 2021 24th September Evening Shift Physics - Gravitation Question 79 English Explanation

Consider mass $$\mathrm{m}$$ at $$\mathrm{A}$$.

The forces exerted on it by the other two masses are given by

$$F_1=G \frac{\mathrm{m}^2}{\mathrm{~L}^2}=\mathrm{F}_2$$

The angle between the two forces is $$60^{\circ}$$.

Hence the resultant force

$$\begin{aligned} & \quad F=\sqrt{\mathrm{F}_1^2+\mathrm{F}_1^2+2 \mathrm{~F}_1^2 \cos 60^{\circ}}=\sqrt{3} \mathrm{~F}_1 \\ & \therefore \quad F=\sqrt{3} \cdot G \frac{\mathrm{m}^2}{\mathrm{~L}^2} \end{aligned}$$

Mass $$\mathrm{m}$$ rotates around the centre $$\mathrm{G}$$. The radius of the circular motion is $$\mathrm{AG}$$.

$$\begin{aligned} \mathrm{AG} & =\frac{2}{3} \mathrm{AD} \\ \mathrm{AD} & =\mathrm{AC} \sin 60^{\circ}=\mathrm{L} \sin 60^{\circ}=\frac{\sqrt{3}}{2} \mathrm{~L} \\ \therefore \quad \mathrm{AG} & =\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \mathrm{~L}=\frac{\mathrm{L}}{\sqrt{3}} \\ \therefore \text { radius } \mathrm{r} & =\frac{\mathrm{L}}{\sqrt{3}} \end{aligned}$$

For uniform circular motion, the gravitational force provides the centripetal force.

$$\begin{aligned} & \therefore \mathrm{mr} \frac{\mathrm{L}}{\sqrt{3}} \cdot \omega^2=\sqrt{3} \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2} \\ & \therefore \omega^2=3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3} \\ & \therefore \quad \omega=\left(3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3}\right)^{1 / 2} \\ & \therefore \frac{2 \pi}{T}=\left(\frac{3 G \mathrm{~m}}{\mathrm{~L}^3}\right)^{1 / 2} \\ & \therefore T=2 \pi\left(\frac{\mathrm{L}^3}{3 \mathrm{Gm}}\right)^{1 / 2} \\ & \therefore T \propto \mathrm{L}^{3 / 2} \end{aligned}$$

Gravitation

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