Question

Four particles each of mass $M$ and equidistant from each other, move along a circle of radius under the acting of their mutual gravitational attraction. The speed of each particle is:

• A. $\sqrt{\frac{GM}{R}\left(1+2\sqrt{2}\right)}$
• B. $\frac{1}{2}\sqrt{\frac{GM}{R}\left(1+2\sqrt{2}\right)}$
• C. $\sqrt{\frac{GM}{R}}$
• D. $\sqrt{2\sqrt{2}\frac{GM}{R}}$

Answer 1

The correct option is B

$\frac{1}{2}\sqrt{\frac{GM}{R}\left(1+2\sqrt{2}\right)}$

Step 1: Given data

Four particles each of mass $M$ and equidistant from each other are move along a circle of radius under the acting of their mutual gravitational attraction.

Step 2: Formula used

Pythagoras theorem,

$a^2+b^2=c^2$

Gravitational force,

$F=\frac{G{M}_1{M}_2}{R^2}$

Centripetal force,

$F=\frac{M{v}^2}{R}$

Step 3: Drawing a figure and calculating the distance between two masses

From the figure, by applying Pythagoras theorem, we get

$$\begin{gathered} A{O}^2+O{B}^2=A{B}^2 \\ \Rightarrow AB=\sqrt{A{O}^2+O{B}^2} \\ \Rightarrow AB=\sqrt{R^2+R^2}\left(\because AO=R,OB=R\right) \\ \Rightarrow AB=\sqrt{2R^2} \\ \Rightarrow AB=\sqrt{2}R \end{gathered}$$

Since the particles are arranged at equal distances symmetrically, the force on all the particles are same

Step 4: Finding gravitational force on each particle

From the figure, it is clear that the component of $F_{AC}and{F}_{AB}$ in the direction of $D$ are $F_{AC}\cos \theta and{F}_{AB}\cos \theta$

Gravitational force on particle $A$,

$$\begin{gathered} F_A=F_{AB}+F_{AC}+F_{AD} \\ \Rightarrow F_G=F_{AB}+F_{AC}+F_{AD} \end{gathered}$$

Since, here AB and AC are same (B and C are symmetrical points)

$$\begin{gathered} F_{AB}=F_{AC} \\ \Rightarrow F_G=F_{AB}\cos 45°+F_{AC}\cos 45°+F_{AD} \\ \Rightarrow F_G=2F\cos 45°+F_{AD}\left(\because F_{AB}=F_{AC}\right) \\ \Rightarrow F_G=2\left(\frac{G{M}^2}{{\left(R\sqrt{2}\right)}^2}\right)\cos 45°+\left(\frac{G{M}^2}{{\left(2R\right)}^2}\right)\left(\because F=\frac{G{M}^2}{{\left(R\sqrt{2}\right)}^2},F_{AD}=\frac{G{M}^2}{{\left(2R\right)}^2}\right) \\ \Rightarrow F_G=2\left(\frac{G{M}^2}{{\left(R\sqrt{2}\right)}^2}\right)\left(\frac{1}{\sqrt{2}}\right)+\left(\frac{G{M}^2}{{\left(2R\right)}^2}\right)\left(\because \cos 45°=\frac{1}{\sqrt{2}}\right) \\ \Rightarrow F_G=\left(\frac{G{M}^2}{\sqrt{2}{R}^2}\right)+\left(\frac{G{M}^2}{4R^2}\right) \\ \Rightarrow F_G=\frac{G{M}^2}{R^2}\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right) \\ \Rightarrow F_G=\frac{G{M}^2}{R^2}\left(\frac{4+\sqrt{2}}{4\sqrt{2}}\right) \\ \Rightarrow F_G=\frac{G{M}^2}{4R^2}\left(2\sqrt{2}+1\right) \end{gathered}$$

Gravitational force on each of the four particle = gravitational force on particle A

Gravitational force on each of the four particle, $F_G=\frac{G{M}^2}{4R^2}\left(2\sqrt{2}+1\right)$

Step 5: Equating centripetal force and gravitational force to find the velocity of the particle

Since the particles also traverse on a circular path, they also experience a centripetal force towards the center. This centripetal force is balanced by the gravitational force experienced by the particle due to other particles.

Therefore,

$$\begin{gathered} F_C=F_G \\ \Rightarrow \frac{M{v}^2}{R}=\frac{G{M}^2}{4R^2}\left(2\sqrt{2}+1\right) \\ \Rightarrow v^2=\frac{GM}{4R}\left(2\sqrt{2}+1\right) \\ \Rightarrow v=\frac{1}{2}\sqrt{\frac{GM}{R}\left(2\sqrt{2}+1\right)} \end{gathered}$$

Therefore, we get the velocity of each particle as $\frac{1}{2}\sqrt{\frac{GM}{R}\left(2\sqrt{2}+1\right)}$.

Hence, Option B is correct.