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Step-by-Step Solution
Step 1: Understand the System
Four identical particles, each of mass 1 kg, are placed at the corners of a circle of radius 1 m. Each particle experiences the gravitational pull due to the other three particles, and they move in a circular path under this mutual gravitational attraction. We need to determine the speed of each particle.
Step 2: Identify the Forces on One Particle
Consider one particle. It is attracted by the other three particles. Because of symmetry, these four particles form a square when viewed along the circle, and the net resultant attraction on each particle points towards the center of the circle.
Step 3: Calculate the Pairwise Gravitational Forces
The gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by:
$$
F = G \frac{m_1 m_2}{r^2}.
$$
Here, each mass is 1 kg, so $m_1 = m_2 = 1\,\text{kg}$. The distances from the chosen particle to the other three are:
The adjacent particles on the circle are at a central angle of $120^\circ$, so the distance between one particle and its adjacent particle is 2 (radius) × sin(60°) = $2 \times \sin(60^\circ) = \sqrt{3}$ if they form an equilateral triangle. However, we should be precise that the four particles are equally spaced on a circle, so each particle is separated by 90° along the circumference from its neighbors. The actual chord length between adjacent particles on the circle is $2R \sin(\theta/2)$; for $\theta = 90^\circ$, this distance is $2 \times 1 \times \sin(45^\circ) = \sqrt{2}.$
The distance to the particle diagonally opposite on the circle is the chord subtending 180° across the circle, which is $2 \times 1 = 2.$
Therefore, each particle sees:
Two particles at distance $\sqrt{2}\,\text{m}$,
One particle at distance $2\,\text{m}.$
Step 4: Calculate the Net Radial Force
(a) Force due to one particle at distance $\sqrt{2}\,\text{m}$:
$$
F_{\sqrt{2}} = G \frac{(1)(1)}{(\sqrt{2})^2} = \frac{G}{2}.
$$
(b) Force due to one particle at distance $2\,\text{m}$:
$$
F_{2} = G \frac{(1)(1)}{(2)^2} = \frac{G}{4}.
$$
There are two particles at a distance $\sqrt{2}\,\text{m}$ each, and one particle at a distance $2\,\text{m}$. We must also sum the vector components toward the center. By symmetry, the net radial force from the two particles at distance $\sqrt{2}\,\text{m}$ plus the one particle at distance $2\,\text{m}$ comes out to:
$$
F_{\text{net}} = 2 \times \frac{G}{2} \cos(45^\circ) + \frac{G}{4}.
$$
Here, $\cos(45^\circ) = \frac{1}{\sqrt{2}}.$ Accounting for directions carefully gives the well-known result:
$$
F_{\text{net}} = G \left(\frac{1 + 2\sqrt{2}}{4}\right) \times \text{(some factor)},
$$
which is often simplified directly through a known geometric argument (refer to the figure below for the exact vector decomposition). The final combined radial force is generally quoted as:
$$
F_{\text{net}} = \frac{G}{4}(1 + 2\sqrt{2}).
$$
This result matches the fact that two forces at distance $\sqrt{2}$ have a component sum in the radial direction plus the single force from distance 2. This is a standard result for four equal masses at the vertices of a square, each attracting a corner mass. The geometry ensures the net direction is radially inwards.
Step 5: Equate Net Force to Centripetal Force
Since each mass moves in a circle of radius $R = 1\,\text{m}$ with speed $v$, the required centripetal force is:
$$
F_{\text{centripetal}} = \frac{m v^2}{R} = \frac{1 \times v^2}{1} = v^2.
$$
Set this equal to the net gravitational force:
$$
v^2 = \frac{G}{4}\,(1 + 2\sqrt{2}).
$$
Step 6: Solve for the Speed
Taking the square root on both sides,
$$
v = \sqrt{ \frac{G}{4}\,(1 + 2\sqrt{2}) } = \frac{1}{2}\sqrt{G(1 + 2\sqrt{2})}.
$$
Step 7: Final Answer
The speed of each particle is:
$$
\boxed{ \frac{1}{2}\sqrt{ G(1 + 2\sqrt{2}) } }.
$$
