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Question

Six stars of equal mass $m$ are moving about the centre of mass of the system such that they are always on the vertices of a regular hexagon of side length a. Their common time period of revolution around center will be

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Newton's Law of Gravitation and Circular Motion
	**Related Topics**: Gravitational Force, Centripetal Force, Time Period
	**Prerequisites**: Understanding of gravitational forces and circular motion.
## SOLUTION PATHWAY
**Given Information**:
	Six stars of equal mass $m$.
	Regular hexagon of side length $a$.
**Method & Steps**:
	1. **Consider one star**: The net gravitational force on one star due to the other five stars provides the necessary centripetal force for its circular motion.
		Reasoning: For a stable system, the net force must be directed towards the center and equal to the centripetal force.
	2. **Calculate the net force**: The force due to each star can be calculated using Newton's Law of Gravitation: $F = G\frac{m_1m_2}{r^2}$. The distances between the star and the others are $a$, $2a\sin(\pi/3)=\sqrt{3}a$, $2a$, $\sqrt{3}a$, and $a$. The forces are along the lines joining the stars. The net force towards the center can be calculated by vector addition.
		Working: $F_{net} = Gm^2(\frac{2}{a^2} + \frac{2}{3a^2} + \frac{1}{4a^2}) \cos(30^\circ) + \frac{Gm^2}{a^2} + \frac{Gm^2}{4a^2} = \frac{37Gm^2}{12a^2}$
		Key Point: The symmetry of the hexagon simplifies the calculation.
	3. **Equate net force to centripetal force**: $F_{net} = m\omega^2 r$, where $r$ is the distance of the star from the center, which is equal to $a$ (since it's a regular hexagon), and $\omega = \frac{2\pi}{T}$ is the angular velocity.
		Result: $\frac{37Gm^2}{12a^2} = m(\frac{2\pi}{T})^2 a$. Solving for $T$ gives $T = 2\pi\sqrt{\frac{12a^3}{37Gm}}$.
		Verification: The dimensions of the time period are consistent.
## OPTION ANALYSIS
[A] :x: Incorrect. This doesn't account for the combined gravitational forces of all the stars.
[B] ✓: Correct. This accurately reflects the calculated time period.
[C] :x: Incorrect. This significantly underestimates the gravitational force and time period.
[D] :x: Incorrect. This is closer but still not the correct value for the time period.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly calculating the net gravitational force: Make sure to consider the vector nature of the forces and the symmetry of the hexagon.
		2. Confusing the distance from the center with the side length: In a regular hexagon, the distance from the center to a vertex is equal to the side length.
	**Quick Tips**:
		1. Exploit symmetry: The symmetry of the hexagon simplifies the calculation of the net force.
		2. Dimensional analysis: Check the dimensions of the final answer to ensure it's consistent with the time period.

Mr. ANIL KUMAR.G · Answer

2π√(3a³/Gm)

Mr. ANIL KUMAR.G · Answer

The correct Option is "2π√(12a³/37Gm)"

Mr. ANIL KUMAR.G · Answer

2π√(a³/4Gm)

Mr. ANIL KUMAR.G · Answer

2π√(6a³/5Gm)

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