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Three identical stars of mass \(M\) form an equilateral triangle that rotates around the triangle's center as the stars move in a common circle about that center. The triangle has edge length \(L\). What is the speed of the stars?

Short Answer

Expert verified
The speed of the stars is \( v = \sqrt{\frac{3GM}{L}} \).

Step by step solution

01

Determine the position of the center of mass

Since the system is symmetrical, the center of mass is at the centroid of the equilateral triangle. For an equilateral triangle, the centroid is equidistant from all vertices.
02

Calculate gravitational force between two stars

For any two stars at a distance of edge length \( L \), the gravitational force \( F \) between them is given by Newton's law of gravitation: \( F = \frac{G M^2}{L^2} \).
03

Calculate net gravitational force towards the center of mass

Each star is attracted to the other two stars. Consider one star: the net gravitational force towards the center is the sum of the components of the gravitational forces between it and the other two stars, directed towards the center of the triangle. Each of these has a component towards the center, given by \( F_{net} = 3 \cdot \frac{G M^2}{L^2} \cdot \cos(30^\circ) \) because the angle between the line connecting any star to the center and the line connecting two stars is \( 30^\circ \).
04

Express the centripetal force on a star

The centripetal force \( F_c \) needed to keep a star in circular motion is given by \( F_c = \frac{M v^2}{R} \), where \( R = \frac{L}{\sqrt{3}} \) is the distance from each star to the center (centroid of the triangle) of the rotating system.
05

Equate gravitational and centripetal forces

For circular motion, the net gravitational force acts as the centripetal force. Equate them: \( \frac{3 G M^2}{L^2} \cdot \cos(30^\circ) = \frac{M v^2}{R} \), using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( R = \frac{L}{\sqrt{3}} \).
06

Solve for speed \( v \)

Substitute \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( R = \frac{L}{\sqrt{3}} \) into the equation from Step 5: \( \frac{3 G M^2}{L^2} \cdot \frac{\sqrt{3}}{2} = \frac{M v^2 \sqrt{3}}{L} \). Solve for \( v \): \[ v = \sqrt{\frac{3 G M}{L}} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
Understanding gravitational force is key when working with celestial bodies, like stars. Gravitational force is a natural phenomenon by which all things with mass are brought toward one another. In this context, the gravitational force between two stars of mass \( M \) can be calculated using Newton's Law of Universal Gravitation: \( F = \frac{G M^2}{L^2} \).

Here, **\( G \)** is the gravitational constant, \( 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \). This constant allows us to compute the attractive force based on the mass of the objects and the distance between them.

Key points to remember about gravitational force:
  • It is always attractive.
  • It depends on the product of the masses involved.
  • It decreases with the square of the distance between the centers of the two masses.
In an equilateral triangle of stars, each star exerts a gravitational pull on the others, contributing to the net force acting towards the center of the formation.
Centripetal Force
Centripetal force is crucial in understanding rotational motion and the behavior of objects moving along a curved path. When any object moves in a circle, it constantly changes direction, which requires a force directed towards the center of the circle, known as centripetal force.

For a star rotating around the center of an equilateral triangle, the centripetal force required to maintain its circular path can be expressed as \( F_c = \frac{M v^2}{R} \), where:
  • \( M \) is the mass of the star.
  • \( v \) is the velocity of the star.
  • \( R \) is the radius of the circular path, specifically \( \frac{L}{\sqrt{3}} \), derived from the geometry of the equilateral triangle.

The gravitational forces between the stars provide the necessary centripetal force for this rotational motion. Understanding how these forces balance allows us to solve for the speed of the stars.
Equilateral Triangle
An equilateral triangle is a geometric shape where all three sides are of equal length, and all three interior angles are \( 60^\circ \). This symmetry plays a significant role in the problem presented, as it dictates the way the forces interact among the stars.

In the context of three stars forming an equilateral triangle, the centroid acts as the point of rotation. The centroid is the spot where all the "mass" of the triangle can be considered to be concentrated, and it is equidistant from all the vertices.

Important properties of an equilateral triangle:
  • All sides, by definition, are the same length, which simplifies calculations of forces.
  • Its symmetry means any vertex is always \( 60^\circ \) apart from another.
  • The distance from the centroid to any vertex (\( R = \frac{L}{\sqrt{3}} \)) is crucial in determining rotational dynamics.
In our scenario, using these geometrical properties allows us to translate the gravitational interactions into useful centripetal forces, leading us to calculate the velocity of the stars in the rotating system.

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Most popular questions from this chapter

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