Question

Three identical stars of mass M are located at the vertices of an equilateral triangle with side L. The speed at which they will move if they all revolve under the influence of one another's gravitational force in a circular orbit circumscribing the triangle while still preserving the equilateral triangle:

• A. $\sqrt{\frac{2GM}{L}}$
• B. $\sqrt{\frac{GM}{L}}$
• C. $2\sqrt{\frac{GM}{L}}$
• D. not possible at all

Answer 1

Answer: (B)

(B)

Answer 2

The problem asks for the speed at which three identical stars, forming an equilateral triangle of side L, will revolve in a circular orbit under their mutual gravitational attraction.

1. Identify the forces acting on one star:
   Consider one star, say star A. It is attracted by the other two stars, B and C.
   The gravitational force between any two stars (e.g., A and B) is given by Newton's Law of Gravitation: $F = \frac{GM \cdot M}{L^2} = \frac{GM^2}{L^2}$
   The force $F_{BA}$ from star B on A acts along the line BA, and the force $F_{CA}$ from star C on A acts along the line CA.

2. Calculate the net gravitational force:
   Since the stars form an equilateral triangle, the angle between $F_{BA}$ and $F_{CA}$ at star A is $60^\circ$.
   The magnitude of both forces is $F = \frac{GM^2}{L^2}$.
   The net gravitational force ($F_{net}$) on star A is the vector sum of $F_{BA}$ and $F_{CA}$. Using the parallelogram law of vector addition: $F_{net} = \sqrt{F^2 + F^2 + 2F \cdot F \cos(60^\circ)}$
   $F_{net} = \sqrt{2F^2 + 2F^2 \cdot \frac{1}{2}}$
   $F_{net} = \sqrt{2F^2 + F^2}$
   $F_{net} = \sqrt{3F^2} = F\sqrt{3}$
   Substitute the value of F: $F_{net} = \sqrt{3} \frac{GM^2}{L^2}$
   This net force is directed towards the center of the equilateral triangle, which is the center of the circular orbit.

3. Determine the radius of the circular orbit:
   The stars revolve in a circular orbit circumscribing the equilateral triangle. The center of this circle is the centroid of the triangle.
   For an equilateral triangle with side L, the distance from the centroid to any vertex (which is the radius R of the circumscribing circle) is: $R = \frac{L}{\sqrt{3}}$

4. Equate the net gravitational force to the centripetal force:
   The net gravitational force provides the necessary centripetal force ($F_c$) for the circular motion.
   The centripetal force for a star of mass M moving with speed v in a circle of radius R is: $F_c = \frac{Mv^2}{R}$
   Equating $F_{net}$ and $F_c$: $\frac{Mv^2}{R} = \sqrt{3} \frac{GM^2}{L^2}$

5. Solve for the speed (v):
   Substitute $R = \frac{L}{\sqrt{3}}$ into the equation: $\frac{Mv^2}{\frac{L}{\sqrt{3}}} = \sqrt{3} \frac{GM^2}{L^2}$
   $\frac{\sqrt{3} Mv^2}{L} = \sqrt{3} \frac{GM^2}{L^2}$
   Cancel $\sqrt{3}$ from both sides: $\frac{Mv^2}{L} = \frac{GM^2}{L^2}$
   Cancel M from both sides and one L from the denominator on both sides: $v^2 = \frac{GM}{L}$
   $v = \sqrt{\frac{GM}{L}}$

The final answer is $\sqrt{\frac{GM}{L}}$.

Explanation of the solution: The net gravitational force on each star is the vector sum of forces from the other two stars. For an equilateral triangle, this net force is $\sqrt{3} \frac{GM^2}{L^2}$ and directed towards the center. This force provides the centripetal force for the circular orbit. The radius of this orbit is the distance from the center of the triangle to a vertex, which is $R = \frac{L}{\sqrt{3}}$. Equating the centripetal force $\frac{Mv^2}{R}$ to the net gravitational force and solving for $v$ yields $v = \sqrt{\frac{GM}{L}}$.