Question

Two particles of equal mass '\(m\)' move in a circle of radius '\(r\)' under the action of their mutual gravitational attraction. The speed of each particle will be ( \(\text{G} =\) Universal gravitational constant)

• A. \(\sqrt{\frac{\text{Gm}}{4r}}\)
• B. \(\sqrt{\frac{\text{Gm}}{r}}\)
• C. \(\sqrt{\frac{\text{Gm}}{2r}}\)
• D. \(\sqrt{\frac{4\text{Gm}}{r}}\)

Hint:

In mutual rotation, distance between masses is \(2r\), but circular radius is \(r\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When two particles of equal mass move in a circle under mutual gravitational attraction, they must always be diametrically opposite to each other to maintain symmetrical motion.
The gravitational force between them provides the necessary centripetal force for their circular motion.
Step 2: Key Formulas or Approach:
Newton's law of gravitation: \( F_g = \frac{Gm_1m_2}{d^2} \).
Centripetal force required: \( F_c = \frac{mv^2}{R} \).
Equating the two forces gives the condition for steady circular motion.
Step 3: Detailed Explanation:
Let the mass of each particle be \( m \) and the radius of the circular path be \( r \).
The distance between the two diametrically opposite particles is \( d = 2r \).
The magnitude of the gravitational force of attraction between them is:
\[ F_g = \frac{G \cdot m \cdot m}{(2r)^2} = \frac{Gm^2}{4r^2} \] This force acts towards the center of the circle and acts as the centripetal force.
For a particle moving in a circle of radius \( r \) with speed \( v \), the required centripetal force is:
\[ F_c = \frac{mv^2}{r} \] Equating the gravitational force to the centripetal force:
\[ \frac{mv^2}{r} = \frac{Gm^2}{4r^2} \] Cancel one \( m \) and one \( r \) from both sides:
\[ v^2 = \frac{Gm}{4r} \] Taking the square root gives the speed of each particle:
\[ v = \sqrt{\frac{Gm}{4r}} \] Step 4: Final Answer:
The speed of each particle is \( \sqrt{\frac{Gm}{4r}} \).