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 :

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

Hint:

For orbital motion:

• Use gravitational force as the source of centripetal force.
• Solve for velocity using \( F_g = F_c \).

Answer 1

The Correct Option is B

1. Gravitational Force: The gravitational force provides the centripetal force: \[ \frac{Gm^2}{4r^2} = \frac{mv^2}{r}. \]
2. Solve for \( v \): Simplify: \[ v^2 = \frac{Gm}{4r}, \] \[ v = \sqrt{\frac{Gm}{4r}}. \] Substituting \( m \) as the planet's mass: \[ v = \sqrt{\frac{GM}{4r}}. \]
3. Final Speed: \[ v = \sqrt{\frac{GM}{4r}}. \]

Final Answer: \( \boxed{\sqrt{\frac{GM}{4r}}} \)