Question

Two identical particles each of mass ‘m’ go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be ______

Options

• `sqrt ((Gm)/(2a^3))`

• `sqrt ((Gm)/(8a^3))`

• `sqrt ((Gm)/(4a^3))`

• `sqrt ((Gm)/a^3)`

Answer 1

Two identical particles each of mass ‘m’ go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be `bbunderline(sqrt ((Gm)/(4a^3)))`.

Explanation:

Two identical masses (m) revolve about their centre of mass, each at radius a.

Distance between them = 2a

Gravitational force provides centripetal force:

F = `(Gm^2)/(2a)^2`

= `(Gm^2)/(4a^2)`    ...(i)

Centripetal force (F) = mω²a

Putting this value of F in equation (i), we get,

mω²a = `(Gm^2)/(4a^2)`

ω² = `(Gm)/(4a^3)`

ω = `sqrt ((Gm)/(4a^3))`