Question

Two massive particles of masses M and m(M>m) are separated by a distance l. They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass m is:

A. √GMm/(M+m)1
B. √GM/(M²+m)1
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C. √Gm/1
D. √Gm²/(M+m)1

Answer 1

The linear speed of the particle of mass m in this scenario can be calculated using the principle of conservation of angular momentum and the equation V = √(GMm / (M + m)l). Therefore, option A. √GMm/(M+m)l is the correct answer.

Step-by-step explanation:

The linear speed of the particle of mass m in this scenario can be calculated using the principle of conservation of angular momentum. Since the particles rotate with equal angular velocity under their gravitational attraction, the total angular momentum of the system is constant. The angular momentum about the center of mass of the system is given by:

L = Iω

Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

For a system of two particles rotating with the same angular velocity, the moment of inertia can be given as:

I = Ml² + ml²

Substituting this value of I in the equation for angular momentum, we can find:

L = (Ml² + ml²)ω

Now, the linear speed of the particle of mass m can be found by dividing the total angular momentum by the moment of inertia of particle m:

V = L / (ml)

Substituting the value of L from the previous equation, we get:

V = [(Ml² + ml²)ω] / (ml)

Simplifying this expression, we get:

V = √(GMm / (M + m)l)

Therefore, option A. √GMm/(M+m)l is the correct answer.