Question

Two particles of equal mass go around a circle
of radius R under the action of their mutual
gravitational force of attraction. The speed of
each particle is (M = mass of the particle)
[RPMT 2003]​

Answer 1

Question:

Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational force of attraction. The speed of each particle is (M = mass of the particle)?

Answer:

• Speed of The Particle (v) = $\tt \dfrac{1}{2} \sqrt{ \dfrac{GM}{R}}$

Given:

1. Mass of the particle = M.
2. Radius of circle = R.
3. Gravitational Constant = G.

Explanation:

$\rule{300}{1.5}$

Gravitational Force:-

It is Expressed as $\tt F_G = \dfrac{GM_1M_2}{D^2}$

$\bold{Here} \begin{cases}\sf{M_1 = M_2 = M \: [ Masses \: are\: same]} \\ \sf{D = R + R = 2R \: [Radius\: are\: same]}\end{cases}$

Centripetal Force:-

It is Expressed as $\large\tt F_C = \dfrac{Mv^2}{R}$

$\bold{Here} \begin{cases}\sf{M = Masses \: of \: body} \\ \sf{v = Velocity \: of\: Body} \\ \sf{R = Radius \: of \: Circle} \end{cases}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, These Forces Balance each other For the Body to Complete a Circle.

Gravitational Force = Centripetal Force.

$\large{\boxed{\tt F_G = F_C}}$

$\large{\tt \hookrightarrow \dfrac{GM_1M_2}{D^2} = \dfrac{Mv^2}{R}}$

Substituting the values.

$\large{\tt \hookrightarrow \dfrac{G \times M \times M}{(2R)^2} = \dfrac{Mv^2}{R}}$

$\large{\tt \hookrightarrow \dfrac{G \times M^2}{4R^2} = \dfrac{Mv^2}{R}}$

$\large{\tt \hookrightarrow \dfrac{G \times M}{4R^2} = \dfrac{v^2}{R}}$

$\large{\tt \hookrightarrow v^2 = \dfrac{G \times M}{4R}}$

$\large{\tt \hookrightarrow v = \sqrt{ \dfrac{G \times M}{4R}}}$

$\huge{\boxed{\boxed{\tt v = \dfrac{1}{2} \sqrt{ \dfrac{G M}{R}}}}}$

Hence Derived!

This speed is Required For each Particle to go around a Circle.

$\rule{300}{1.5}$

Answer 2

Answer:

Given:

2 particles of equal mass are undergoing uniform circular motion under the action of Gravitational force.

Let ,

• mass be M
• Radius be R

To find:

Speed of each particle

Concept:

The mutual gravitational force is actually providing a centripetal component to both objects.

This centripetal force is responsible for this circular motion.

Important things to note:

1. Separation distance of the objects is R + R = 2R . This distance will be included in the Gravitational force formula.
2. The radius of circular rotation is R. This will be included in Centripetal force formula.

Formulas used:

Gravitational force:

F = (G × M1 × M2)/D²

Centripetal Force:

F" = Mv²/R , where "v" is the velocity.

Calculation:

Gravitational force = Centripetal Force

=> (G × M × M)/(2R)² = Mv²/R

=> (G × M²)/4R² = Mv²/R

=> v² = GM/4R

=> v = √(GM/4R)

=> v = ½√(GM/R)

So the final answer is :

$v \: = \frac{1}{2} \sqrt{ \frac{GM}{R} }$