Gravitation

The problem describes two identical particles of mass m=1kg each moving in a circle of radius R under the action of their mutual gravitational attraction. This means that the particles are moving around a common center, and the distance between the particles is 2R (as the diameter of the circle).

In this case, the force providing the centripetal force for each particle to move in a circular path is the gravitational force between the particles.

The gravitational force between two masses m1 and m2 separated by a distance r is given by Newton's law of universal gravitation:

$F = G \frac{{m_1 m_2}}{{r^2}}$

Since the two particles are identical, m1=m2=m=1kg. And the distance between them r is 2R. Substituting these into the above equation gives the gravitational force between the two particles:

$F = G \frac{{m^2}}{{(2R)^2}} = G \frac{{1}}{{4R^2}}$

This gravitational force is also equal to the centripetal force needed for each particle to move in a circular path of radius R. The centripetal force is given by:

$F = m R ω^2$

Setting these two equations equal to each other gives:

$G \frac{{1}}{{4R^2}} = 1 \cdot R \cdot ω^2$

Rearranging this to solve for ω (the angular speed) gives:

$ω = \frac{1}{2} \sqrt{\frac{G}{R^3}}$

So, the angular speed of each particle is: $\frac{1}{2} \sqrt{\frac{G}{R^3}}$.

In a binary star system, the two stars orbit around a common center of mass. When considering periods of revolution, Kepler's Third Law comes into play. This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit. It's often written in the following form for a single object orbiting another:

T² ∝ r³

For a binary star system, this would still hold true. The periods of revolution for both stars A and B will be the same because they are both orbiting the same common center of mass, regardless of their individual masses or individual orbital radii. In other words, star A and star B complete one orbit in the same amount of time.

So, Option B: T_A = T_B is correct.

The other options (Option A, C, and D) would not be correct. Kepler's Third Law is not a ratio between the periods and radii of two different bodies, and the periods do not depend on the masses of the individual stars or their individual distances from the center of mass.

Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:

$T^2 \propto r^3$

where:

• $T$ is the orbital period
• $r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)

Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T'$ and radius $9R$, we can form an equation:

$\frac{{T'}^2}{T^2} = \frac{(9R)^3}{R^3}$

This simplifies to:

$\frac{{T'}^2}{T^2} = 729$

Taking the square root of both sides, we get:

$T' = T \times \sqrt{729} = T \times 27$

So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.

According to Kepler’s second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.

Net force on mass M at position B towards centre of circle is

${F_{BO\,net}} = {F_{BD}} + {F_{BA}}\sin 45^\circ + {F_{BC}}\sin 45^\circ $

$ = {{G{M^2}} \over {{{(\sqrt 2 a)}^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right) + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right)$

[where, diagonal length BD is $\sqrt2$a]

$ = {{G{M^2}} \over {2{a^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{2 \over {\sqrt 2 }}} \right) = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

This force will act as centripetal force.

Distance of particle from centre of circle is ${a \over {\sqrt 2 }}$.

Here, ${F_{centripetal}} = {{M{v^2}} \over r} = {{M{v^2}} \over {{a \over {\sqrt 2 }}}} = {{\sqrt 2 M{v^2}} \over a}$ ($\because$ $r = {a \over {\sqrt 2 }}$)

So, for rotation about the centre,

${F_{centripetal}} = {F_{BO(net)}}$

$ \Rightarrow \sqrt 2 {{M{v^2}} \over a} = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

$ \Rightarrow {v^2} = {{GM} \over a}\left( {1 + {1 \over {2\sqrt 2 }}} \right) = {{GM} \over a}(1.35)$

$ \Rightarrow v = 1.16\sqrt {{{GM} \over a}} $