Gravitation

Concept. Two equal masses attracting each other go round their common centre of mass, which is the centre of the circle. The gravitational pull of one on the other supplies the centripetal force. Be careful: the separation is the diameter $2a$, not $a$.

Step 1. Force between them. $F=\dfrac{Gm^2}{(2a)^2}=\dfrac{Gm^2}{4a^2}$.

Step 2. Centripetal condition (each moves on radius $a$): $$\frac{Gm^2}{4a^2}=m\omega^2a\ \Rightarrow\ \omega^2=\frac{Gm}{4a^3}$$

Answer. $\omega=\sqrt{\dfrac{Gm}{4a^3}}$ (Option C).

Concept. In a binary system both stars orbit the common centre of mass with the same period. Write the centripetal equation for one star using its own orbit radius; the result depends only on the total mass and the separation.

Step 1. Orbit radii. The centre of mass divides $d$ inversely to the masses: the star of mass $m$ is at $r_1=\dfrac{2m}{3m}d=\dfrac{2d}{3}$ from it.

Step 2. Centripetal equation for mass $m$. $$\frac{G(m)(2m)}{d^2}=m\omega^2\cdot\frac{2d}{3}\ \Rightarrow\ \omega^2=\frac{3Gm}{d^3}$$

Step 3. Period. $T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{d^3}{3Gm}}$.

Answer. Option A. In general $T=2\pi\sqrt{\dfrac{d^3}{G(m_1+m_2)}}$.

Concept. Each particle is pulled by the other three. The pulls' components along the radius add up to the centripetal force; the tangential components cancel by symmetry.

Step 1. Geometry. Four particles on a circle of radius $1$ m form a square of side $\sqrt2$ m and diagonal $2$ m.

Step 2. Pull of the two neighbours. Each is $\dfrac{G(1)(1)}{(\sqrt2)^2}=\dfrac G2$, at $45^\circ$ to the radius. Their radial parts add: $$2\cdot\frac G2\cos45^\circ=\frac{G}{\sqrt2}$$

Step 3. Pull of the opposite particle, straight along the radius: $\dfrac{G}{2^2}=\dfrac G4$.

Step 4. Centripetal condition ($m=1$, $r=1$): $$\frac{v^2}{1}=\frac{G}{\sqrt2}+\frac G4=\frac{G(2\sqrt2+1)}{4}\ \Rightarrow\ v=\frac{\sqrt{(1+2\sqrt2)G}}{2}$$

Answer. Option D.