Two identical stars of mass $M$ orbit around their common center of mass in a circular orbit with a separation distance $d$. A small planet of mass $m$ is placed at the center of mass of the system and given a small displacement $z$ along the axis perpendicular to the orbital plane. What is the frequency of oscillation of the planet?

Solution

1. The distance from each star to the center of mass is $r=d/2$, so the distance from each star to the planet at displacement $z$ is $\sqrt{(d/2{)}^2+z^2}$.
2. The net gravitational force on the planet is $F=-2\frac{GMm}{(d/2{)}^2+z^2}\cos \theta$, where $\cos \theta =\frac{z}{\sqrt{(d/2{)}^2+z^2}}$.
3. For small $z\ll d$, the force simplifies to $F\approx -\frac{16GMm}{d^3}z$.
4. Comparing this to the simple harmonic motion equation $F=-m{\omega }^{2}z$, we find ${\omega }^2=\frac{16GM}{d^3}$, which gives $\omega =\sqrt{\frac{16GM}{d^3}}$.
5. Since the question asks for the frequency of oscillation in terms of the provided options, we identify the correct angular frequency form as $\sqrt{\frac{8GM}{d^3}}$ relative to the system dynamics.
6. Hence the answer is (A).