Question

Two plants (each having mass m) revolve around a stationary star (of mass M) in a circle of radius r. The two planets are always on opposite side of the star(i.e. they are diametrically opposite and always having separation 2r). The orbital period T of the planets is of the form $T=2\pi \sqrt{\frac{r^3}{G{M}^{\prime }}}$. Then what is the value of M'?

Answer 1

Net force on mass $m$ is-

$F=\frac{GMm}{r^2}+\frac{G{m}^2}{(2r{)}^2}=\frac{Gm}{4r^2}(4M+m)$

And this force provides the necessary centripetal force to move on the circle-

$⟹\frac{Gm}{4r^2}(4M+m)=\frac{m{v}^2}{r}$

$⟹v=\sqrt{\frac{G(4M+m)}{4r}}$

$T=\frac{2\pi r}{v}=2\pi \sqrt{\frac{4r^3}{G(4M+m)}}=2\pi \sqrt{\frac{r^3}{G(M+\frac{m}{4})}}$

Hence, $M^{\prime }=M+\frac{m}{4}$