Two planets are in a circular orbit of radius R and 4R about a star. At a specific time, the two planets and the star are in a straight line. If the period of the closest plane is T, then the star and planets will again be in a straight line after a minimum time

The time period of a revolving body in circular path is given by,
$T^2=\frac{4{\pi }^2}{GM}{R}^3$

where,
R is the radius of the orbit,
M is the mass of central body.
$T^2\propto R^3$

$\Rightarrow \frac{{\mathrm{T}}^{\mathrm{\prime }2}}{{\mathrm{T}}^2}=\frac{(4\mathrm{R}{)}^3}{{\mathrm{R}}^3}$

$\Rightarrow {\mathrm{T}}^{\mathrm{\prime }2}=64{\mathrm{T}}^2$

$\Rightarrow {\mathrm{T}}^{\mathrm{\prime }2}=64{\mathrm{T}}^2$
Minimum time after which the same confi guration will happen is,
Case I: Moving in the same sense.
$\mathrm{t}=\frac{2\pi }{{\omega }_{\text{rel}}}=\frac{2\pi }{\frac{2\pi }{\mathrm{T}}-\frac{2\pi }{{\mathrm{T}}^{\mathrm{\prime }}}}=\frac{2\pi }{\frac{2\pi }{\mathrm{T}}-\frac{2\pi }{8\mathrm{T}}}=\frac{8\mathrm{T}}{7}$

$\text{or}\frac{4T}{7}\text{(when they are diametrically opposite)}$
Case II: Moving in opposite sense.
$\mathrm{t}=\frac{2\pi }{{\omega }_{\text{rel}}}=\frac{2\pi }{\frac{2\pi }{\mathrm{T}}+\frac{2\pi }{{\mathrm{T}}^{\mathrm{\prime }}}}=\frac{2\pi }{\frac{2\pi }{\mathrm{T}}+\frac{2\pi }{8\mathrm{T}}}=\frac{8\mathrm{T}}{9}$

$\text{or}\frac{4T}{9}\text{(when they are diametrically opposite)}$
Note: Among the given options, they will align after 8T also, but this will not be the minimum time.