Two planets revolve with the same angular velocity about a star. The radius of orbit of the outer planet is twice the radius of orbit of the inner planet. If $T$ is the time period of the revolution of the outer planet, find the time in which the inner planet will fall into the star. If it suddenly stopped.
A. $\dfrac{{T\sqrt 2 }}{8} \\ $
B. $\dfrac{{T\sqrt 2 }}{{16}} \\ $
C. $\dfrac{{T\sqrt 2 }}{4} \\ $
D. $\dfrac{{T\sqrt 2 }}{{32}}$

Hint: In this case, when a problem is based on Gravitation, we know that various quantities such as radius, time period, mass, etc., play a significant role in establishing a relationship between different parameters hence, use Kepler's law for calculating time period and use the given conditions precisely in order to provide an accurate solution.

Formula used:
The expression of kepler’s law is,
$T^2=kr^3$
Here, $T$ is the time period and $r$ is the length of the semi major axis.

Complete step by step solution:
Angular velocity of two planets is same i.e., $\omega $ (given).
The radius of the orbit of an outer planet is twice the radius of the orbit of the inner planet i.e., ${r_o} = 2{r_i}$ (given).
Let the time period of an outer planet is denoted by ${T_o} = T$ (given)
Now, The Time period is defined as: -
$T = \dfrac{{2\pi r}}{\omega }$
For an outer planet,
${T_o} = \dfrac{{2\pi {r_o}}}{\omega }$
And for the inner planet,
${T_i} = \dfrac{{2\pi {r_i}}}{\omega }$
$\Rightarrow {T_i} = \dfrac{{2\pi {r_o}}}{{2\omega }} = \dfrac{{{T_o}}}{2}$....…(1).........$\left( {\therefore {r_i} = \dfrac{{{r_o}}}{2}} \right)$

According to Kepler’s Law,
$\dfrac{T'^2}{(r/4)^3}=\dfrac{T_i^2}{(r/4)^3} \\ $
$ \Rightarrow T{'^2} = \dfrac{8}{{64}}{T_i}^2 \\ $
Substitute the value ${T_i}$ from eq. (1) in the above expression, we get
$T{'^2} = \dfrac{8}{{64}}{\left( {\dfrac{{{T_o}}}{2}} \right)^2} \\ $
$ \Rightarrow T{'^2} = \dfrac{{{T_o}^2}}{{32}} \\ $
Taking Square roots both sides, we get
$ \Rightarrow T' = \dfrac{{{T_o}}}{{4\sqrt 2 }} \\ $
Since, ${T_i} = \dfrac{{{T_o}}}{2}$

Therefore, $\dfrac{{T'}}{2}$ will represent the time in which the inner planet will fall into the star,
$\left( {\dfrac{{T'}}{2}} \right) = \dfrac{{\dfrac{{{T_o}}}{{4\sqrt 2 }}}}{2} \\
\Rightarrow \left( {\dfrac{{T'}}{2}} \right) = \dfrac{{{T_o}}}{{8\sqrt 2 }}$
Also, ${T_o} = T$ (given)
$ \Rightarrow \left( {\dfrac{{T'}}{2}} \right) = \dfrac{T}{{8\sqrt 2 }} \\
\therefore \left( {\dfrac{{T'}}{2}} \right)= \dfrac{{T\sqrt 2 }}{{16}} \\ $
Thus, the time in which the inner planet will fall into the star, if it was suddenly stopped, will be $\dfrac{{T\sqrt 2 }}{{16}}$.

Hence, the correct option is B.

Note: Since this is a problem related to the calculation of time period hence, given conditions must be analysed very carefully, and then Kepler’s law should be applied. Quantities that are required to calculate the time period, must be identified on a prior basis as it gives a better understanding of the problem.