If gravitational attraction between two point masses be given by $F = \dfrac{{G{m_1}{m_2}}}{{{r^n}}}$. Then the period of the satellite in a circular orbit is proportional to
A) $r$
B) ${r^2}$
C) ${r^{\dfrac{1}{2}}}$
D) Independent of n

Hint: The satellite revolving around the planet would have both centripetal force and force of gravitation. The period of the satellite can be determined from the linear velocity of the satellite. The linear and angular velocities are related.

Complete step by step answer:
We have given according to the Universal law of gravitation, the force of attraction between two point masses as,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^n}}}$
Where, ${m_1}$and ${m_2}$ are the masses and $r$ is the distance between them, and $G$ is the gravitational constant.
We can take the mass of the planet as ${m_1}$ and the mass of the satellite as ${m_2}$.
For the satellite to orbit around the planet the force of gravitation between the planet and satellite will be equal to the centripetal force of the satellite. Thus the satellite is maintained in a circular orbit revolving around the planet.
The expression for the centripetal force of the satellite is given as,
${F_C} = \dfrac{{{m_2}{v^2}}}{r}$
Where ${m_2}$ mass of the satellite, $v$ is the linear velocity of the satellite and $r$ is the radius of the orbit. it will be equal to the distance between the planet and satellite.
Therefore for the orbital motion, equate the force of attraction and centripetal force.
$
\Rightarrow \dfrac{{{m_2}{v^2}}}{r} = \dfrac{{G{m_1}{m_2}}}{{{r^n}}} \\
\Rightarrow {v^2} = \dfrac{{G{m_1}r}}{{{r^n}}} \\
\Rightarrow {v^2} = \dfrac{{G{m_1}}}{{{r^{n - 1}}}} \\
\Rightarrow v = \sqrt {\dfrac{{G{m_1}}}{{{r^{n - 1}}}}} \\
$
Thus the linear velocity of the satellite is found.
The linear and angular velocities are related by the expression as,
$v = r\omega $
Where $\omega $ is the angular velocity.
Therefore, $\omega = \dfrac{v}{r}$
The time period of the satellite is given as,
$T = \dfrac{{2\pi }}{\omega }$
Substitute for the angular velocity $\omega $ in the above expression.
$
\Rightarrow T = \dfrac{{2\pi r}}{v} \\
\Rightarrow T = 2\pi r\sqrt {\dfrac{{{r^{n - 1}}}}{{G{m_1}}}} \\
\Rightarrow T = 2\pi \sqrt {\dfrac{{{r^2} \times {r^{n - 1}}}}{{G{m_1}}}} \\
\Rightarrow T = 2\pi \sqrt {\dfrac{{{r^{n + 1}}}}{{G{m_1}}}} \\
$
Therefore,
$
\Rightarrow T\propto {\left( {{r^{n + 1}}} \right)^{\dfrac{1}{2}}} \\
\Rightarrow T\propto {r^{\dfrac{{n + 1}}{2}}} \\
\Rightarrow T\propto {r^{\dfrac{n}{2}}}.{r^{\dfrac{1}{2}}} \\
$

We can say that the time period of the satellite is proportional to ${r^{\dfrac{1}{2}}}$. The correct answer is option C.

Note:
When the angular velocity is greater than, the time period will be a small value. Also, the angular velocity is proportional to the distance. If the distance increases, gradually the time period decreases.