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**Hint:**We can recall that the period of revolution of an orbiting is the time taken to complete a cycle of its orbit. The centripetal force required to keep any object in orbit about a planet is provided by the gravitational force of attraction which pulls the object towards the planet

Formula used: In this solution we will be using the following formulae;

\[KE = \dfrac{1}{2}m{v^2}\] where \[KE\] is the kinetic energy of a body, \[m\] is the mass of the body, and \[v\] is the velocity of the body.

\[R = \dfrac{{GM}}{{{v^2}}}\] where \[R\] is the radius of an object orbiting a planet, \[G\] is the universal gravitational constant, \[M\] is the mass of the planet and \[v\] is the velocity of the orbiting object at that specific radius.

\[T = \dfrac{{2\pi R}}{v}\] where \[T\] is the period of revolution of the orbit.

**Complete Step-by-Step solution:**

To solve this we note that

\[KE = \dfrac{1}{2}m{v^2}\] where \[KE\] is the kinetic energy of a body, \[m\] is the mass of the body, and \[v\] is the velocity of the body.

Hence, \[x = \dfrac{1}{2}m{v^2}\]

\[ \Rightarrow v = {\left( {\dfrac{{2x}}{m}} \right)^{\dfrac{1}{2}}}\]

Now, recalling that the period can be defined as

\[T = \dfrac{{2\pi R}}{v}\] where \[T\] is the period of revolution of the orbit, where \[R\] is the radius of an object satellite the planet.

But the radius can be given as

\[R = \dfrac{{GM}}{{{v^2}}}\]

Hence, inserting into \[T = \dfrac{{2\pi R}}{v}\]

\[T = \dfrac{{2\pi \left( {\dfrac{{GM}}{{{v^2}}}} \right)}}{v} = \dfrac{{2GM\pi }}{{{v^3}}}\]

But \[v = {\left( {\dfrac{{2x}}{m}} \right)^{\dfrac{1}{2}}}\]

Hence,

\[T = \dfrac{{2GM\pi }}{{{{\left( {{{\left( {\dfrac{{2x}}{m}} \right)}^{\dfrac{1}{2}}}} \right)}^3}}}\]

\[T \Rightarrow \dfrac{{2GM\pi {m^{\dfrac{3}{2}}}}}{{{8^{\dfrac{3}{2}}}{x^{\dfrac{3}{2}}}}}\]

Hence, is inversely proportional to \[{x^{\dfrac{3}{2}}}\] or proportional to \[{x^{ - \dfrac{3}{2}}}\]

**Thus, the correct option is B**

**Note:**For clarity, the formula for the radius of the orbiting object \[R = \dfrac{{Gm}}{{{v^2}}}\] can be gotten from the knowledge that the centripetal force for revolution is provided by the gravitational force of attraction. Hence,

\[\dfrac{{m{v^2}}}{R} = \dfrac{{GmM}}{{{R^2}}}\] where all quantities take the value as specified previously.

Hence, by dividing both sides by \[m\] and multiplying by \[R\], we have

\[{v^2} = \dfrac{{GM}}{R}\]

\[ \Rightarrow R = \dfrac{{GM}}{{{v^2}}}\]

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