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# Let the kinetic energy of a satellite is $x$, then its time of revolution $T$ is proportional to (A) ${x^{ - 3}}$(B) ${x^{ - \dfrac{3}{2}}}$(C) ${x^{ - 1}}$(D) $\sqrt x$

Last updated date: 20th Jun 2024
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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
${v^2} = \dfrac{{GM}}{R}$
$\Rightarrow R = \dfrac{{GM}}{{{v^2}}}$