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(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $K.E. = \frac{GMm}{2r}$.Given $

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$x^{-3/2}$

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(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $K.E. = \frac{GMm}{2r}$.Given $K.E. = x$,we have $x = \frac{GMm}{2r}$.This implies $r = \frac{GMm}{2x}$.Since $GMm/2$ is a constant,we can write $r \propto \frac{1}{x}$.According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$,i.e.,$T^2 \propto r^3$.Substituting $r \propto x^{-1}$ into the relation,we get $T^2 \propto (x^{-1})^3 = x^{-3}$.Taking the square root on both sides,we get $T \propto (x^{-3})^{1/2} = x^{-3/2}$.Thus,the time of revolution $T$ is proportional to $x^{-3/2}$.

Let the kinetic energy of a satellite be $x$,then its time of revolution $T$ is proportional to ..............

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