Answer
The spacecraft's orbital period is $57,600~seconds$
Work Step by Step
When a spacecraft orbits a planet, the gravitational force provides the centripetal force to keep the spacecraft moving in a circle. Let $M_p$ be the mass of the planet and let $M_s$ be the mass of the spacecraft. We can find the angular speed of a spacecraft:
$\frac{G~M_p~M_s}{(3.0~R_J)^2} = M_s~\omega^2~(3.0~R_J)$
$\frac{G~M_p}{R_J^2} = 27~\omega^2~R_J$
$\omega^2 = (\frac{1}{27~R_J})(\frac{G~M_p}{R_J^2})$
$\omega = \sqrt{(\frac{1}{27~R_J})(\frac{G~M_p}{R_J^2})}$
$\omega = \sqrt{(\frac{1}{(27)(7.15\times 10^7~m)})(23~N/kg)}$
$\omega = 1.09\times 10^{-4}~rad/s$
We can find the spacecraft's orbital period $P$:
$P = \frac{2\pi}{\omega}$
$P = \frac{2\pi}{1.09\times 10^{-4}~rad/s}$
$P = 57,600~s$
The spacecraft's orbital period is $57,600~seconds$
