Answer
The orbital period of the moon at a distance of 48,000 km is 24.5 days.
The orbital period of the moon at a distance of 64,000 km is 37.7 days.
Work Step by Step
Let $P_C$ be Charon's period.
Let $R_C$ be Charon's orbital radius.
Let $P_1$ be period of the moon at a distance of 48,000 km.
Let $R_1$ be the orbital radius of the moon at a distance of 48,000 km.
We can use Kepler's third law to find the period $P_1$.
$(\frac{P_1}{P_C})^2 = (\frac{R_1}{R_C})^3$
$P_1 = P_C~(\frac{R_1}{R_C})^{3/2}$
$P_1 = (6.39~days)~(\frac{48,000~km}{19,600~km})^{3/2}$
$P_1 = 24.5~days$
The orbital period of the moon at a distance of 48,000 km is 24.5 days.
Let $P_2$ be period of the moon at a distance of 64,000 km.
Let $R_2$ be the orbital radius of the moon at a distance of 64,000 km.
We can use Kepler's third law to find the period $P_2$.
$(\frac{P_2}{P_C})^2 = (\frac{R_2}{R_C})^3$
$P_2 = P_C~(\frac{R_2}{R_C})^{3/2}$
$P_2 = (6.39~days)~(\frac{64,000~km}{19,600~km})^{3/2}$
$P_2 = 37.7~days$
The orbital period of the moon at a distance of 64,000 km is 37.7 days.
