# Ganymede is a moon of Jupiter mass = 1.90 \times 10^{27} kg. it orbits Jupiter in a circle of...

## Question:

Ganymede is a moon of Jupiter mass {eq}= 1.90 \times 10^{27} {/eq} kg. it orbits Jupiter in a circle of radius {eq}1.07 \times 10^9 {/eq} m. What is Ganymede's period in hours?

## The time period of the object:

According to Kepler law, the square fo the time period of the satellite is directly proportional to the cube of the radius of the orbit in which the satellite is moving.

Given Data

Mass of the Jupiter {eq}(M) = 1.9 \times 10^{27} \ kg {/eq}

Radius of the orbit {eq}(R) = 1.07 \times 10^{9} \ m {/eq}

Now, the time period is given by

{eq}T^{2} = \dfrac{4\pi^{2}R^{3}}{GM} \\ T^{2} = \dfrac{4\pi^{2}(1.07 \times 10^{9})^{3}}{(6.67 \times 10^{-11}) \times (1.9 \times 10^{27})} \\ T = 617754.36 \ s \\ T = (617754.36 \ s) \dfrac{1 \ hr}{3600 \ s} \\ T = 171.598 \ hr {/eq}