Answer: 14.62 Earth days
Explanation:
This problem can be solved by Kepler’s Third Law of Planetary motion:
[tex]T=2 \pi \sqrt{\frac{a^{3}}{GM}}[/tex]
Where:
[tex]T[/tex] is the period of the probe
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant
[tex]M=1(10)^{13} kg[/tex] is the mass of the comet Churyumov–Gerasimenko
[tex]a=30 km \frac{1000 m}{1 km}=30000 m[/tex] is the semimajor axis of the orbit the probe described around the comet (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
[tex]T=2 \pi \sqrt{\frac{(30000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1(10)^{13} kg)}}[/tex]
[tex]T=1,263,771.768 s \frac{1 h}{3600 s} \frac{1 Earth-day}{24 h}=14.62 Earth-days[/tex]
Hence, the orbital period of the probe is 14.62 Earth days.
