Question

In 2014, a space probe approached the rocky core of the comet Churyumov–Gerasimenko, which is only a few km in diameter. The probe then entered orbit around the comet at a distance of 30 km. The comet was found to have a mass of 1.0 * 10^13 kg. What was the orbital period of the probe around the comet, in earth days?

Answer 1

Answer: 14.62 Earth days

Explanation:

This problem can be solved by Kepler’s Third Law of Planetary motion:

$T=2 \pi \sqrt{\frac{a^{3}}{GM}}$

Where:

$T$ is the period of the probe

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=1(10)^{13} kg$ is the mass of the comet Churyumov–Gerasimenko

$a=30 km \frac{1000 m}{1 km}=30000 m$ 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)

$T=2 \pi \sqrt{\frac{(30000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1(10)^{13} kg)}}$

$T=1,263,771.768 s \frac{1 h}{3600 s} \frac{1 Earth-day}{24 h}=14.62 Earth-days$

Hence, the orbital period of the probe is 14.62 Earth days.