Answer:
16.63min
Explanation:
The question is about the period of the comet in its orbit.
To find the period you can use one of the Kepler's law:
[tex]T^2=\frac{4\pi}{GM}r^3[/tex]
T: period
G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2
r: average distance = 1UA = 1.5*10^11m
M: mass of the sun = 1.99*10^30 kg
By replacing you obtain:
[tex]T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min[/tex]
the comet takes around 16.63min
