Question

Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyond Mars when furthest from the Sun (aphelion), but with an average distance (semi-major axis) of 1 AU. How long will it take to complete an orbit and where will it spend most of its time

Answer 1

16.63min

Step-by-step 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:

`T^2=(4\pi)/(GM)r^3`

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:

`T=\sqrt{(4\pi)/(GM)r^3}=\sqrt{(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`

the comet takes around 16.63min