Answer:
Orbital period, T = 1.42 years
Explanation:
It is given that,
Orbital period of a solar system planet, [tex]T=8\ years=2.52\times 10^{8}\ s[/tex]
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
M is the mass of the sun
[tex]r^3=\dfrac{T^2GM}{4\pi^2}[/tex]
[tex]r^3=\dfrac{(2.52\times 10^{8})^2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}}{4\pi^2}[/tex]
[tex]r^3=2.134\times 10^{35}[/tex]
[tex]r=5.975\times 10^{11}\ m[/tex]
r = 1.42 AU
So, the solar-system planet that has an orbital period of 8 years would have an orbital radius of about 1.42 AU.
