Explanation:
Orbital radius, [tex]r=1\ AU=1.496\times 10^{11}\ m[/tex]
According to Kepler's law :
[tex]T^2\propto r^3[/tex]
[tex]T^2=\dfrac{4\pi^2r^3}{GM}[/tex]
Where
M is the mass of sun, [tex]M=1.98\times 10^{30}\ kg[/tex]
[tex]T^2=\dfrac{4\pi^2\times (1.496\times 10^{11})^3}{6.67\times 10^{-11}\times 1.98\times 10^{30}}[/tex]
[tex]T=\sqrt{1.0008\times 10^{15}}[/tex]
T = 31635423.18 s
or
T = 1.0003 years
So, a solar-system planet that has an orbital radius of 1 AU would have an orbital period of about 1.0003 year(s).
