Question

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 1 AU would have an orbital period of about __________ year(s).

Answer 1

Explanation:

Orbital radius, $r=1\ AU=1.496\times 10^{11}\ m$

According to Kepler's law :

$T^2\propto r^3$

$T^2=\dfrac{4\pi^2r^3}{GM}$

Where

M is the mass of sun, $M=1.98\times 10^{30}\ kg$

$T^2=\dfrac{4\pi^2\times (1.496\times 10^{11})^3}{6.67\times 10^{-11}\times 1.98\times 10^{30}}$

$T=\sqrt{1.0008\times 10^{15}}$

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).