Kepler’s Third Law states that the square of the period ${\displaystyle T}$ of any planet is proportional to the cube of the radius ${\displaystyle r}$ of its orbit.

${\displaystyle F_{gravity}=F_{centripetal}}$

${\displaystyle G{\frac {mM}{r^{2}}}\ ={\frac {mv^{2}}{r}}\ }$

${\displaystyle G{\frac {mM}{r^{2}}}\ =mr({\frac {2\pi }{T}})^{2}}$

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

${\displaystyle T^{2}\propto r^{3}}$

for example for Earth,

${\displaystyle T^{2}={\frac {4\pi ^{2}}{(6.67x10^{-11})(1.989x10^{30})}}(150x10^{9})^{3}}$

${\displaystyle T=3.16x10^{7}seconds={\frac {3.16x10^{7}}{(365)(24)(60)(60)}}year=1.00year}$

for example for Mars,

${\displaystyle T^{2}={\frac {4\pi ^{2}}{(6.67x10^{-11})(1.989x10^{30})}}(230x10^{9})^{3}}$

${\displaystyle T=6.00x10^{7}seconds={\frac {6.00x10^{7}}{(365)(24)(60)(60)}}year=1.90year}$