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.

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

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

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

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

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

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,

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

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