As per the question the distance of venus from sun is given as 0.723 AU
We have been asked to calculate the time period of the planet venus.
As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically
[tex]T^{2} \alpha R^{3}[/tex]
⇒ [tex]T^{2} = KR^{3}[/tex] where is k is the proportionality constant
We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days
Hence [tex]T_{1} =365.5 days[/tex]
The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun
Hence [tex]R_{1} =1 AU[/tex]
The distance of venus from sun is 0.723 AU i.e[tex]R_{2} =0.723[/tex]
From keplers law we know that-[tex]\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }[/tex]
⇒[tex]T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }[/tex]
Putting the values mentioned above we get-
[tex]T_{2} ^{2} =50,350.132851075[/tex]
⇒ [tex]T_{2} =\sqrt{50,350.132851075}[/tex]
⇒[tex]T_{2} = 224.388352752710 days.[/tex]
Hence the time period of venus is 224.388352752710 days
