Question

Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital period of Venus, and then calculate it.

Answer 1

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

                                        $T^{2} \alpha R^{3}$

                                         ⇒ $T^{2} = KR^{3}$ 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 $T_{1} =365.5 days$

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence $R_{1} =1 AU$

The distance of venus from sun is 0.723 AU i.e$R_{2} =0.723$

From keplers law we know that-$\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }$

                            ⇒$T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }$

Putting the values mentioned above we get-

                                      $T_{2} ^{2} =50,350.132851075$

                                         ⇒ $T_{2} =\sqrt{50,350.132851075}$

                                        ⇒$T_{2} = 224.388352752710 days.$

Hence the time period of venus is 224.388352752710 days

                                         

                     

                           

Answer 2

Orbital period of Venus can be calculated using Kepler third law which states that period squared is equal to distance cubed, that is, (T1/T2)^2 = (a1/a2)^3 Where T1 equals the orbital period of venus, T2 is the orbital period of the earth and is equal to 365.25 days, a1 equals 0.723 AU and a2 equals 1AU. Using the equation and the values above, the orbital period of Venus is 224.54 days.