On a

Someday, the Sun, Venus, and the Earth will be in eclipse, i.e., Venus will be between the Sun and the Earth. Venus orbits the Sun in 225 days. In how many years will all three bodies be in alignment again?

Correct answer:

x =  1.6071

Step-by-step explanation:

$$\begin{gathered} T_1=225\text{d} \\ {T}_2=365\text{d} \\ 225=3^2\cdot 5^2 \\ 365=5\cdot 73 \\ LCM(225,365)=3^2\cdot 5^2\cdot 73=16425 \\ T=LCM(T_1,T_2)=LCM(225,365)=16425 \\ {x}_0=T/T_2=16425/365=45 \\ {\omega }_1=360/T_1=360/225=\frac{8}{5}=1\frac{3}{5}=1.6\text{°/d} \\ {\omega }_2=360/T_2=360/365=\frac{72}{73}\doteq 0.9863\text{°/d} \\ x\cdot {\omega }_1=x\cdot {\omega }_2+k\cdot 360 \\ {x}_1\cdot ({\omega }_1-{\omega }_2)=1\cdot 360 \\ {x}_1\cdot (1.6-0.98630136986301)=1\cdot 360 \\ 0.613699x_1=360 \\ {x}_1=\frac{360}{0.61369863}=586.60714286 \\ {x}_1=\frac{16425}{28}\doteq 586.607143 \\ {x}_2\cdot ({\omega }_1-{\omega }_2)=2\cdot 360 \\ {x}_2\cdot (1.6-0.98630136986301)=2\cdot 360 \\ 0.613699x_2=720 \\ {x}_2=\frac{720}{0.61369863}=1173.21428571 \\ {x}_2=\frac{16425}{14}\doteq 1173.214286 \\ {x}_3\cdot ({\omega }_1-{\omega }_2)=3\cdot 360 \\ {x}_3\cdot (1.6-0.98630136986301)=3\cdot 360 \\ 0.613699x_3=1080 \\ {x}_3=\frac{1080}{0.61369863}=1759.82142857 \\ {x}_3=\frac{49275}{28}\doteq 1759.821429 \\ x=x_1/T_2=586.6071/365=1.6071 \end{gathered}$$

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