Answer:
[tex]T_1=0.24y[/tex]
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
[tex](\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3[/tex]
Where [tex]T_1[/tex] and [tex]T_2[/tex] are the orbital periods of Mercury and Earth respectively. We have [tex]D_1=0.39D_2[/tex] and [tex]T_2=1y[/tex]. Replacing this and solving for
[tex]T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y[/tex]
