Respuesta :
From a series of derivations in the equation, I got C. 2^2/3 though I'm not very sure with that. Hope that helps.
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Answer:
[tex]2\sqrt2[/tex] or 2.828
Step-by-step explanation:
The given equation is
[tex]T^2=A^3[/tex]
Let the mean distance from the sun of planet X is A then
mean distance from the sun of planet Y is 2A.
Therefore, we have
For planet X-
[tex]T_x^2=A^3....(i)[/tex]
For planet Y-
[tex]T_y^2=(2A)^3\\\\T_y^2=8A^3....(ii)[/tex]
Divide equation (i) and (ii)
[tex]\frac{T_x^2}{T_y^2}=\frac{A^3}{8A^3}\\\\\frac{T_x^2}{T_y^2}=\frac{1}{8}\\\\\frac{T_x}{T_y}=\frac{1}{2\sqrt2}\\\\T_y=2\sqrt2T_x[/tex]
Therefore, we can conclude that orbital period is increase by a factor of [tex]2\sqrt2[/tex] or 2.828