Answer:
The period of Y increases by a factor of [tex]k^ {3/2}[/tex] with respect to the period of X
Step-by-step explanation:
The equation [tex]T ^ 2 = a ^ 3[/tex] shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?
For the planet Y:
[tex]T_y ^ 2 = a_y ^ 3[/tex]
For planet X:
[tex]T_x ^ 2 = a_x ^ 3[/tex]
To know the factor of aumeto we compared [tex]T_x[/tex] with [tex]T_y[/tex]
We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:
[tex]a_y ^ 3 = (a_xk) ^ 3[/tex]
[tex]\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x[/tex]
Finally the period of Y increases by a factor of [tex]k^ {3/2}[/tex] with respect to the period of X
