the equation $t^{2}=a^{3}$ shows the relationship between a planets orbital period, $t$, and the planets…

the equation $t^{2}=a^{3}$ shows the relationship between a planets orbital period, $t$, and the planets mean distance from the sun, $a$, in astronomical units, au. if planet y is $k$ times the mean distance from the sun as planet x, by what factor is the orbital period increased?\n$\\frac{1}{k^{\\frac{1}{3}}}$\n$\\frac{1}{k^{\\frac{1}{2}}}$\n$k^{\\frac{2}{3}}$\n$k^{\\frac{3}{2}}$

the equation $t^{2}=a^{3}$ shows the relationship between a planets orbital period, $t$, and the planets mean distance from the sun, $a$, in astronomical units, au. if planet y is $k$ times the mean distance from the sun as planet x, by what factor is the orbital period increased?\n$\\frac{1}{k^{\\frac{1}{3}}}$\n$\\frac{1}{k^{\\frac{1}{2}}}$\n$k^{\\frac{2}{3}}$\n$k^{\\frac{3}{2}}$

Answer

Explanation:

Step1: Let distances and periods

Let the mean - distance of planet X from the sun be $A_X$ and its orbital period be $T_X$, so $T_X^{2}=A_X^{3}$. Let the mean - distance of planet Y from the sun be $A_Y = kA_X$ and its orbital period be $T_Y$, so $T_Y^{2}=A_Y^{3}$.

Step2: Substitute $A_Y$ into the equation

Substitute $A_Y = kA_X$ into $T_Y^{2}=A_Y^{3}$, we get $T_Y^{2}=(kA_X)^{3}=k^{3}A_X^{3}$.

Step3: Express $T_Y$ in terms of $T_X$

Since $T_X^{2}=A_X^{3}$, then $T_Y^{2}=k^{3}T_X^{2}$. Take the square - root of both sides: $T_Y=\sqrt{k^{3}}T_X = k^{\frac{3}{2}}T_X$.

Answer:

$k^{\frac{3}{2}}$