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$k^{\frac{1}{3}}$\n$k^{\frac{1}{2}}$\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$k^{\frac{1}{3}}$\n$k^{\frac{1}{2}}$\n$k^{\frac{2}{3}}$\n$k^{\frac{3}{2}}$

Answer

Answer:

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

Explanation:

Step1: Define variables for planets X and Y

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

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 $A_X^{3}$ in terms of $T_X$

Since $T_X^{2}=A_X^{3}$, we can substitute $A_X^{3}=T_X^{2}$ into the equation for $T_Y^{2}$. So, $T_Y^{2}=k^{3}T_X^{2}$.

Step4: Solve for the ratio of $T_Y$ to $T_X$

Take the square - root of both sides of the equation $T_Y^{2}=k^{3}T_X^{2}$. We have $T_Y=\sqrt{k^{3}}T_X = k^{\frac{3}{2}}T_X$. So the orbital period of planet Y is $k^{\frac{3}{2}}$ times the orbital period of planet X, which means the orbital period is increased by a factor of $k^{\frac{3}{2}}$.