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

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, so $T_Y^{2}=A_Y^{3}$.

Step2: Substitute $A_Y$ into the equation for planet Y

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 $T_X^{2}=A_X^{3}$, then $T_Y^{2}=k^{3}T_X^{2}$.

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

Take the square - root of both sides of $T_Y^{2}=k^{3}T_X^{2}$. We have $T_Y=\sqrt{k^{3}}T_X = k^{\frac{3}{2}}T_X$. The factor by which the orbital period is increased is $\frac{T_Y}{T_X}=k^{\frac{3}{2}}$.

Answer:

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