Question

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 the orbital period of planet y is twice the orbital period of planet x, by what factor is the mean distance increased? $2^{\frac{4}{3}}$ $2^{\frac{2}{3}}$ $2^{\frac{1}{2}}$ $2^{\frac{1}{3}}$

Explanation:

Step1: Let orbital period of X be $T_X$

Let $T_X$ be the orbital - period of planet X and $A_X$ be its mean distance from the sun. So, $T_X^2 = A_X^3$.

Step2: Define orbital period of Y

The orbital period of planet Y is $T_Y = 2T_X$. Let $A_Y$ be the mean distance of planet Y from the sun. Then $T_Y^2 = A_Y^3$.

Step3: Substitute $T_Y$ into its equation

Substitute $T_Y = 2T_X$ into $T_Y^2 = A_Y^3$. We get $(2T_X)^2=A_Y^3$, which simplifies to $4T_X^2 = A_Y^3$.

Step4: Substitute $T_X^2 = A_X^3$

Since $T_X^2 = A_X^3$, substitute it into $4T_X^2 = A_Y^3$. We have $4A_X^3 = A_Y^3$.

Step5: Solve for the ratio $\frac{A_Y}{A_X}$

From $4A_X^3 = A_Y^3$, we can write $\frac{A_Y^3}{A_X^3}=4$. Then $\frac{A_Y}{A_X}=4^{\frac{1}{3}}=(2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}}$.

Answer:

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