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

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

Answer

Explanation:

Step1: Let orbital - period and distance for planet X

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

Step2: Use the given relationship between $T_X$ and $T_Y$

We know that $T_Y = 2T_X$. Substitute $T_Y$ into the equation $T_Y^{2}=A_Y^{3}$: $(2T_X)^{2}=A_Y^{3}$.

Step3: Substitute $T_X^{2}=A_X^{3}$

Since $T_X^{2}=A_X^{3}$, then $(2T_X)^{2}=4T_X^{2}=4A_X^{3}=A_Y^{3}$.

Step4: 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:

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