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: Define variables for planets X and Y

Let $T_X$ and $A_X$ be the orbital - period and mean distance of planet X, and $T_Y$ and $A_Y$ be those of planet Y. We know that $T_Y = 2T_X$, and from the equation $T^{2}=A^{3}$, for planet X: $T_X^{2}=A_X^{3}$, and for planet Y: $T_Y^{2}=A_Y^{3}$.

Step2: Substitute $T_Y = 2T_X$ into the equation for planet Y

Substitute $T_Y$ into $T_Y^{2}=A_Y^{3}$, we get $(2T_X)^{2}=A_Y^{3}$. Since $T_X^{2}=A_X^{3}$, then $4T_X^{2}=A_Y^{3}$. Replace $T_X^{2}$ with $A_X^{3}$, so $4A_X^{3}=A_Y^{3}$.

Step3: Solve for the ratio $\frac{A_Y}{A_X}$

From $4A_X^{3}=A_Y^{3}$, we can rewrite it as $\frac{A_Y^{3}}{A_X^{3}} = 4$. Then $\left(\frac{A_Y}{A_X}\right)^{3}=4$. Taking the cube - root of both sides, $\frac{A_Y}{A_X}=4^{\frac{1}{3}}=(2^{2})^{\frac{1}{3}} = 2^{\frac{2}{3}}$.

Answer:

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