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 twice the mean distance from the sun as planet x, by what factor is the orbital period 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 planet y is twice the mean distance from the sun as planet x, by what factor is the orbital period 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 distances and periods

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

Step2: Substitute $A_Y$ into the formula

Substitute $A_Y = 2A_X$ into $T_Y^{2}=A_Y^{3}$, we get $T_Y^{2}=(2A_X)^{3}=8A_X^{3}$.

Step3: Express $T_Y$ in terms of $T_X$

Since $T_X^{2}=A_X^{3}$, then $A_X^{3}=T_X^{2}$. Substitute $A_X^{3}=T_X^{2}$ into $T_Y^{2}=8A_X^{3}$, we have $T_Y^{2}=8T_X^{2}$.

Step4: Solve for the ratio $\frac{T_Y}{T_X}$

Take the square - root of both sides of $T_Y^{2}=8T_X^{2}$: $T_Y=\sqrt{8}T_X = 2^{\frac{3}{2}}T_X$. So the factor by which the orbital period is increased is $2^{\frac{3}{2}}$.

Answer:

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