College Algebra, Chapter 2, 2.5, Section 2.5, Problem 42

According to the Kepler's Third Law, the square of the period $T$ of a planet is directly proportional to the cube of its average distance $d$ from the Sun. a.) Express Kepler's Third Law as an equation.
$T^2 = kd^3$

b.) Find the constant of proportionality by using the fact that for our planet, the period is about 365 days and the average distance is about 93 million miles.

$
\begin{equation}
\begin{aligned}
\left( 365 \text{days} \right)^2 &= k \left( 93 \times 10^6 \text{mi} \right)^3\\
\\
k &= \frac{(365)^2}{\left( 93 \times 10^6 \right)^3 } \frac{\text{days}^2}{\text{mi}^3}
\end{aligned}
\end{equation}
$

c.) The planet Neptune is about $2.79 \times 10^9$ miles from the Sun. Find the period of Neptune.

$
\begin{equation}
\begin{aligned}
T^2 &= kd^3\\
\\
T^2 &= \frac{(365)^2}{\left( 93 \times 10^6 \right)^3 } \frac{\text{days}^2}{\text{mi}^3} \left( 2.79 \times 10^9 \right)^3 \text{mi}^3\\
\\
T^2 &= 3597075000 \text{ days}^2\\
\\
T &= 59975.62 \text{ days}
\end{aligned}
\end{equation}
$

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