Single Variable Calculus, Chapter 1, 1.2, Section 1.2, Problem 26

The average distances d of the planets from the sun and their periods T (time of revolution in years) are tabulated as follows.


$
\begin{array}{|c|c|c|}
\hline
\text{Planet} & d & t\\
\hline
\text{Mercury} & 0.387& 0.241\\
\text{Venus} &0.723 &0.615\\
\text{Earth} &1.000 &1.000\\
\text{Mars} &1.523 &1.881\\
\text{Jupiter} &5.203& 11.861\\
\text{Saturn} &9.541 &29.457\\
\text{Uranus} &19.190 &84.008\\
\text{Neptune} &30.086 &164.784\\
\hline
\end{array}
$


a.) To fit a power model, we will use a 2 sets of the given data, Mercury and Venus.

$
\begin{equation}
\begin{aligned}
\text{Having the Power Function: }T = a(d)^{b} ;\text{ where }T &= \text{ period }(\text{Time of revolution in years}) \\
d &= \text{ distances of the planets from the sun}
\end{aligned}
\end{equation}
$


Then, we need to find the values of $a$ and $b$ that will satisfy all the given data.

First, we will use the data of planet Mercury.



$
\begin{equation}
\begin{aligned}
T &= a(d)^b; \quad \text{ where } \quad T = 0.241, \quad d = 0.387\\
0.241 &= a(0.387)^b \quad (\text{solving for }a)\\
a &= \displaystyle \frac{0.241}{(0.387)^b} \quad \text{Equation 1}
\end{aligned}
\end{equation}
$


Next, we will use the data of planet Venus.


$
\begin{equation}
\begin{aligned}
T &= a (d)^b; \quad \text{ where } \quad T = 0.615 , \quad d = 0.723\\
0.615 &= a \left( 0.723\right)^b\\
a &= \displaystyle \frac{0.615}{(0.723)^b} \quad \text{Equation 2}
\end{aligned}
\end{equation}
$


Solve the value of b using equations 1 and 2.



$
\begin{equation}
\begin{aligned}
\displaystyle \frac{0.241}{(0.387)^b} &= \displaystyle \frac{0.615}{(0.723)^b} \qquad \text{(applying cross multiplication)}\\
\displaystyle \left(\frac{0.723}{0.387}\right)^b &= \displaystyle \frac{0.615}{0.241} \qquad \quad \text{(tabing the natural logarithm of both sides)}\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
b &\left[ ln \left( \displaystyle \frac{0.723}{0.387}\right)\right] = ln \left(\displaystyle \frac{0.615}{0.241} \right) \qquad \text{(using the property of natural logarithm we can now solve for the value of } b)\\

b &= \displaystyle
\frac{
\text{ln} \displaystyle \left(\frac{0.615}{0.241}\right)
}
{
\text{ln} \displaystyle \left(\frac{0.725}{0.387}\right)
}\\

\end{aligned}
\end{equation}\\
\boxed{b = 1.4990}
$

Now we can solve for the value of $a$ by substituting the value of $b$ in either equation 1 and 2



$
\begin{equation}
\begin{aligned}
a &= \displaystyle \frac{0.241}{(0.387)} b \quad \text{ where } \quad b = 1.4990\\
a &= \displaystyle \frac{0.241}{(0.387)} 1.4900\\
a &= 1
\end{aligned}
\end{equation}
$


Now, we have:


$
\begin{equation}
\begin{aligned}
T & = 1 (d)^{1.4990} \quad \text{ where } \quad 1.4990 \approx 1.50\\
T & = d^{1.5}\\
T &= d^{3/2} \qquad \text{(by squaring both sides of the equation)}\\
T^2 &= d^3
\end{aligned}
\end{equation}
$


Therefore, our model corroborates the Kepler's Third Law.
$T^2 = k d^3$