Exercise 1.2.34

Question

The table shows the mean (average) distances $d$ of the planets from the sun (taking the unit of measurement to be the distance from the earth to the sun) and their periods $T$ (time of revolution in years).
    \begin{enumerate}[label=(\alph*)]
        \item Fit a power model to the data.
        \item Kepler’s Third Law of Planetary Motion states that “The square of the period of revolution of a planet is proportional to the cube of its mean distance from the sun.” Does your model corroborate Kepler’s Third Law?
    \end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)]

\item Using \href{https://www.wolframalpha.com/input?i=power+fit+%280.387%2C+0.241%29%2C+%280.723%2C+0.615%29%2C+%281%2C1%29%2C+%281.523%2C1.881%29%2C+%285.203%2C11.861%29%2C+%289.541%2C+29.457%29%2C+%2819.19%2C+84.008%29%2C+%2830.086%2C+164.784%29}{WolframAlpha}, we get:
    \begin{equation}\label{b1}
        \forall d\in (0,+\infty):\quad T(d)=1.00253 x^{1.49884}.
    \end{equation}
    We sketch the graph of $T$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=left, xmin=0, xmax=31, ymax=180, ymin=0, xlabel={Distance from the sun, }, ylabel={Time of revolution in years}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=0:31, samples=50, mark=none, color=green!60!black, ultra thick] {1.00253*x^(1.49884)};
         \addplot[mark=*, mark options={scale=1, fill=black}](0.387,0.241);
         \addplot[mark=*, mark options={scale=1, fill=black}](0.723,0.615);
         \addplot[mark=*, mark options={scale=1, fill=black}](1,1);
         \addplot[mark=*, mark options={scale=1, fill=black}](1.523,1.881);
         \addplot[mark=*, mark options={scale=1, fill=black}](5.203,11.861);
         \addplot[mark=*, mark options={scale=1, fill=black}](9.541,29.547);
         \addplot[mark=*, mark options={scale=1, fill=black}](19.19,84.008);
         %\addplot +[dashed,violet,mark=none] coordinates {(17,0) (17,79242.52) (0,79242.52)};
         %\addplot +[dashed,green!60!black,mark=none] coordinates {(32,0) (32,96115.42) (0,96115.42)};
         %\addplot[mark=*, mark options={scale=1, fill=yellow!70!black}](32,96115.42);
         \addplot[mark=*, mark options={scale=1, fill=black}](30.086,164.784);
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    
    \item The square of the period of revolution of a planet being proportional to the cube of its mean distance means that for some $k\in R_+,$ we have
    \begin{equation}\label{a2}
        \forall d\in \mathbb{R}_+:\quad  T(d)^2=kd^3.
    \end{equation}
    Statement ef{a2} is equivalent to:
    %$$	ext{For some $k\in \mathbb{R}_+$ }\forall d\in \mathbb{R}_+:\quad T(d)=k^{\frac{3}{2}}.$$
    $$T(d)=\sqrt{k}\ d^{3/2}.$$
    Since the function from ef{b1} is approximately equal to $$T(d)=1.00253 d^{1.49884}\approx 1.0 d^{1.5},$$ within a certain error margin, our model is in accordance with Kepler's Third Law.
    
\end{enumerate}

Exercise 1.2.34

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Exercise 1.2.34

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