Exercise 3.1.7

Kepler found that the orbital period \(\tau\) of a planet depends on its mean distance \(R\) from the sun according to \(\tau=c R^{\alpha}\) for a simple rational number \(\alpha\). Perform a linear least-squares fit from the following table in order to determine the most likely simple rational value of \(\alpha\).

tau = [87.99, 224.7, 365.26, 686.98, 4332.4, 10759, 30684, 60188];
R = [57.59, 108.11, 149.57, 227.84, 778.14, 1427, 2870.3, 4499.9];
scatter(R,tau,title="Orbital Period (days)", label = "data",
    xlabel=L"R (Mkm)",ylabel=L"tau")
figure 4
figure 4

Using the log-log transformation:

\[ \log \tau = \log c + \alpha \log R \]

V = [R.^0 log.(R)];
c = V \ log.(tau);
@show c[2]

# c[2] = 1.4986486620150812

So the exponent is close to 3/2, which matches Keplar’s third law (usually expressed as \(\tau^2 \propto R^3\))