Question

According to Kepler’s first law, a comet should have an elliptic, parabolic, or hyperbolic orbit (with gravitational attractions from the planets ignored). In suitable polar coordinates, the position \(\left( {r,\vartheta } \right)\) of a comet satisfies an equation of the form

\(r = \beta + e\left( {r \cdot \cos \vartheta } \right)\)

Where \(\beta \) is a constant and \(e\) is the eccentricity of the orbit, with \(0 \le e < 1\) for an ellipse, \(e = 1\) for a parabola, and \(e > 1\) for a hyperbola. Suppose observations of a newly discovered comet provide the data below. Determine the type of orbit, and predict where the comet will be when \(\vartheta = 4.6\left( {{\rm{radians}}} \right)\).

\(\begin{array}{{}{}} \vartheta & & {.88}&{1.10}&{1.42}&{1.77}&{2.14} \\ \hline r& {3.00}&{2.30}&{1.65}&{1.25}&{1.01} \end{array}\)

Short answer

The orbit of the comet is elliptical and \(r = 1.33\).

Step-by-step solution

The General Linear Model

The equation of the general linear model is defined as:

\({\bf{y}} = X\beta + \in \)

Here, \({\bf{y}} = \left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right)\) is an observational vector, \(X = \left( {\begin{aligned}{{}}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right)\) is the design matrix, \(\beta = \left( {\begin{aligned}{{}}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right)\) is parameter vector, and \( \in = \left( {\begin{aligned}{{}}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is a residual vector.

Find design matrix, observation vector, parameter vector for given data

The given equation is\(r = \beta + e\left( {r \cdot \cos \vartheta } \right)\), and the given table is shown as:

\(\begin{array}{{}{}}\vartheta & & {.88}&{1.10}&{1.42}&{1.77}&{2.14} \\ \hline r& & {3.00}&{2.30}&{1.65}&{1.25}&{1.01} \end{array}\)

Write the Design matrix, observational vector, and the parameter vector for the given equation and data set using the information provided in step 1.

Design matrix:

\(X = \left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)\)

Observational vector:

\({\bf{y}} = \left( {\begin{aligned}{{}}3\\{2.3}\\{1.65}\\{1.25}\\{1.01}\end{aligned}} \right)\)

And the parameter vectorfor the given equation is,

\({\bf{\beta }} = \left( {\begin{aligned}{{}}\beta \\e\end{aligned}} \right)\)

These are the best fit for the given data set and equation.

Normal equation

The normal equation is given by,

\({X^T}X\beta = {X^T}{\bf{y}}\)

Find the least-squares curve

The general least-squares equation is given by \(r = \beta + e\left( {r \cdot \cos \vartheta } \right)\), and to find the associated least-squares curve, the values of \(\beta ,e\) are required, so find the values of \(\beta ,e\) by using normal equation.

By using the obtained information from step 2, the normal equation will be,

\(\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}{\bf{y}}\)

That implies:

\(\left( {\begin{aligned}{{}}\beta \\e\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}3\\{2.3}\\{1.65}\\{1.25}\\{1.01}\end{aligned}} \right)\)

Use the following steps to find the associated values for the obtained data in MATLAB.

1. Enter the data \(\left( {\begin{aligned}{{}}\beta \\e\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}3\\{2.3}\\{1.65}\\{1.25}\\{1.01}\end{aligned}} \right)\) in the tab in the form of \({\left( {{{\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{}}1&{3\cos .88}\\1&{2.3\cos 1.1}\\1&{1.65\cos 1.42}\\1&{1.25\cos 1.77}\\1&{1.01\cos 2.14}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}3\\{2.3}\\{1.65}\\{1.25}\\{1.01}\end{aligned}} \right)\).
2. Use colons after that and press ENTER.

So, the value of \(\left( {\begin{aligned}{{}}\beta \\e\end{aligned}} \right)\) is \(\left( {\begin{aligned}{{}}{1.45}\\{0.811}\end{aligned}} \right)\).

As, \(e = 0.811\) which is less than 1, so the orbit is of elliptical form.

Now, substitute the obtained values into \(r = \beta + e\left( {r \cdot \cos \vartheta } \right)\).

\(r = 1.45 + 0.811\left( {r \cdot \cos \vartheta } \right)\)

For \(\vartheta = 4.6\), \(r = 1.33\).

So, the required valueof\(r\) is \(1.33\).