Q5.3-28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Lunar Orbit. The motion of a moon moving in a planar orbit about a planet is governed by the equations $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathbf{G}\frac{\mathbf{mx}}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{,}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathbf{G}\frac{\mathbf{my}}{{\mathbf{r}}^{\mathbf{3}}}$ where $\mathbf{r}\mathbf{=}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}{\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{2}}}$, G is the gravitational constant, and m is the mass of the planet. Assume Gm = 1. When  $\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ the motion is a circular orbit of radius 1 and period $\mathbf{2}\mathbf{\pi }$.

(a) The setting ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{x}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{y}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{y}\mathbf{\text{'}}$ expresses the governing equations as a first-order system in normal form.

(b) Using $\mathbf{h}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{100}}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{0628318}$, compute one orbit of this moon (i.e., do N = 100 steps.). Do your approximations agree with the fact that the orbit is a circle of radius 1?

Step-by-Step Solution

Answer

At the value of a complete orbit, it’s a circle of radius 1.

Transform the equation

Here the equation is $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-\mathrm{G}\frac{\mathrm{mx}}{{\mathrm{r}}^3},\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=-\mathrm{G}\frac{\mathrm{my}}{{\mathrm{r}}^3}$.

 

The system can be written as:

 

$$\begin{gathered} {\mathrm{x}}_1=\mathrm{x} \\ {\mathrm{x}}_2=\mathrm{x}\text{'} \\ =\mathrm{x}{\text{'}}_1 \\ {\mathrm{x}}_3=\mathrm{y} \\ {\mathrm{x}}_4=\mathrm{y}\text{'} \\ =\mathrm{x}{\text{'}}_3 \end{gathered}$$

 

The transform equation is

$$\begin{gathered} \mathrm{x}{\text{'}}_1={\mathrm{x}}_2 \\ \mathrm{x}{\text{'}}_2=-\mathrm{G}\frac{\mathrm{mx}}{{\mathrm{r}}^3} \\ =-\frac{{\mathrm{x}}_1}{{({{\mathrm{x}}^2}_1+{{\mathrm{x}}^2}_3)}^{\frac{3}{2}}} \\ \mathrm{x}{\text{'}}_3={\mathrm{x}}_4 \\ \mathrm{x}{\text{'}}_4=-\mathrm{G}\frac{\mathrm{my}}{{\mathrm{r}}^3} \\ =-\frac{{\mathrm{x}}_3}{{({{\mathrm{x}}^2}_1+{{\mathrm{x}}^2}_3)}^{\frac{3}{2}}} \end{gathered}$$

The initial conditions are;

$$\begin{gathered} {\mathrm{x}}_1\left(0\right)=\mathrm{x}\left(0\right)=1 \\ {\mathrm{x}}_2\left(0\right)=\mathrm{x}\text{'}\left(0\right)=0 \\ {\mathrm{x}}_3\left(0\right)=\mathrm{y}\left(0\right)=0 \\ {\mathrm{x}}_4\left(0\right)=\mathrm{y}\text{'}\left(0\right)=1 \end{gathered}$$

Apply the Runge-Kutta method

T	X	Y
0	1	0
0.06	0.9980	0.0627
0.12	0.992	0.1253
0.25	0.9685	0.248
0.50	0.876	0.4817
0.75	0.7289	0.6845
1	0.5358	0.8443
2.01	-0.4257	0.904
2.51	-0.8090	0.5877
3.01	-0.992	0.125
3.14	-0.9999	-1.2383

Graph

At the value of a complete orbit, it’s a circle of radius 1.

 

This is the required result.