The equations of motion for the position x,y of a planet in its orbital plane are...

 

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The equations of motion for the position x,y of a planet in its orbital plane are the same as those for any orbiting body: dx dt2 I = GM- dy dt2 =-GM GM/ Y where G = 6.6738 x 10-11m3kg s 2 is Newton's gravitational constant, M = the mass of the Sun, and r = x + y. 1.9891 x 1030kg is Let us first solve these equations for the orbit of the Earth. The Earth's orbit is not perfectly circular, but rather slightly elliptical. When it is at its closest approach to the Sun, its perihelion, it is moving precisely tangentially (i.e., perpendicular to the line between itself and the Sun) and it has distance 1.4710 x 10m from the Sun and linear velocity 3.0287 x 10 m/s. a) Write a program to calculate the orbit of the Earth and find out its aphelion using the Bulirsch-Stoer method to a positional accuracy of 1 km per year. Divide the orbit into intervals of length H = 1 week and then calculate the solution for each interval using the combined modified midpoint method. Make a plot of the orbit, showing at least one complete revolution about the Sun. (Remember to use ax.set_aspect ("equal"), or the result may look ugly) b) Modify your program to calculate the orbit of the dwarf planet [1] Pluto and find out its aphelion. The distance between the Sun and Pluto at perihelion is 4.4368 x 1012 m and the linear velocity is 6.1218 x 103 m/s. Choose a suitable value for H to make your calculation run in reason- able time, while once again giving a solution accurate to 1 km per year. You should find that the orbit of Pluto is significantly elliptical much moreso than the orbit of the Earth. Pluto is a Kuiper belt object [2], similar to a comet, and (unlike true planets) it's typical for such objects to have quite elliptical orbits. The equations of motion for the position x,y of a planet in its orbital plane are the same as those for any orbiting body: dx dt2 I = GM- dy dt2 =-GM GM/ Y where G = 6.6738 x 10-11m3kg s 2 is Newton's gravitational constant, M = the mass of the Sun, and r = x + y. 1.9891 x 1030kg is Let us first solve these equations for the orbit of the Earth. The Earth's orbit is not perfectly circular, but rather slightly elliptical. When it is at its closest approach to the Sun, its perihelion, it is moving precisely tangentially (i.e., perpendicular to the line between itself and the Sun) and it has distance 1.4710 x 10m from the Sun and linear velocity 3.0287 x 10 m/s. a) Write a program to calculate the orbit of the Earth and find out its aphelion using the Bulirsch-Stoer method to a positional accuracy of 1 km per year. Divide the orbit into intervals of length H = 1 week and then calculate the solution for each interval using the combined modified midpoint method. Make a plot of the orbit, showing at least one complete revolution about the Sun. (Remember to use ax.set_aspect ("equal"), or the result may look ugly) b) Modify your program to calculate the orbit of the dwarf planet [1] Pluto and find out its aphelion. The distance between the Sun and Pluto at perihelion is 4.4368 x 1012 m and the linear velocity is 6.1218 x 103 m/s. Choose a suitable value for H to make your calculation run in reason- able time, while once again giving a solution accurate to 1 km per year. You should find that the orbit of Pluto is significantly elliptical much moreso than the orbit of the Earth. Pluto is a Kuiper belt object [2], similar to a comet, and (unlike true planets) it's typical for such objects to have quite elliptical orbits.