Modeling Periodic Phenomena Lab Report

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Modeling Periodic Phenomena: The Orbit of Mars

Introduction
Billions of years old, Mars is the 4th planet from the sun in our solar system and the last of the terrestrial planets before the asteroid belt. The planet neighbors Earth and has seen a plethora of exploration craft from our planet. Mars orbits around the Sun on an ellipse, as dictated by Kepler’s laws (which state that all orbits are ellipses). Mars’ elliptical shape is a lot less circular than Earth, due to its high orbital eccentricity, which is over 5 Earth’s. Because of the elliptical shape of orbits, the distance between Mars and the Sun changes throughout the Martian year, which is 687 Earth days or 668.5991 Martian days (sols). Once per Martian year, Mars is at its farthest
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It uses the same x and y values as the graph. I chose to use a positive cosine equation to represent a scenario where Mars is at its aphelion at day 0. This is because a positive cosine function starts at the highest value, where Mars is farthest away from the Sun (aka the aphelion). This equation is a form of the parent function y = a cos(b(x-c)) + d. The “a” value in my equation represents half the distance from the aphelion to the perihelion, or the distance from the aphelion to the average distance from Mars to the Sun (the midline of my graph). The “a” value represents how far away from the average distance to the Sun Mars will get (in either direction) throughout its year. The “b” value in my equation represents the number of times that Mars completes an orbit of the Sun that are performed within 2훑 days. As one might expect, that number is a small fraction, due to the fact that it takes 687 days for Mars to complete an orbit of the Sun and 687 is greater than 2훑. This b value can be more easily understood in relation to the scenario by using the equation b=2훑/p, where p is the period. In this scenario, the period, or length to complete a cycle is 687 days (on Earth). This is because 687 days is the time that Mars takes to complete a single orbit around the Sun. By using the 687 days value as p in our equation, we can see that b=2훑/687. By simplifying this, we can achieve

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