/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 The orbit of Mars around the sun... [FREE SOLUTION] | 91Ó°ÊÓ

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The orbit of Mars around the sun is an ellipse with eccentricity 0.093 and semimajor axis \(2.28 \times 10^{8} \mathrm{km} .\) Find a polar equation for the orbit.

Short Answer

Expert verified
The polar equation of Mars's orbit is \( r = \frac{2.25907308 \times 10^8}{1 + 0.093 \cos \theta} \).

Step by step solution

01

Understanding Ellipse Parameters

The orbit of Mars is an elliptical shape. An ellipse is characterized by its semi-major axis, which is the longest diameter, and the eccentricity, which measures the deviation of the ellipse from being circular. Here, the semi-major axis, denoted as \(a\), is \(2.28 \times 10^8\) km, and the eccentricity \(e\) is 0.093.
02

Using the Formula for Polar Equations of Ellipses

In polar coordinates, the equation for an ellipse with the sun at one focus is given by the formula \( r = \frac{a(1 - e^2)}{1 + e \cos \theta} \), where \(r\) is the distance from the focus (the sun) to a point on the ellipse, \( \theta \) is the angle from the semi-major axis, \(a\) is the semi-major axis, and \(e\) is the eccentricity.
03

Substitute Known Values into the Formula

We substitute \(a = 2.28 \times 10^8\) km and \(e = 0.093\) into the polar equation. Thus, the polar equation of Mars's orbit becomes:\[ r = \frac{2.28 \times 10^8 (1 - 0.093^2)}{1 + 0.093 \cos \theta} \]
04

Simplifying the Expression

To simplify, first calculate \(1 - e^2\). Since \(e = 0.093\), then \(e^2 = 0.093^2 = 0.008649\). Thus, \(1 - e^2 = 0.991351\). The equation simplifies to:\[ r = \frac{2.28 \times 10^8 \times 0.991351}{1 + 0.093 \cos \theta} \]
05

Final Polar Equation for Mars' Orbit

Perform the multiplication in the numerator: \(2.28 \times 10^8 \times 0.991351 \approx 2.25907308 \times 10^8\). Therefore, the final polar equation is:\[ r = \frac{2.25907308 \times 10^8}{1 + 0.093 \cos \theta} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipse Eccentricity
Eccentricity is a crucial concept when discussing ellipses, particularly in the context of celestial orbits. It measures how much an ellipse deviates from being a perfect circle. This is a dimensionless number that ranges from 0 to 1.
  • An eccentricity of 0 indicates a perfect circle.
  • A value closer to 1 signifies a more elongated ellipse.

In the case of Mars, its orbit has an eccentricity of 0.093. This suggests that the orbit is almost circular but slightly elongated. A lower eccentricity affects how far a planet can move from the sun during its orbit. Understanding eccentricity helps in visualizing planetary orbits and predicting their characteristics. It plays a key role in determining seasonal variations and climate conditions on planets due to the changing distances from their stars.
Semi-Major Axis
The semi-major axis is another fundamental parameter that defines an ellipse. It is the longest radius of the ellipse and extends from its center to its farthest edge. This is also the same axis that contains the foci.For planetary orbits, the semi-major axis serves as a measure of the orbit's size. In the exercise provided, the semi-major axis for Mars is given as \(2.28 \times 10^8\) kilometers.The semi-major axis is intimately tied to the orbital mechanics of a planet. It dictates the orbital period of the planet through Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis. Thus, the semi-major axis not only tells us how stretched the orbit is but also gives crucial information about the time it takes for a complete orbit.
Orbital Mechanics
Orbital mechanics, often referred to as celestial mechanics, involves the study of the motions of objects in space, primarily under the influence of gravitational forces. This can include the orbits of planets, moons, and artificial satellites. Understanding these mechanics begins with Newton's laws of motion and his law of universal gravitation.
  • The gravitational force maintains planets in their elliptical orbits around stars.
  • These orbits can be predicted and described using mathematical equations, one of which is the polar equation of an ellipse.

In Mars's case, its orbit around the Sun is a particular focus of study. Such study helps in aspects like determining the best times for launching missions to Mars, predicting solar eclipses, or understanding seasonal changes on Mars.
Polar Coordinates
Polar coordinates are a useful tool in the analysis of planetary orbits, especially when combined with the properties of ellipses.Unlike Cartesian coordinates, which use \((x, y)\) positions on a flat plane, polar coordinates are based on the distance and angle from a fixed point, known as the pole.For orbit calculations, the pole is typically the primary focus of the ellipse, such as the sun in the case of a planet's orbit. The distance \(r\) from the sun to a point on the ellipse is expressed based on the angle \(\theta\), measured from the semi-major axis.The polar equation of an ellipse, as shown in the exercise, derives from this coordinate system and helps describe how the orbit looks relative to its central body. Polar coordinates make it easier to translate the properties of orbits into usable data for orbital predictions and simulations.

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Most popular questions from this chapter

The LORAN (LOng RAnge Navigation) radio navigation system was widely used until the 1990 s when it was superseded by the GPS system. In the LORAN system, two radio stations located at \(A\) and \(B\) transmit simultaneous signals to a ship or an aircraft located at \(P\). The onboard computer converts the time difference in receiving these signals into a distance difference \(|P A|-|P B|\), and this, according to the definition of a hyperbola, locates the ship or aircraft on one branch of a hyperbola (see the figure). Suppose that station B is located \(400 \mathrm{mi}\) due east of station A on a coastline. A ship received the signal from B 1200 microseconds (\mus) before it received the signal from A. $$ \begin{array}{l}{\text { (a) Assuming that radio signals travel at a speed of } 980 \mathrm{ft} / \mu \mathrm{s} \text { , }} \\ {\text { find an equation of the hyperbola on which the ship lies. }} \\ {\text { (b) If the ship is due north of } B \text { , how far off the coastline is }} \\ {\text { the ship? }}\end{array} $$

Let \(P\left(x_{1}, y_{1}\right)\) be a point on the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) with foci \(F_{1}\) and \(F_{2}\) and let \(\alpha\) and \(\beta\) be the angles between the lines \(P F_{1}, P F_{2}\) and the hyperbola as shown in the figure. Prove that \(\alpha=\beta\). This is the reflection property of the hyperbola. It shows that light aimed at a focus \(F_{2}\) of a hyperbolic mirror is reflected toward the other focus \(F_{1} .\).)

Find the area of the region that lies inside both curves. $$ r=a \sin \theta, \quad r=b \cos \theta, \quad a>0, b>0 $$

Find all points of intersection of the given curves. $$ r=\cos 3 \theta, \quad r=\sin 3 \theta $$

Find an equation for the conic that satisfies the given conditions. $$ \begin{array}{l}{\text { Hyperbola, vertices }(-3,-4),(-3,6)} \\ {\text { foci }(-3,-7),(-3,9)}\end{array} $$

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