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Chapter 11: Problem 37

You are given the eccentricity e and the length a of the semimajor axis for the orbits of the planets Pluto and Mars. Compute the distance of each planet from the Sun at perihelion and at aphelion (as in Example 7 ). For Pluto, round the final answers to two decimal places; for Mars, round to three decimal places. $$\text { Pluto: } e=0.2484 ; a=39.44 \mathrm{AU}$$

Short Answer

Expert verified
Pluto's perihelion is 29.63 AU, aphelion is 49.26 AU.

Step by step solution

01

Understanding the Terms

First, let's understand what perihelion and aphelion are. Perihelion is the point in the orbit of a planet where it is closest to the Sun, while aphelion is the farthest point from the Sun. The formula to calculate the perihelion distance is given by \(d_{\text{perihelion}} = a(1 - e)\) and for aphelion distance by \(d_{\text{aphelion}} = a(1 + e)\) where \(a\) is the semi-major axis and \(e\) is the eccentricity.
02

Calculate Perihelion for Pluto

Using the formula \(d_{\text{perihelion}} = a(1 - e)\), substitute Pluto's values: \(a = 39.44\) AU and \(e = 0.2484\). Calculate as follows: \[d_{\text{perihelion}} = 39.44 \times (1 - 0.2484)\]\[d_{\text{perihelion}} = 39.44 \times 0.7516\]\[d_{\text{perihelion}} \approx 29.63\] Round to two decimal places to get 29.63 AU.
03

Calculate Aphelion for Pluto

Using the formula \(d_{\text{aphelion}} = a(1 + e)\), substitute Pluto's values: \(a = 39.44\) AU and \(e = 0.2484\). Calculate as follows: \[d_{\text{aphelion}} = 39.44 \times (1 + 0.2484)\]\[d_{\text{aphelion}} = 39.44 \times 1.2484\]\[d_{\text{aphelion}} \approx 49.26\] Round to two decimal places to get 49.26 AU.

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Key Concepts

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

Eccentricity
In orbital mechanics, eccentricity is a fundamental parameter that describes the shape of an orbit. It tells us how elongated an orbit is compared to a perfect circle. When the eccentricity (\(e\)) is 0, the orbit is perfectly circular. Any value between 0 and 1 indicates an elliptical orbit, which is common for planets in our solar system.
Eccentricity can have significant impacts on a planet's climate and seasons. The higher the eccentricity, the more variation there will be in the planet's distance from the Sun over the course of its orbit.
  • Lower eccentricity means the orbit is more circular.
  • Higher eccentricity indicates a more elongated ellipse.
For example, Pluto鈥檚 eccentricity is 0.2484, which means its orbit is more elongated compared to Earth's orbit (eccentricity of about 0.0167). This greater elongation affects how far Pluto is from the Sun at various points in its orbit.
Perihelion and Aphelion Distances
Perihelion and aphelion are key terms in understanding planetary orbits. They describe the points at which a planet is, respectively, closest to and farthest from the Sun during its orbit.
To calculate these distances, you can use the semi-major axis (\(a\)) and eccentricity (\(e\)) of the orbit:
  • Perihelion Distance: \(d_{\text{perihelion}} = a(1 - e)\)
  • Aphelion Distance: \(d_{\text{aphelion}} = a(1 + e)\)
The formula uses the semi-major axis adjusted by eccentricity to find these distances. For Pluto, with \(a = 39.44\text{ AU}\) and \(e = 0.2484\):
  • Perihelion: \(39.44 \times (1 - 0.2484) = 29.63\text{ AU}\)
  • Aphelion: \(39.44 \times (1 + 0.2484) = 49.26\text{ AU}\)
Understanding perihelion and aphelion helps explain why planets experience different levels of solar energy throughout their orbits, influencing temperatures and climate.
Semi-Major Axis
The semi-major axis is another crucial element in the study of orbital mechanics. It represents half of the longest diameter of the elliptical orbit, effectively setting the scale of the orbit. It is a measure from the center of the ellipse to the farthest point along the ellipse, providing a baseline for calculating the overall shape and size.
The length of the semi-major axis helps determine the orbit's period as dictated by Kepler's third law, where larger orbits have longer periods.
  • Determines the size of the orbit.
  • Affects the orbital period.
  • Used in calculating both perihelion and aphelion distances with the eccentricity.
Pluto鈥檚 semi-major axis is \(39.44\text{ AU}\), illustrating how it fits within the wide range of planetary orbits. In combination with its eccentricity, you can understand the extent of its distance variation from the Sun during its orbit. The greater the semi-major axis, the further and longer the orbit.

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Most popular questions from this chapter

(a) Verify that the point \(P(6,4 \sqrt{3})\) lies on the hyperbola \(16 x^{2}-9 y^{2}=144\) (b) In Example \(1,\) we found that the foci of this hyperbola were \(F_{1}(-5,0)\) and \(F_{2}(5,0) .\) Compute the lengths \(F_{1} P\) and \(F_{2} P\), where \(P\) is the point \((6,4 \sqrt{3})\) (c) Verify that \(\left|F_{1} P-F_{2} P\right|=2 a\)

Consider the equation \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\) (a) Use the symmetry tests from Section 1.7 to explain why the graph of this equation must be symmetric about both the \(x\) -axis and the \(y\) -axis. (b) Show that solving the equation for \(y\) yields \(y=\pm \frac{1}{3} \sqrt{36-4 x^{2}}\) (c) Let \(y=\frac{1}{3} \sqrt{36-4 x^{2}} .\) Use the techniques of Section 2.4 to find the domain of this function. (You need to solve the inequality \(36-4 x^{2} \geq 0 .\) ) You should find that the domain is the closed interval [-3,3] (d) Use a calculator to complete the following table. Then plot the resulting points and connect them with a smooth curve. This gives you a sketch of the first quadrant portion of the graph of \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\) $$\begin{array}{llllllll}x & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 \\\\\hline y=\frac{1}{3} \sqrt{36-4 x^{2}} & 2 & & & & & & 0 \\\\\hline\end{array}$$ (e) Use your graph in part (d) along with the symmetry results in part (a) to sketch a complete graph of \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\).

Graph the equations. $$3 x^{2}+4 x y+6 y^{2}=7$$

Let \(\overline{P Q}\) be a focal chord of the parabola \(y^{2}=4 p x,\) and let \(M\) be the midpoint of \(\overline{P Q} .\) A perpendicular is drawn from \(M\) to the \(x\) -axis, meeting the \(x\) -axis at \(S .\) Also from \(M,\) a line segment is drawn that is perpendicular to \(\overline{P Q}\) and that meets the \(x\) -axis at \(T\). Show that the length of \(\overline{S T}\) is onehalf the focal width of the parabola.

Find the equation of the line that is tangent to the hyperbola at the given point. Write your answer in the form \(y=m x+b\). $$16 x^{2}-25 y^{2}=400 ;(10,4 \sqrt{3})$$
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