/*! 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 79 Use the fact that the orbit of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 79

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. The mean distance of Earth from the Sun is 93 million miles. If the aphelion of Earth is 94.5 million miles, what is the perihelion? Find an equation for the orbit of Earth around the Sun.

Short Answer

Expert verified
Perihelion is 91.5 million miles. The equation of Earth's orbit is \frac{x^2}{93^2} + \frac{y^2}{8646.75} = 1.

Step by step solution

01

- Identify the variables

Let the semimajor axis (mean distance from the Sun) be denoted as \(a = 93\) million miles. The aphelion, which is the farthest distance from the Sun, is given as \(A = 94.5\) million miles. We need to find the perihelion distance \(P\).
02

- Establish the relationship between aphelion, perihelion, and semimajor axis

For an elliptical orbit, the relationship between aphelion \(A\), perihelion \(P\), and the semimajor axis \(a\) is given by:a = \frac{A + P}{2}Solve this equation for \(P\).
03

- Solve for perihelion

Plug in the values of \(a\) and \(A\) into the equation:a = \frac{94.5 + P}{2} \implies 93 = \frac{94.5 + P}{2}Multiply both sides by 2:186 = 94.5 + PSubtract 94.5 from both sides:P = 91.5Therefore, the perihelion distance \(P\) is 91.5 million miles.
04

- Write the equation of the orbit

The equation of an ellipse in standard form, with the Sun at one of the foci, can be written as:\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1Here, \(a\) is the semimajor axis, and \(b\) is the semiminor axis. The distance between the center of the ellipse and the focus is given by \(c = \frac{A-P}{2} = \frac{94.5-91.5}{2} = 1.5\) million miles.Since \(c^2 = a^2 - b^2\), we can solve for \(b\):c = \frac{94.5 - 91.5}{2} = 1.5c^2 = 1.5^2 = 2.25b^2 = a^2 - c^2 = 93^2 - 2.25 = 8649 - 2.25 = 8646.75Therefore, the equation of Earth's orbit around the Sun is:\frac{x^2}{93^2} + \frac{y^2}{8646.75} = 1

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

ellipse
An ellipse is a geometric shape that looks like a flattened circle. It has two focal points. For planetary orbits, one of these foci is the Sun. Ellipses have interesting properties: the sum of the distances from any point on the ellipse to the two foci is constant. This explains the varying distance of planets from the Sun as they move in their orbits.
Ellipses are important in astronomy as they describe the shape of planetary orbits. Unlike circular orbits, ellipses account for the different distances a planet can have from the Sun over time.
aphelion
The aphelion is the point in a planet's orbit where it is farthest from the Sun. For Earth, this distance is about 94.5 million miles. This term comes from Greek, where 'apo' means away and 'helios' means sun.
Understanding the aphelion helps scientists determine the Earth's maximum distance from the Sun, providing insights into variations in solar energy the Earth receives during its orbit. This is connected to seasonal changes and climate patterns on Earth.
perihelion
Conversely, the perihelion is the point in a planet's orbit where it is closest to the Sun. For Earth, this distance is approximately 91.5 million miles.
The term perihelion is derived from Greek, with 'peri' meaning near and 'helios' meaning sun. Knowing the perihelion is crucial for understanding how close the Earth gets to the Sun, which affects temperature and weather patterns due to the increased solar intensity.
semimajor axis
The semimajor axis of an elliptical orbit represents the planet's average distance from the Sun. For Earth, this distance is about 93 million miles.
The semimajor axis is half the length of the longest diameter of the ellipse and is a crucial element in describing the size and shape of an orbit. This measurement helps astronomers to calculate the orbital period and other key parameters of a planet鈥檚 motion around the Sun. The formula relating aphelion (A), perihelion (P), and semimajor axis (a) is given by:
\[a = \frac{A + P}{2}\]
astronomical mathematics
Astronomical mathematics involves calculations based on principles like Kepler's laws of planetary motion. These calculations help us describe and predict the movements of celestial bodies.
For Earth's orbit, knowing the values of the aphelion, perihelion, and semimajor axis, we can derive important details such as the equation of the orbit. For instance, to find the orbit鈥檚 equation:
  • Calculate the distance between the center of the ellipse and the focus (c):
    • \[c = \frac{94.5 - 91.5}{2} = 1.5 \text{ million miles}\]
  • Use the formula \[c^2 = a^2 - b^2\] to find \(b\):
    • \[b^2 = a^2 - c^2 = 93^2 - 1.5^2 = 8646.75\]
  • The orbit equation then is:

    • \[\frac{x^2}{93^2} + \frac{y^2}{8646.75} = 1\]
This equation represents Earth's elliptical path around the Sun, integrating fundamental concepts of astronomical mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Lamp Shadow The light from a lamp creates a shadow on a wall with a hyperbolic border. Find the equation of the border if the distance between the vertices is 18 inches and the foci are 4 inches from the vertices. Assume the center of the hyperbola is at the origin.

Nuclear Power Plant Some nuclear power plants utilize "natural draft" cooling towers in the shape of a hyperboloid, a solid obtained by rotating a hyperbola about its conjugate axis. Suppose that such a cooling tower has a base diameter of 400 feet, and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. If the diameter at the top of the tower is 300 feet, how tall is the tower?

Multiple Choice If a circle rolls along a horizontal line without slipping, a fixed point \(P\) on the circle will trace out a curve called \(\mathrm{a}(\mathrm{n})\) ________. (a) cycloid (b) epitrochoid (c) hyptrochoid (d) pendulum

Arch in St. Louis is often mistaken to be parabolic in shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 630 feet high and 630 feet wide at its base. (a) Find the equation of a parabola with the same dimensions. Let \(x\) equal the horizontal distance from the center of the arch. (b) The table below gives the height of the Arch at various widths; find the corresponding heights for the parabola found in (a). $$ \begin{array}{|cc|} \hline \text { Width (ft) } & \text { Height (ft) } \\ \hline 567 & 100 \\ 478 & 312.5 \\ 308 & 525 \\ \hline \end{array} $$ (c) Do the data support the notion that the Arch is in the shape of a parabola?

The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 400 feet apart and 100 feet high. If the cables are at a height of 10 feet midway between the towers, what is the height of the cable at a point 50 feet from the center of the bridge?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.