Comet Orbit Halley's comet has an elliptical orbit with the...

Comet Orbit Halley's comet has an elliptical orbit with the sun at one focus. The eccentricity of the orbit is approximately $0.967 .$ The length of the major axis of the orbit is approximately 35.88 astronomical units. (An astronomical unit is about 93 million miles.) (a) Find an equation of the orbit. Place the center of the orbit at the origin and place the major axis on the $x$ -axis. (b) Use a graphing utility to graph the equation of the orbit. (c) Find the greatest (aphelion) and least (perihelion) distances from the sun's center to the comet's center.

Comet Orbit Halley's comet has an elliptical orbit
with the sun at one focus. The eccentricity of the orbit
is approximately $0.967 .$ The length of the major axis of
the orbit is approximately 35.88 astronomical units. (An
astronomical unit is about 93 million miles.)
(a) Find an equation of the orbit. Place the center of the
orbit at the origin and place the major axis on the
$x$ -axis.
(b) Use a graphing utility to graph the equation of the
orbit.
(c) Find the greatest (aphelion) and least (perihelion)
distances from the sun's center to the comet's center.

Key Concepts

Focus

The focus (plural: foci) of an ellipse is one of two fixed points used in its definition. In planetary orbits, the gravitational center (such as the sun) is located at one of these foci. The position of the foci relative to the center of the ellipse defines the orbit's characteristics, including its eccentricity.

Major Axis

The major axis of an ellipse is the longest line segment that passes through the center of the ellipse, running through both foci. Its length is twice the semi-major axis. This axis is critical in defining the shape of the ellipse and is used to calculate distances such as aphelion and perihelion in orbital mechanics.

Aphelion and Perihelion Distances

Aphelion is the point on an orbit at which the orbiting object is farthest from the central body, and perihelion is the point where it is closest. These distances are determined by the properties of the ellipse, specifically the semi-major axis and the eccentricity, and are crucial for understanding the variations in an object’s distance from the sun during its orbit.

Ellipse

An ellipse is a geometric shape that represents all points for which the sum of the distances to two fixed points (foci) is constant. In the context of celestial mechanics, ellipses describe the orbits of planets and comets, with one of the foci typically occupied by a central massive body like the sun.

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a circle. It is defined as the ratio of the distance from the center of the ellipse to a focus over the semi-major axis. An eccentricity close to 0 corresponds to a nearly circular orbit, whereas an eccentricity approaching 1 indicates a highly elongated ellipse.

Transcript

00:02 We are given a question about comet's orbit, and we're asked to answer this using ellipses.

00:13 So we're told that haley's comet has an elliptical orbit, the sun at one focus.

00:29 We're told that the eccentricity of the orbit is approximately .967.

00:42 We're told that the length of the major axis of the orbit is approximately 35 .88 astronomical units.

01:07 In part a, we're asked to find an equation of the orbit of haley's comet.

01:14 We're told to place the center of the orbit at the origin and place the major axis on the x axis.

01:23 So the center is at zero -zero, and the major axis is on the x -axis.

01:32 Of course, this means that the major axis is horizontal, and therefore, this, along with the fact that our centers at zero -zero means our equation must be of the form x squared over a squared plus y squared over b squared equals 1, where a is greater than b.

01:58 Now to find a, we know that the length of the major axis, 35 .88 astronomical units, is equal to 2 times a.

02:11 So the a is equal to 35 .88 divided by 2.

02:16 I'll leave this in this form for now.

02:28 Now we want to find b.

02:34 To find b, well, we need to find c.

02:38 So we know the eccentricity is about 0 .967.

02:49 And the eccentricity is c over a, which we know is 35 .88 over 2 approximately.

03:01 Therefore, c is about equal to 35 .88 divided by 2 times .967.

03:13 So now we have both a and c, and we know that a, b, and c are related by b squared equals a squared minus c squared.

03:24 And so plugging this in, this is 35 .88 divided by 2 squared.

03:34 And then c is 35 .88 times .967 over 2 squared.

03:46 So in other words, this is 35 .88 divided by 2 squared times 1 minus .967 squared.

04:00 So now we know a and we know b squared.

04:03 And so our equation becomes x squared over a squared, which we know is 35 .88 over 2 squared plus y squared over b squared, which we know is 35 .88 over 2 squared times 1 minus 0 .967, our eccentricity, squared equals 1...

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