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Chapter 3: Problem 40

The elliptical orbit of a comet recently visited by a spacecraft is \(1.24 \mathrm{AU}\) from the Sun at its closest approach and \(5.68 \mathrm{AU}\) from the Sun at its farthest. a. Sketch the orbit of the comet. When is it moving fastest? When is it moving slowest? b. What is the semimajor axis of its orbit? How long does it take to go around the Sun? c. What is the distance from the Sun to the "center" of the ellipse? What is the eccentricity of the comet's orbit?

Short Answer

Expert verified
The comet moves fastest at perihelion and slowest at aphelion. The semimajor axis is 3.46 AU, the orbital period is 6.43 years, the distance from the Sun to the center of the ellipse is 2.22 AU, and the eccentricity is 0.64.

Step by step solution

01

- Identify the distances from the Sun

The closest approach (perihelion) of the comet to the Sun is given as 1.24 AU. The farthest distance (aphelion) from the Sun is given as 5.68 AU.
02

- Sketch the elliptical orbit

Draw an ellipse with the Sun located at one of the foci. The perihelion is the shortest distance from the focus to the ellipse, and the aphelion is the longest distance from the focus to the ellipse. The comet moves fastest at perihelion (closest approach) and slowest at aphelion (farthest distance).
03

- Calculate the semimajor axis

The semimajor axis, denoted by 'a', is the average of the perihelion (rp) and aphelion (ra) distances.Use the formula: \[ a = \frac{(r_p + r_a)}{2} \]Plugging in the values: \[ a = \frac{(1.24 + 5.68)}{2} = 3.46 \mathrm{AU} \]
04

- Calculate the orbital period

The orbital period 'T' is related to the semimajor axis by Kepler's Third Law. For an orbit around the Sun, the period in years can be calculated using: \[ T^2 = a^3 Therefore: T = \sqrt{a^3} \]Substitute the value of 'a': \[ T = \sqrt{3.46^3} \approx 6.43 \text{ years} \]
05

- Determine the distance from the Sun to the 'center' of the ellipse

The distance from the Sun to the center of the ellipse is given by the formula for the distance from the focus to the center, which is named 'c'. \[ c = \sqrt{a^2 - b^2} \]However, here we directly use: \[ c = a - \text{perihelion distance} c = 3.46 - 1.24 = 2.22 \mathrm{AU} \]
06

- Calculate the eccentricity

The eccentricity 'e' of the ellipse is given by: \[ e = \frac{c}{a} \]Using the values: \[ e = \frac{2.22}{3.46} \approx 0.64 \]

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Key Concepts

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

Elliptical Orbit
The path of a comet around the Sun is often an elliptical orbit. This shape is like a stretched circle, with two focal points instead of one center point. The Sun is situated at one of these foci. In such orbits, the comet's distance from the Sun varies over time.

Key points of an elliptical orbit include:
  • Various distances from the focal point during its orbit.
  • Two main points of interest: the closest (perihelion) and the farthest (aphelion) distances from the focal point, which is the Sun in this case.
The shape of the orbit influences the speed of the comet. The comet moves fastest when it is nearest to the Sun (at perihelion) and slowest at its farthest point (aphelion). This variation is due to the gravitational forces applied.
Semimajor Axis
The semimajor axis of an elliptical orbit is half of the longest diameter of the ellipse. It represents the average distance between the comet and the Sun over one complete orbit.

Here's how the semimajor axis is calculated:
The ellipse's longest diameter spans from the perihelion to the aphelion.
For our example, the perihelion is 1.24 AU and the aphelion is 5.68 AU.
Using the formula:
\( a = \frac{(r_p + r_a)}{2} \)
where \(r_p\) is the perihelion distance and \(r_a\) is the aphelion distance, we get:
\( a = \frac{(1.24 + 5.68)}{2} = 3.46 \mathrm{AU} \)
This value is essential because it helps to determine other orbital characteristics like the orbital period.
Orbital Period
The orbital period is the time a comet takes to complete one full orbit around the Sun. For objects orbiting the Sun, Kepler's Third Law helps us establish the relationship between the orbital period and the semimajor axis.

Kepler's Third Law states: \( T^2 = a^3 \)
where \( T \) is the orbital period in years, and \( a \) is the semimajor axis in astronomical units (AU).
Rewriting this as:
\( T = \sqrt{a^3} \)
For the given semimajor axis of 3.46 AU, he formula becomes:
\( T = \sqrt{3.46^3} \approx 6.43 \text{ years} \)
This result shows the time for the comet to make a single loop around the Sun.
Eccentricity
Eccentricity measures how 'stretched out' an elliptical orbit is. It ranges from 0 (perfect circle) to just below 1 (parabolic trajectory). A higher eccentricity means a more elongated shape.

The formula for eccentricity 'e' is:
\( e = \frac{c}{a} \)
where \( c \) is the distance from the center of the ellipse to a focal point, and \( a \) is the semimajor axis.
For our example:
\( a = 3.46 \ \text{AU} \) and \( c = 2.22 \ \text{AU} \)
Thus:
\( e = \frac{2.22}{3.46} \approx 0.64 \)
An eccentricity of 0.64 indicates a significantly elongated orbit compared to a circle. This shape affects both the comet's speed and distance at various points in its orbit.
Perihelion
The perihelion is the point in the orbit of a comet where it is closest to the Sun.
This results in the highest speed of the comet due to the stronger gravitational pull. Kepler's laws explain this faster movement.

In our exercise, the perihelion distance is 1.24 AU. At perihelion, the Sun's gravitational force pulls the comet strongly, accelerating it to its maximum speed within the orbit.
This proximity and speed contrast significantly with what happens at the comet's furthest point from the Sun, or aphelion.
Aphelion
The aphelion is the point where the comet is farthest from the Sun. Unlike perihelion, at aphelion, the comet moves at its slowest.

In our example, the aphelion distance is 5.68 AU. The reduced gravitational pull from the Sun at this point leads to a decrease in the comet's speed.
These contrasting speeds at perihelion and aphelion are a direct result of the elliptical orbit's varying distances from the focal point.
Kepler's laws again clarify why such variations occur within elliptical orbits, demonstrating the balance between distance and speed throughout an orbit.

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Most popular questions from this chapter

Suppose you read in the newspaper that a new planet has been found. Its average speed in orbit is \(33 \mathrm{km} / \mathrm{s}\). When it is closest to its star it moves at \(31 \mathrm{km} / \mathrm{s}\), and when it is farthest from its star it moves at \(35 \mathrm{km} / \mathrm{s}\). This story is in error because a. the average speed is far too fast. b. Kepler's third law says the planet has to sweep out equal areas in equal times, so the speed of the planet cannot change. c. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away. d. using these numbers, the square of the orbital period will not be equal to the cube of the semimajor axis.

A net force must be acting when an object a. accelerates. b. changes direction but not speed. c. changes speed but not direction. d. all of the above

Go to the online "Extrasolar Planets Encyclopedia" (http:// exoplanet.eu/catalog a. Find a planet with an orbital period similar to that of Earth. What is the semimajor axis of its orbit? If it is very different from \(1 \mathrm{AU}\), then the mass of the star is different from that of the Sun. Click on the star name in the first column to see the star's mass. What is the orbital eccentricity? b. Click on "Planet" to sort by name, and select a star with multiple planets. Verify that Kepler's third law applies by showing that the value of \(P^{2} / A^{3}\) is about the same for each of the planets of this star. How eccentric are the orbits of the multiple planets?

Planets with high eccentricity may be unlikely candidates for life because a. the speed varies too much. b. the period varies too much. c. the temperature varies too much. d. the orbit varies too much.

Suppose a new dwarf planet is discovered orbiting the Sun with a semimajor axis of 50 AU. What would be the orbital period of this new dwarf planet?
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