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Chapter 3: Problem 32

The distance that Neptune has to travel in its orbit around the Sun is approximately 30 times greater than the distance that Earth must travel. Yet it takes nearly 165 years for Neptune to complete one trip around the Sun. Explain why.

Short Answer

Expert verified
Neptune takes 165 years to orbit the Sun because its orbit is 30 times larger, and according to Kepler's Third Law, greater distances result in much longer orbital periods.

Step by step solution

01

- Understand the problem

Neptune travels a much larger distance around the Sun compared to Earth. Despite this, it takes significantly longer to complete one orbit. The goal is to explain why.
02

- Kepler's Third Law

Recall Kepler's Third Law which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, this is expressed as: \[\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} \] where \( T \) is the orbital period and \( R \) is the distance from the Sun (semi-major axis).
03

- Apply Kepler's Third Law

Given that Neptune's distance from the Sun is 30 times that of Earth's, we let \( R_2 = 30 R_1 \). Substituting in Kepler's law we get: \[\frac{T_1^2}{R_1^3} = \frac{T_2^2}{(30 R_1)^3} \] Which simplifies to: \[\frac{T_1^2}{R_1^3} = \frac{T_2^2}{27,000 R_1^3} \]
04

- Solve for Neptune's Period

Since Earth鈥檚 orbital period \( T_1 \) is 1 year, let's solve for Neptune鈥檚 orbital period \( T_2 \): \[\frac{1^2}{1^3} = \frac{T_2^2}{27,000} \] Thus: \[\frac{1}{1} = \frac{T_2^2}{27,000} \] And: \ T_2 = \sqrt{27,000} = 164.32 \text{ years} So it takes Neptune about 165 years to complete one orbit around the Sun.

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Key Concepts

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

Orbital Period
The orbital period is the time a planet takes to complete one full orbit around the Sun. It is an important concept in astronomy, helping us understand the dynamics of planetary movement.
Kepler's Third Law plays a crucial role in determining the orbital period. According to this law, the square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit.
This means that if we know the distance of a planet from the Sun, we can calculate how long it takes to orbit the Sun.
For example, Earth takes about one year to orbit the Sun because its distance, or semi-major axis, is defined as 1 Astronomical Unit (AU).
Semi-Major Axis
The semi-major axis is the longest radius of an ellipse, which is the shape of a planet's orbit around the Sun. It is essentially the average distance between the planet and the Sun.
In mathematical terms, the semi-major axis is symbolized by \( R \).
According to Kepler's Third Law, the cube of the semi-major axis is directly related to the square of the orbital period.
For Neptune, the semi-major axis is 30 times greater than that of Earth. This gives us a clear understanding of why Neptune鈥檚 orbit takes so much longer.
Planetary Orbits
Planetary orbits are the paths that planets follow around the Sun. These paths are not perfect circles; they are ellipses.
Kepler's laws offer a deep understanding of these orbits. His Third Law, in particular, is vital to grasp why different planets have varying orbital periods.
The further a planet is from the Sun, the larger its orbit, and consequently, the longer its orbital period. This relationship is mathematically captured in the expression:
\( \frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} \)
Neptune's Orbit
Neptune's orbit around the Sun is a fascinating example of how Kepler's Third Law works. Despite being 30 times farther away from the Sun than Earth, Neptune takes nearly 165 years to complete one orbit.
To understand this, we apply Kepler's Third Law. Given that Earth's distance from the Sun is defined as 1 AU, Neptune's distance is 30 AU.
By substituting into Kepler's formula, we see that if Earth's orbital period is 1 year, Neptune's orbital period, \( T_2 \), becomes:
\( \frac{1^2}{1^3} = \frac{T_2^2}{30^3} \)
Simplifying, we get \( T_2 = \sqrt{27,000} \approx 164.32 \text{ years} \). This tells us why Neptune takes almost 165 years to orbit the Sun.

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Most popular questions from this chapter

You are driving down a straight road at a speed of \(90 \mathrm{km} / \mathrm{h},\) and you see another car approaching you at a speed of \(110 \mathrm{km} / \mathrm{h}\) along the road. a. Relative to your own frame of reference, how fast is the other car approaching you? b. Relative to the other driver's frame of reference, how fast are you approaching the other driver's car?

Galileo observed that Venus had phases that correlated with its size in his telescope. From this information, you may conclude that Venus a. is the center of the Solar System. b. orbits the Sun. c. orbits Earth. d. orbits the Moon.

Imagine you are walking along a forest path. Which of the following is not an action-reaction pair in this situation? a. the gravitational force between you and Earth; the gravitational force between Earth and you. b. your shoe pushing back on Earth; Earth pushing forward on your shoe c. your foot pushing back on the inside of your shoe; your shoe pushing forward on your foot d. you pushing down on Earth; Earth pushing you forward 1

Go to the online "Extrasolar Planets Encyclopedia" (http://exoplanet.eu/catalog-all.php), and 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 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? Now 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?

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. planets stay at a constant distance from their stars; they don't move closer or farther away. d. Kepler's second law says the planet must move fastest when it's closest, not when it is farthest away. e. using these numbers, the square of the orbital period will not be equal to the cube of the semimajor axis.
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