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Chapter 3: Problem 36

Suppose you read online that "experts have discovered a new planet with a distance from the Sun of 2 AU and a period of 3 years." Use Kepler's third law to argue that this is impossible.

Short Answer

Expert verified
It is impossible as Kepler's Third Law shows a contradiction: \( 9 eq 8 \).

Step by step solution

01

Understand Kepler's Third Law

Kepler's Third Law states that the square of the period of any planet is proportional to the cube of the semi-major axis of its orbit. The formula is given by: \[ T^2 \propto a^3 \] where \( T \) is the orbital period and \( a \) is the semi-major axis (distance from the Sun in astronomical units, AU).
02

Write the proportion

For planets orbiting the Sun, we can write Kepler's Third Law as: \[ \left( \frac{T}{T_{Earth}} \right)^2 = \left( \frac{a}{a_{Earth}} \right)^3 \] where \( T_{Earth} = 1 \) year and \( a_{Earth} = 1 \) AU for Earth's orbit.
03

Plug in the given values

Given the data: \( T_{new} = 3 \) years and \( a_{new} = 2 \) AU. Plugging these values into the equation gives us: \[ \left( \frac{3}{1} \right)^2 = \left( \frac{2}{1} \right)^3 \] Simplifying this: \[ 9 = 8 \]
04

Analyze the result

The equation \( 9 = 8 \) is a contradiction. This means that, according to Kepler's Third Law, it is impossible for a planet with a 2 AU distance from the Sun to have a period of 3 years.

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Key Concepts

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

Astronomical Units
An astronomical unit (AU) is the average distance from the Earth to the Sun, which equals approximately 150 million kilometers (93 million miles). Astronomical units are useful for expressing distances within our solar system due to their convenience in scale.

For example, Earth's orbit around the Sun is defined as 1 AU. This means if a planet's distance is said to be 2 AU, it is twice as far from the Sun as the Earth. Using AU simplifies the comparison of planetary distances. Understanding AU is a vital part of applying Kepler's Third Law since the law often uses this unit to express semi-major axes.
Orbital Period
The orbital period is the time a planet takes to complete one full orbit around the Sun. This period is usually measured in Earth years for simplicity.

Kepler's Third Law links the orbital period and the semi-major axis. Specifically, the square of the orbital period (T^2) is proportional to the cube of the semi-major axis (a^3). This means you can determine one if you know the other.

For instance, the Earth’s orbital period is 1 year and this is used as a baseline in Kepler’s Third Law to compare with other celestial bodies. If a planet has an orbital period of 3 years, Kepler’s Law helps us find its relative distance from the Sun.
Semi-Major Axis
The semi-major axis is half of the longest diameter of an elliptical orbit. It is effectively the average distance from a planet to the Sun in its orbit, expressed in astronomical units (AU).

In Kepler’s Third Law, the semi-major axis (a) is central to calculating the relationship between a planet's distance from the Sun and its orbital period. According to the law, the cube of the semi-major axis is proportional to the square of the orbital period.

For example, if the semi-major axis (a) of a new planet is 2 AU, using Kepler’s Third Law, you can evaluate whether this distance matches a given orbital period, ensuring the consistency of their values.

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Most popular questions from this chapter

An empirical science is one that is based on a. hypothesis. b. calculus. c. computer models. d. observed data.

During the latter half of the 19 th century, a few astronomers thought there might be a planet circling the Sun inside Mercury's orbit. They even gave it a name: Vulcan. We now know that Vulcan does not exist. If a planet with an orbit one-fourth the size of Mercury's actually existed, what would be its orbital period relative to that of Mercury?

Galileo came up with the concept of inertia. What do we mean by inertia?

For Earth, \(P^{2} / A^{3}=1.0\) (in appropriate units). Suppose a new dwarf planet is discovered that is 14 times as far from the Sun as Earth is. For this planet, a. \(P^{2} / A^{3}=1.0\) b. \(P^{2} / A^{3} > 1.0\) c. \(P^{2} / A^{3} < 1.0\) d. one can't know the value of \(P^{2} / A^{3}\) without more information.

When Earth catches up to a slower-moving outer planet and passes it in its orbit, the planet a. exhibits retrograde motion. b. slows down because it feels Earth's gravitational pull. c. decreases in brightness as it passes through Earth's shadow. d. moves into a more elliptical orbit.
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