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.

Short Answer

Expert verified
a. \( \frac{P^2}{A^3} = 1.0 \)

Step by step solution

01

Understanding the given relationship for Earth

The given relationship for Earth is \( \frac{P^2}{A^3} = 1.0 \), where \( P \) is the orbital period and \( A \) is the average distance from the Sun (semimajor axis). This indicates that Earth's orbital period and distance follow Kepler's third law.
02

Expressing the relationship for the new dwarf planet

According to Kepler's third law, for any planet, \( \frac{P^2}{A^3} \) should theoretically equal the same constant (which is 1.0 in this scenario). Thus, the relationship should apply to the new dwarf planet as well.
03

Determining the distance of the new dwarf planet

The problem states that the new dwarf planet is 14 times as far from the Sun as Earth is. Therefore, \( A_{new} = 14A_{Earth} \).
04

Verifying the relationship \( \frac{P^2}{A^3} \) for the new dwarf planet

Using Kepler's third law, we still get the same constant: \( \frac{P_{new}^2}{(14A_{Earth})^3} = 1.0 \). Therefore, \( \frac{P_{new}^2}{A_{new}^3} = 1.0 \).
05

Concluding the result

For the new dwarf planet, the value of \( \frac{P^2}{A^3} = 1.0 \), just like Earth.

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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 it takes for a planet or any celestial body to complete one full orbit around a star, such as the Sun. In calculations, it's usually represented by the letter \(P\). For Earth, one complete orbit around the Sun takes about 365.25 days, which is our year.

Orbital periods can vary greatly depending on the distance from the star. The farther away the body is, the longer the orbital period. This relationship is crucial in understanding planetary motion.

Kepler's third law, which we will discuss later, provides a straightforward way to calculate the orbital period if we know the distance. For example, if a dwarf planet is found farther from the Sun than Earth, it will have a longer orbital period.
  • For a planet closer to the Sun, the orbital period is shorter.
  • For a planet farther from the Sun, the orbital period is longer.
  • Orbital periods help in understanding the timing of planetary events like transits and eclipses.
Semimajor Axis
The semimajor axis, often denoted as \(A\), is a measure of the size of an orbit. Specifically, it's the average distance from the orbiting body to the star it revolves around. For Earth, this distance is about 93 million miles or 150 million kilometers, which defines one Astronomical Unit (AU).

When Kepler's third law mentions \(A^3\), it is referring to the cube of this distance. For instance, if a newly discovered dwarf planet is said to be 14 times farther from the Sun than Earth, its semimajor axis would be \(14A_{Earth}\).

The distance affects various factors like gravitational forces and orbital periods. Here are some highlights:
  • The larger the semimajor axis, the longer the orbital path.
  • For Earth, this value is 1 AU, setting a standard unit for measuring other planets' distances.
  • Knowing the semimajor axis is essential for calculating other orbital properties.
Planetary Motion
Planetary motion describes how planets and other celestial bodies move in their orbits around a star. Kepler's laws of planetary motion provide a foundational understanding of these movements.

Kepler's third law states that the square of a planet's orbital period \((P^2)\) is directly proportional to the cube of the semimajor axis of its orbit \((A^3)\). This relationship is expressed as:
\[ \frac{P^2}{A^3}=1.0 \]

For Earth, \((P^2 / A^3 = 1.0)\). This same ratio applies to any planet orbiting the same star, like a new dwarf planet discovered 14 times farther from the Sun than Earth. Regardless of the distance, the ratio \((P^2 / A^3)\) remains constant.

With the discovery of a new body in our solar system:
  • Its motion can be predicted using Kepler’s laws.
  • The semimajor axis of 14 times the distance of Earth leads to an equivalent orbital period ratio.
  • Understanding planetary motion helps us predict planetary positions at various times.

Kepler's third law aids not just in understanding distant planets but also moons and artificial satellites.

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Most popular questions from this chapter

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?

Suppose a planet is discovered orbiting a star in a highly elliptical orbit. While the planet is close to the star it moves________, but while it is far away it moves________. a. faster; slower b. slower; faster c. retrograde; prograde d. prograde; retrograde

An inferior planet is one that is a. smaller than Earth. b. larger than Earth. c. closer to the Sun than Earth is d. farther from the Sun than Earth is.

Go to the Museo Galileo website and view the exhibit on Galileo's telescope (http://www.museogalileo.it/en/explore/ exhibitions/pastexhibitions/galileostelescope.html). What did his telescope look like? What other instruments did he use? From the museum page you can link to short videos (in English) on his science and his trial (http://catalogue museogalileo.it/index/VideoIndexByThematicArea.html#s7). For example, click on "Galileo's micrometer": How did he measure the separation of the moons from Jupiter? How did this measurement allow him to show that the moons obeyed Kepler's law? Why is Galileo often considered the first modern scientist? Why is his middle finger on display in the museum?

Suppose you are pushing a small refrigerator of mass \(50 \mathrm{kg}\) on wheels. You push with a force of \(100 \mathrm{N}\) a. What is the refrigerator's acceleration? b. Assume the refrigerator starts at rest. How long will the refrigerator accelerate at this rate before it gets away from you (that is, before it is moving faster than you can run-of the order \(10 \mathrm{m} / \mathrm{s}\) )?

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