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. you 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

- Understand Kepler's Third Law

Kepler's Third Law states that the square of the orbital period (P) of a planet is directly proportional to the cube of the semi-major axis (A) of its orbit. Mathematically, this is expressed as \[ \frac{P^2}{A^3} = k \] where \( k \) is a constant that has the same value for all objects orbiting the Sun.
02

- Identify Given Information

Given for Earth: \[ \frac{P^2}{A^3} = 1.0 \] Also, the new dwarf planet is 14 times as far from the Sun as Earth, so \( A_{planet} = 14 \).
03

- Apply Given Information

Using the provided constant for Earth, \[ \frac{P^2}{A^3} = 1.0 \] we can determine that this constant applies to the dwarf planet as well since gravitational law applies equally.
04

- Confirm the Relationship

Let's confirm if \[ \frac{P_{\text{planet}}^2}{A_{\text{planet}}^3} = 1.0 \] Since \( A_{\text{planet}} = 14 \) (which is 14 times greater than Earth's semi-major axis), and given that the constant is the same for both, \[ \frac{P^2}{(14A_{\text{Earth}})^3} \] must be calculated.
05

- Simplify the Relationship

Given Earth’s equation: \[ \frac{P_{\text{Earth}}^2}{A_{\text{Earth}}^3} = 1.0 \] then for the new planet: \[ A_{\text{planet}} = 14A_{\text{Earth}} \] therefore, \[ \frac{P_{\text{planet}}^2}{(14A_{\text{Earth}})^3} = 1.0 \]. Simplifying further, we see that \[ (14^3) = 2744 \].
06

- Calculate and Conclude

Since \[ \frac{P_{\text{planet}}^2}{2744A_{\text{Earth}}^3} = 1.0 \] multiplying both sides by \( 2744 \): \[ P_{\text{planet}}^2 = 2744 \] Thus, \[ P_{\text{planet}} = \sqrt{2744} = 52.4A_{\text{Earth}}^{3/2} \], implying \[ \frac{P_{\text{planet}}^2}{A_{\text{planet}}^3} = 1.0 \]

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Key Concepts

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

Orbital Period
Orbital period refers to the time it takes for a celestial object, like a planet or moon, to complete one full orbit around another object, such as a star or planet. This period is essential in understanding how objects move in space. The formula for calculating the orbital period (P) of a planet is derived from Kepler's Third Law. This law tells us that the square of the orbital period (P²) is directly proportional to the cube of the semi-major axis (A³) of its orbit.

In simple terms, if you know how far an object is from the star it orbits, you can predict how long it will take to complete one orbit. For planets in our solar system, this relationship is especially useful as it shows us that further away planets take longer to orbit the Sun compared to those closer to it.
Semi-Major Axis
The semi-major axis (A) is essentially half of the longest diameter of an elliptical orbit. This axis is crucial in calculating the shape and size of a planet's orbit around the Sun or any other star. When Kepler's Third Law is applied, the semi-major axis helps us determine the orbital period. The relationship is represented as \[ \frac{P^2}{A^3}=k \] where \( k \) is a constant.

In the context of the exercise, if a new dwarf planet is discovered to be 14 times further from the Sun than Earth, its semi-major axis will be 14 times that of Earth (denoted as \( 14A_{Earth} \)). This directly impacts its orbital period, as we earlier established that the further the distance (or greater the semi-major axis), the longer the object takes to orbit the Sun.
Gravitational Law
Gravitational law governs how objects attract each other in space. Kepler's Third Law is a direct application of these gravitational principles. The law states that the ratio \[ \frac{P^2}{A^3} = 1.0 \] is constant for all objects orbiting the Sun, assuming it's in Earth years and astronomical units. This ratio remains constant because of the predictable nature of gravitational forces exerted by massive objects like the Sun.

For the newly discovered dwarf planet, regardless of how far it is, the value of \[ \frac{P^2}{A^3} \] will still equal 1.0, proving that gravitational laws apply uniformly. This universality of gravitational influence makes it possible to understand and predict the movement and relationships of celestial bodies throughout the solar system, and even beyond.

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