Leyes de Kepler y Modelos del Universo

Early Cosmological Models and the Transition to Heliocentrism

The Aristotelian model defined the cosmos with the Tierra (Earth) as an immobile body at the center of the Universe. In this geocentric framework, all celestial bodies, including the Sol (Sun), were believed to describe perfectly circular orbits around the Earth. This view persisted until the work of Nicolás Copérnico (1473 to 1543), who developed the heliocentric theory. Copérnico proposed that the Sol is at the center of the Universe and that the Tierra and other planets revolve around it.

Giordano Bruno (1548 to 1600) further expanded upon the Copernican heliocentric model. He accepted that the Tierra revolves around the Sol but proposed much more radical ideas for his time. Bruno argued that the Universe is infinite and that the Sol is merely one star among many. He also hypothesized the existence of innumerable worlds or planets inhabited by other living beings, defending and broadening the scope of heliocentrism beyond the local planetary system.

Kepler's First Law: The Law of Ellipses

Johannes Kepler established three fundamental laws of planetary motion, starting with the 1.° Ley de Kepler, also known as the "Ley de las Formas" (Law of Shapes) or the Law of Ellipses. This law states that the orbits described by planets as they travel around the Sol are ellipses characterized by low eccentricity. In this elliptical model, the Sol is not at the center but is located at one of the two foci of the ellipse.

Kepler's Second Law: The Law of Equal Areas

The 2.° Ley de Kepler, known as the "Ley de las Áreas" (Law of the Areas), states that the radius vector of each planet sweeps out equal areas in equal intervals of time. This implies that the speed of a planet is not constant; it moves faster when it is closer to the Sun and slower when it is further away.

Mathematically, this is expressed as $A_1 = A_2$ when the time intervals are equal ($t_1 = t_2$). Therefore, the ratio of the area to time remains constant: $\frac{A}{t} = \text{constant}$. This principle allows for the calculation of orbital periods based on the geometric area covered by the planet's path relative to the Sun.

Application Exercise: Calculating Orbital Period via the Law of Areas

In a typical exercise, a planet is observed to take 8 months to sweep the area designated as $SDAB$. It is further specified that the area $SOD$ is exactly half of the area $SAB$. To find the total orbital period of the planet around the Sol ($S$), the following steps are taken:

First, variables are assigned to the areas to establish proportions. If the area $SOD = x$, then based on the problem statement, the area $SAB = 2x$. By analyzing the diagram and completing the segments of the ellipse, the area swept in 8 months ($SDAB$) is identified as $4x$. The total area of the entire ellipse is determined to be $12x$.

Applying the formula from Kepler's Second Law: $\frac{A_1}{t_1} = \frac{A_{total}}{T}$$\frac{4x}{8} = \frac{12x}{T}$

To solve for the period $T$: $T = \frac{(12x) \times 8}{4x}$$T = \frac{96x}{4x}$$T = 24 \text{ months}$

Kepler's Third Law: The Law of Periods

The 3.° Ley de Kepler, or the "Ley de los períodos" (Law of Periods), establishes a precise mathematical relationship between a planet's distance from the Sun and its orbital duration. It states that the square of the period of translation ($T$) of any planet is proportional to the cube of the mean distance ($d$ or $R$) between the planet and the Sol. In simple terms, this means that the further a planet is from the Sun, the more slowly it moves and the longer its orbital period becomes.

The relationship is defined by the formula: $\frac{T^2}{R^3} = K$ Where: $T$ = Period of translation around the Sol. $d$ or $R$ = Mean distance between a planet and the Sol. $K$ = Constant of proportionality.

In the context of our Solar System, when using the Earth as a reference ($T = 1 \text{ year}$ and $R = 1 \text{ UA}$), the constant $K$ is equal to 1. Consequently, for any planet in our system, the relationship can be simplified to $T^2 = R^3$ when units are in years and Astronomical Units (UA).

Planetary Data and Validation of the Third Law

The following table provides the orbital radius and period for the planets in our solar system, demonstrating the consistency of Kepler's Third Law:

• Mercurio: Radio $0.38 \text{ UA}$, Periodo $0.241 \text{ años}$
• Venus: Radio $0.72 \text{ UA}$, Periodo $0.615 \text{ años}$
• Tierra: Radio $1.00 \text{ UA}$, Periodo $1.00 \text{ años}$
• Marte: Radio $1.52 \text{ UA}$, Periodo $1.88 \text{ años}$
• Júpiter: Radio $5.20 \text{ UA}$, Periodo $11.86 \text{ años}$
• Saturno: Radio $9.54 \text{ UA}$, Periodo $29.46 \text{ años}$
• Urano: Radio $19.22 \text{ UA}$, Periodo $84.01 \text{ años}$
• Neptuno: Radio $30.06 \text{ UA}$, Periodo $164.79 \text{ años}$

To validate the law using Júpiter as an example: $\frac{T^2}{R^3} = \frac{11.86^2}{5.2^3} = \frac{140.6596}{140.608} \approx 1.001$

Calculation Exercise: Estimating Orbital Period for a New Planet

If a new planet is discovered orbiting at a distance of $60 \text{ UA}$ from the Sol, its orbital period can be estimated using the constant $K = 1$ for the Solar System:

Formula: $\frac{T^2}{R^3} = 1$

Given $R = 60 \text{ UA}$, we substitute the value: $\frac{T^2}{60^3} = 1$$T^2 = 60^3$$T^2 = 216,000$

To find $T$, we take the square root: $T = \sqrt{216,000}$$T \approx 464.7 \text{ años}$

Thus, a planet located at $60 \text{ UA}$ would take approximately 464.7 years to complete one full orbit around the Sol.