This page delves into Kepler's three laws of planetary motion, providing a comprehensive overview of each law and its implications for celestial mechanics.

Highlight: Kepler's laws describe the motion of planets around the Sun and are crucial for understanding orbital dynamics.

First Law of Kepler $Law of Ellipses$:

Definition: In the heliocentric reference frame, the trajectory of a planet's or satellite's center around a celestial body is an ellipse with the Sun at one of its foci.

Second Law of Kepler $Law of Equal Areas$:

Definition: The line segment connecting the Sun to a planet sweeps out equal areas during equal intervals of time.

This law implies that planets move faster when they are closer to the Sun $perihelion$ and slower when they are farther away $aphelion$.

Third Law of Kepler $Law of Periods$:

Definition: The ratio of the square of the orbital period to the cube of the semi-major axis is constant for all planets.

The page provides the mathematical formulation of the third law: T² / a³ = 4π² / $G · Ms$

Where:

• T is the orbital period
• a is the semi-major axis
• G is the gravitational constant
• Ms is the mass of the Sun

This page focuses on the universal law of gravitation and its application to satellite motion around Earth.

Definition: The universal law of gravitation states that the gravitational force between two masses is proportional to the product of their masses and inversely proportional to the square of the distance between them.

The mathematical expression for the gravitational force is given as: F = G · $M1 · M2$ / r²

Where:

• G is the gravitational constant $6.67 × 10⁻¹¹ N·m²/kg²$
• M1 and M2 are the masses of the two objects
• r is the distance between their centers

The page then applies Newton's Second Law to derive the acceleration of a satellite orbiting Earth:

a = -G · MT / R² · ur

Where:

• MT is the mass of Earth
• R is the distance from the center of Earth to the satellite
• ur is the radial unit vector

Example: This derivation is crucial for understanding the motion of artificial satellites and natural moons around planets.

This page continues the analysis of satellite motion, focusing on deriving expressions for satellite velocity and orbital period.

The velocity of a satellite in circular orbit is derived as: v = √$G · MT / R$

Where:

• G is the gravitational constant
• MT is the mass of Earth
• R is the orbital radius

Highlight: This equation shows that satellite velocity depends on the mass of the central body and the orbital radius.

The page then derives the expression for the orbital period of a satellite, which is a direct application of Kepler's Third Law:

T = 2π · √$R³ / (G · MT$)

Example: This formula is used to calculate the orbital periods of satellites and planets, demonstrating the practical application of Kepler's laws.

The derivation of these formulas provides a deep understanding of the relationship between a satellite's orbital characteristics and the properties of the central body it orbits.

This page discusses the characteristics of geostationary satellites and applies Kepler's laws to planetary motion.

Definition: A geostationary satellite is a satellite that remains stationary relative to Earth's surface, orbiting in the equatorial plane with the same rotational period as Earth.

Key characteristics of geostationary satellites:

1. Orbital plane coincides with Earth's equatorial plane
2. Rotates in the same direction as Earth
3. Orbital period equals Earth's rotational period $24 hours$
4. Altitude is approximately 36,000 km

Example: The page provides a practical application by calculating the mass of Jupiter using the orbital period of one of its moons.

Given:

• Jupiter's radius: 69,911 km
• Satellite's orbital period: 1 day, 18 hours, 20 minutes

Using Kepler's Third Law, the mass of Jupiter is calculated to be approximately 1.94 × 10²⁷ kg.

This example demonstrates how Kepler's laws can be applied to determine the properties of celestial bodies based on the motion of their satellites.

This final page focuses on deriving and calculating the altitude of a geostationary satellite.

The derivation starts with Kepler's Third Law and the expression for orbital velocity:

R³ = $T² · G · MT$ / $4π²$

Where:

• R is the orbital radius
• T is the orbital period $24 hours for a geostationary satellite$
• G is the gravitational constant
• MT is the mass of Earth

Highlight: The calculation demonstrates that the altitude of a geostationary satellite is approximately 36,000 km above Earth's surface.

Given:

• Earth's radius: 6,371 km
• Earth's mass: 6.0 × 10²⁴ kg

The page shows the step-by-step calculation to arrive at the final result:

h = R - RT ≈ 36,000 km

Example: This calculation is crucial for understanding the placement of communication satellites and other space-based technologies that require a fixed position relative to Earth's surface.

The derivation and calculation of geostationary satellite altitude serve as a practical application of Kepler's laws and orbital mechanics, demonstrating the power of these principles in modern space technology.

This page introduces the concept of circular motion and its key components. It explains the trajectory and reference frame used in analyzing circular motion.

Vocabulary: Circular motion refers to the movement of an object in a circular path.

The page describes the unit vectors used in circular motion analysis:

• Ut: Tangential unit vector
• Un: Normal unit vector $perpendicular to the trajectory$

It also presents the velocity and acceleration formulas for circular motion:

• v = v · ut
• a = an · un + at · ut

Highlight: For uniform circular motion, velocity magnitude is constant, and acceleration is purely centripetal $a = v²/R · un$.

The page concludes with a statement of Newton's Second Law, which is fundamental to understanding the forces involved in circular motion.

Definition: Newton's Second Law states that the sum of forces applied to a solid is equal to the product of its mass and acceleration.