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Kepler's laws

Kepler's laws

Kepler's area law

[chatgpt]

Kepler's Area Law, also known as the Second Law of Planetary Motion, describes how a planet orbits around the Sun. It states:

"A line segment joining a planet and the Sun sweeps out equal areas in equal intervals of time."

Key Points:

Orbital Motion:
- Planets move in elliptical orbits, with the Sun at one of the two foci.
Varying Speeds:
- The planet moves faster when it is closer to the Sun (at perihelion) and slower when it is farther from the Sun (at aphelion).
Equal Area, Equal Time:
- If you imagine a line connecting the planet to the Sun, the area swept by this line over a certain time interval is the same, regardless of where the planet is in its orbit.
- This implies that the orbital speed of the planet varies depending on its distance from the Sun.

Mathematical Implication:

Kepler's Area Law is a consequence of the conservation of angular momentum. Mathematically: \[ \text{Area swept per unit time} = \frac{1}{2} r^2 \omega \] where:

\( r \): Distance of the planet from the Sun.
\( \omega \): Angular velocity of the planet.

The law reflects that the product of \( r^2 \omega \) (angular momentum) remains constant for the planet's orbit.

Visual Understanding:

Imagine tracing the planet's movement over equal time intervals. Near perihelion, the planet covers a longer arc to sweep out the same area as it does over a shorter arc near aphelion.

This law is crucial in understanding orbital dynamics and has applications in satellite mechanics and space exploration.

Kepler's area law for Europa

[chatgpt] In the case of circular orbits, Kepler’s second law (area law) is easy to verify because the orbital motion is uniform, and the areas swept out by the radius vector are proportional to the time elapsed.

Here’s a step-by-step explanation and example calculation for one of Jupiter’s moons, Europa, which has a nearly circular orbit:

Data for Europa

Orbital radius (\(r\)): \(670,900 \, \text{km}\) (approximately)
Orbital period (\(T\)): \(3.551 \, \text{days}\) = \(3.551 \times 86400 \, \text{seconds}\)
Orbital speed (\(v\)): Calculated using \(v = \frac{2\pi r}{T}\)

Step 1: Compute Orbital Speed

\[ v = \frac{2\pi r}{T} = \frac{2\pi \times 670,900 \times 10^3}{3.551 \times 86400} \, \text{m/s} \] \[ v \approx 13,740 \, \text{m/s} \]

Step 2: Compute Area Swept in Time \(\Delta t\)

The area swept in time \(\Delta t\) is given by: \[ \Delta A = \frac{1}{2} r^2 \Delta \theta \] For circular motion, the angular velocity \(\omega\) is constant and can be calculated as: \[ \omega = \frac{2\pi}{T} \] \[ \Delta \theta = \omega \Delta t \] Substituting, we get: \[ \Delta A = \frac{1}{2} r^2 \omega \Delta t \]

Step 3: Compute for a Specific Time Interval

Let \(\Delta t = 1 \, \text{hour} = 3600 \, \text{seconds}\). Compute \(\Delta A\): \[ \omega = \frac{2\pi}{T} = \frac{2\pi}{3.551 \times 86400} \, \text{rad/s} \] \[ \omega \approx 2.045 \times 10^{-5} \, \text{rad/s} \] \[ \Delta A = \frac{1}{2} (670,900 \times 10^3)^2 \times (2.045 \times 10^{-5}) \times 3600 \] \[ \Delta A \approx 2.592 \times 10^{12} \, \text{m}^2 \]

Step 4: Check for Equal Areas in Successive Time Intervals

Since the motion is uniform, \(\Delta A\) will be the same for any equal time interval (e.g., the next hour). This confirms Kepler's area law for circular orbits.

Key Insight

For circular orbits:

The area swept is proportional to the time (\(\Delta A \propto \Delta t\)).
The angular velocity \(\omega\) and radius \(r\) are constant, making the computation straightforward.

What happens with circular orbits?

This law becomes trivial for circular orbits, right?
And we circularize even elliptical orbits by taking as our "radius" the semi-major axis of the ellipse, which is the mean.

Kepler's first two laws are descriptive…

…but third law is quantitative. Chatgpt confirmed this idea:

Yes, that's a good way to look at it. Kepler's first two laws are descriptive because they tell us how planets move but don’t directly allow us to compute their orbits.

First Law (Elliptical Orbits): Planets move in ellipses with the Sun at one focus. This describes the shape of the orbit but doesn’t give a way to calculate positions over time.
Second Law (Equal Areas in Equal Time): A planet sweeps out equal areas in equal times. This describes variations in orbital speed but doesn’t give an explicit formula for predicting positions at arbitrary times.

However, Kepler’s Rule (Third Law) is different because it is quantitative and allows calculations. Since it states that

\[ T^2 \propto a^3 \]

it directly relates the orbital period \(T\) and the semi-major axis \(a\), making it useful for predicting and comparing orbits. With additional information (such as the gravitational constant and mass of the central body), Kepler’s Rule can be used to derive precise orbital parameters.

So, you could say the first two laws are mainly descriptive, while the third is computationally useful for determining orbits.