# Introduction During the Renaissance period, many, many people began to look up to the sky, creating some of the first catalogues of "stars". I say this as at this point, Venus was believed to be an incredibly bright star! Moving forward, one of these astronomers was Johannes Kepler, who took measurements of the orbits of planets and threw them into a table. Yes, I swear this was a person that contributed to creating our "scientific model". So, he took a very, very detailed look at his observations [^1], noting down conclusions about the motion of the planets. This is what we now define as **Kepler's Laws**, which we still use until today. [^1]: I say his, when in fact it was Kepler's contemporary Tycho Brahe that had taken most of the observations. Oh, academia! %%DRAW DIAGRAMS ESP. FOR K'S SECOND LAW%% # Kepler's First Law Otherwise known as the *Law of Orbits*, this dictates that **every planet orbiting the sun orbits in an ellipse.** That means there is no true circular orbit, no matter the conditions. We'll be diving into a conics-heavy section for the time being. For more information on Conic Sections, check out our introductory Conics' document: [[Conics - Circles, Parabolas & More! (Maths)]], though this might be a sort of 'practical review' for that, if you get my groove. Let's start with defining a random orbit. For this one, we'll take a really eccentric one, say, a comet. # Kepler's Second Law This denotes that the area that an object sweeps out per unit time must be equal, an offshoot of [[Angular Momentum - Rotational Motion Extended (Mechanics)#Conservation of Angular Momentum|conservation of angular momentum]]. Notice that means that an orbiting body will orbit slowest at [[Celestial Coordinates (Astro)#Aphelion - Attributes|aphelion]] and fastest at [[Celestial Coordinates (Astro)#Perihelion - Attributes|perihelion]] - which explains certain orbital simulations! (personal one to be created soon!) >[!Abstract]- Calculus-Based Proof >Take a highly elliptical orbit (e.g a comet) and draw it out, with semi-major axis $a$ and semi-minor axis $b$. >Given the description of Kepler's Second Law, we can construct a diagram to express this relationship and help us visualise it: > >![[Pasted image 20230822142721.png]] > >Now extend this to an arbitrarily small $\Delta A$. The area of a sector in radians is $\frac{1}{2}r^2\theta$. As the change in the area decreases, the change in the angle decreases proportionally, giving us: >$\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt}$ >This $\frac{d\theta}{dt}$ is just the angular velocity $\omega$. Let's now pan over to the momentum diagram: > >![[Pasted image 20230822143344.png]] > >Notice that the momentum due to the centrifugal force is at a 90 degree angle to the angular momentum $L$. This gives us the equation: >$L = rp_{\perp} = rmv_{\perp} = mr^2\omega$ >We can substitute the [[Angular Momentum - Rotational Motion Extended (Mechanics)#Angular Momentum for Rigid Bodies|angular momentum expression]] to get a [[Ordinary Differential Equations (Maths)#An ODE in the form $ frac {dy}{dx} = f(x)$|differential equation]] for $\frac{dA}{dt}$: >$\frac{dA}{dt} = \frac{L}{2m}$ >Therefore, for conservation of angular momentum to be true (the $L$ term must be constant!), $dA$ must be constant as well. Refreshing! If we need to find the speed of an orbiting object, it's also now possible for us to use this equation as a means to find its angular momentum. ![[Pasted image 20240304090656.png]] *Areas swept out during Aphelion and Perihelion over a time t. You'll usually find these types of orbits in comics!* In reality, there's going to be loads and loads of unknown factors (asteroids, other planets!) that affect the orbital speed of a planet - take the two saturnian moons Janus and Epimetheus, who swap orbits due to fluke in their orbits! # Kepler's Third Law >[!Success]- Kepler's Empirical Discovery >Though we've got (thanks to [[Angular Momentum - Rotational Motion Extended (Mechanics)|rotational motion]]) a mathematical proof for this, this was actually first proven empirically by none other than [Kepler]() himself. > Simply put, this is an empirical formula, so it was found without any proof whatsoever. Thank goodness for Newton and Halley! Given by: $T^2 = \frac{4\pi^2}{GM}r^3$ Where: - $T$ is the orbital period of the body - $G$ is the gravitational constant - $M$ is the mass of the orbiting body - $r$ is the radius of a (relatively) circular orbit, which can be substituted by the semi-major axis of the orbit $a$. More commonly known as: $T^2 \propto r^3$ >[!Success]- Basic Proof >Using Newton's law $F = ma$ substitute the gravitational force and the value for the centripetal acceleration: >$\frac{GMm}{r^2} = m \frac{v^2}{r}$ >Note that centripetal acceleration can also be written as: $a_{r} = \omega^2r^2$ This means the right hand side of the equation is equal to: >$\frac{GMm}{r^2} = m\omega^2r$ >Angular velocity can be written as $\frac{2\pi}{T}$ where $T$ is the period of the equation. Substitute and reduce to isolate the period $T$ to get our expression: >$\frac{GMm}{r^2} = m\left( \frac{4\pi^2}{T^2} \right)r$ >$\frac{GM}{r^3} = \frac{4\pi^2}{T^2}$ >$T^2 = \frac{4\pi^2}{GM}r^3$ >Note that $M$ refers to the mass of the central body. Since no orbit is going to be a perfect circle, we can just extend this definition of Kepler's Third Law to the [[Conics - Circles, Parabolas & More! (Maths)#Ellipses|ellipse]] - as shown: $T^2 = \frac{4\pi^2}{GM}a^3$ Instead of a radius $r$, we're going to use the semi-major axis of the orbit $a$ - the longest axis from the centre of the ellipse. I'll prove it - using **Newton's formulation of the Third Law**! >[!Abstract]- Newton's Formulation >Since Kepler's formulation of his own law treated the planet as a small mass compared to the star, Newton decided to create a more generalised system where the masses of **both** objects in a system are considered! > >Consider the following binary system: >![[binary-diagram-2.png]] >By Newton's third law, the star $m_{2}$ must remain at the opposite phase in its orbit compared to the star $m_1$ for the system to be stable. Therefore, the periods of each star's orbit, and by extension their centripetal forces, must be equal. >$F_{1} = F_{2}$ >$m_{1}\frac{4\pi^2 r_{1 }}{T^2} = m_{2}\frac{4\pi^2r_{2}}{T^2}$ >Therefore: >$r_{2} = \frac{m_{1}}{m_{2}} r_{1}$ > >Now, we can write the total distance separating the two stars $d$ as $d= r_1 + r_2$. If we substitute our expression for $r_1$ back into this equation, we get: > >$d - r_{2} = r_{1}$ >$\frac{m_{2}d - m_{1}r_{1}}{m_{2}} = r_{1}$ >$m_{2}d = r_{1}\left( 1 + \frac{m_{1}}{m_{2}} \right)$ >$\frac{m_{2}d}{m_{1} + m_{2}} = r_{1}$ >Now, we can substitute this value for the radius back into the equation for the centripetal force in terms of $T^2$, and equate it to the gravitational force: >$\frac{Gm_{1}m_{2}}{d^2} = m_{ 1}\frac{4\pi^2 r_{1}}{T^2}$ >Note how Newton's Law of Gravitation takes $d$ as its input - remember that it describes the forces of the bodies from each other, not the force due to the centre of mass. Rearrange: >$T^2 = \frac{4\pi^2}{Gm_{1}m_{2}} d^2 \frac{m_{2}d}{m_{1}+m_{2}}$ >$T^2 = \frac{4\pi^2}{G} \frac{d^3}{m_{1}+m_{2}}$ >$T^2 = \frac{4\pi^2}{G(m_{1}+m_{2})} d^3$ >Remember that this version of Kepler's Third Law is proportional to the **total distance** between the binary stars! With enough rearranging, it's possible for us to find the total mass of the binary system - and maybe even the invididual components as a result. >[!Success]- A Kepler's Law Trick! >There are certain values we can substitute to get quick ways to manipulate Kepler's Third Law! > >Let's say that we're taking the equation for the Earth's orbit. Okay, we already know the values for that. One Earth Period is one year, and one Earth orbit takes place one astronomical unit away from the sun. > >However, Kepler's Third Law means that this must be equal for the same system. As a result, we can write the equation for the Earth's orbit as: >$T^2 = a^3$ >It's important to note that for the equation to be valid, both the period and the semi-major axis have to be 1. As a result, to use this equation the period must be in terms of *years* and the semi-major axis in terms of *astronomical units - AU.* > >If a planet's orbiting around another star, we can take the mass ratios of the sun and the other star to correct for our equation: >$T^2 = \frac{M_{\odot}}{M_{S}} a^3$ >Where $M_{\odot}$ is the mass of the sun *in solar masses* and $M_S$ is the mass of the reference star *in solar masses.* # Example Questions :( >[!Example]- Module 13-6 Problem 45 > # Continuing down Orbital Mechanics - What's Next? Congratulations, you're now a master at Kepler's laws! This could belong in an orbital mechanics' textbook, but I felt it belonged better in gravitation, allowing for me to inch us towards OrbMech proper. Now, go forth and swamp your humanities' major friends through the power of friendship and mathematics! For the next logical step if you're here for physics, head to the fluids chapter jumping point here: [[Fluid Principles (Mechanics)]]. Alternatively, if you don't understand the content provided make sure to take the time for a quick review at [[Newton's Laws and SUVAT#Pressure|pressure]] and [[Newton's Laws and SUVAT#Density|density]]. For those with the mathematical background and are looking for a better rundown on orbital mechanics, go here: [[Energy along an Orbit]] Navigate to other topics by going back to the main menu: [[The (Incomplete) Physics Almanac]]