https://faculty.fiu.edu/~vanhamme/ast3213/orbits.pdf read through this! look at the astroOly discord for more discussion! perfeido! # Intro to the Standard Gravitational Parameter relative acceleration is given by: $a = -\mu\left( \frac{\overrightarrow{r}}{r^3} \right)$ where $\mu$ is the standard gravitatonal parameter, a generalisation of the force between two particles: $\mu = G(M+m)$ # specific angular momentum given by the cross product between the radius and the velocity. 'specific' in our case means that we're taking the angular momentum of a single point (particle) in a larger body, as the entire bloody thing's moving at basically the same angular momentum! $h = r \times v$ however this value is going to equal 0 because $v$ is antiparallel with $r$. that's why we take the corrected version of the expression, which uses [[Vector Operations (Maths)#Cross Product Identities|cross product identities]]: $a \times h = -\frac{\mu}{r^3}((r \odot v)r - r^2v)$ you can substitute $r \times v$ with $r^2$ as: $\frac{d}{dt} (r \cdot r) = \frac{d}{dt} (r^2)$ by product rule... $2r \cdot v= 2rv$ $r \cdot v = rv$ and after a bit of a fever dream later you get: understood. make sure you stick to you're fundamentals! $a \times h = \mu \frac{d}{dt}\left( \frac{\overrightarrow{r}}{r} \right)$ $$ $$ # kepler's laws - revised for this! second law (you might need specific angular momenta): $\frac{h}{2\mu} = \frac{dA}{dt}$ add this quick QuestionCheat: >[!Example]- Example Question (USAAAO 2020 Q18) >*beautify the answer later* > >![[usaaao q.png]] >ans: >$\left( {\frac{11.19}{365}}^2 \right) = \frac{1}{0.122}a^3$ >$\frac{11.19}{365}^2 \times 0.122 = a^3$ >$a \approx 0.048582\dots \approx 0.05 $ >Hence answer is B # 'Orbital Anomalies' and more! *Taken from the 4th year aerospace notes by Anil Rao (why do i do this to myself)* When we launch spacecraft to another planet, we typically want to know where this object will be when it reaches it. It's not a great feeling knowing that your probe's heading off to some galaxy a cool 50 million light years away, and it's even worse knowing that you missed the planet's gravity well by a margin of millions of kilometers. Naturally, we need to find a way to estimate the distance to an object, and that means we'll have to find the actual position of the object at any time. And yes, we'll have to learn what an 'anomaly' is to do so. Let's hop in! ## Beginner's Luck - The True Anomaly --> add true anomaly basics here - this isn't too difficult to grasp. ## The Eccentric Anomaly *You need to be comfortable with ellipses for this. Brush up on them* [[Conics - Circles, Parabolas & More! (Maths)#Ellipses|here.]] Unfortunately, actually finding these anomalies isn't the most intuitive thing, meaning we'll have to use our definitions to our advantage. Let's start by drawing an elliptical orbit. ![[Pasted image 20240324174251.png]] You'll need your conic section variables for this next part. We're going to draw a circle with a radius $b$ (semi-minor axis) and a circle with radius $a$ (semi-minor axis). Intersection points labelled. ![[Pasted image 20240324191948.png]] Our goal is to find $R$ - the distance to one focus. This'll allow us to relate the anomaly to the time, which will let us find the position of the object directly. To do this, we can use Pythagoras' theorem, which we'll write in terms of side lengths as: $R^2 = \overrightarrow{{OD}^2} + \overrightarrow{{CD}^2}$ >[!Success]- Resolving CD >Let's overlay $CD$ onto a triangle - note that we can just move it forward to make a triangle with the radius of the semi-minor axis. We therefore get an expression: >$CD = b\sin E$ >[!Success]- Resolving OD >Let the distance from O to the midpoint of the circles be $c$, as per [[Conics - Circles, Parabolas & More! (Maths)#Ellipses|ellipse definitions.]] This gives us: >${OD}^2 = a\cos E - OM$ >Where $M$ is the midpoint. Ellipse definitions also tell us that the eccentricity is equal to $\frac{c}{a}$, so: >$c = ea$ >$OD = a\cos E - ae$ Plug the derivations in to get: $R^2 = (a\cos E-ae)^2 + b^2\sin^2 E$ In addition, we can rewrite $b^2$ as: $b^2 = a^2(1-e^2)$ Which gives us: $R^2 = a^2(\cos^2E - 2e\cos E + e^2) + a^2(1-e^2)\sin^2E$ $R^2 = a^2(\cos^2 E - 2e\cos E + e^2 + (1-e^2)(1-\cos^2E))$ $R^2 = a^2(1 + e^2\cos^2 E - 2e\cos E + e^2 - e^2)$ $R^2 = a^2(1- 2e\cos E + e^2\cos^2E)$ $R^2 = a^2(1-e\cos E)^2$ $R = a(1-e\cos E)$ Before we forge on, let's define some lengths. Let $x$ be the distance from point $D$ to the midpoint $M$, or $\overrightarrow{DM}$ for short. With our value for $R$, we now have two different ways to write this equation. Let's equate them and see what happens! ![[Pasted image 20240324200137.png]] $x = a\cos E = R\cos \nu + ae$ Solve for $\cos \nu$ to get: $\cos \nu = \frac{{a\cos E} - ae}{R}$ Since $R$ is $a(1-e \cos E)$ we can rewrite the expression to get: $\cos \nu = \frac{{\cos E} - e}{(1-e\cos E)}$ Which is how we find the [[true anomaly]]! Example question down below shows you how you can use the anomaly to find the distance! >[!Example]- Example Problem - Bombom's POTD Day 40 > ## EXTENSION: The Eccentric Anomaly's Formula ## The Mean Anomaly ## Kepler's Equation