intuition: https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Celestial_Mechanics_(Tatum)/02%3A_Conic_Sections/2.02%3A_The_Ellipse#:~:text=a(1−e2),11%2Becosθ.
# '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
For circles and conics, it's usually just easier to take displacements in terms of angles - there are so many angle rules that simplify the process that it's just easier keeping displacement to angles.
Simply put, the *true anomaly* is the angle between the transverse axis (semi-major axis) and the object. It's just a way to see how far away it is from the perigee/apogee.
![[true anom.png]]
## 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]]! Yes, this was absolutely terrible. Don't toot your horns just yet - it'll get worse soon enough, somehow.
Example question down below shows you how you can use the anomaly to find the distance! This appears to be the limit the IOAA takes this topic to, so if you're here for that your journey ends here.
## EXTENSION: The Eccentric Anomaly's Formula
TBA because I don't know the half angle formulae :(
## The Mean Anomaly
Mean anomaly - the average angular displacement per time as the body moves through its orbit!
![[Pasted image 20240330104625.png]]
So as mean anomaly just denotes the mean displacement of an object as seen on a circle of radius $a$:
$\frac{t}{T} * 2\pi = M$
As $M$ doesn't change, it's possible for us to do this!
Also, the trig derivation. The equation above basically states that the average distance travelled during the orbit per time is the mean anomaly. Remember K's third law:
$T^2 = \frac{4\pi^2}{\mu}a^3$
so $T = \frac{2\pi}{\sqrt{ \mu }}\sqrt{ a^3 }$
to find the mean distance per time travelled for the orbit, we need to divide the period by the perimeter of the ellipse. we use angular coordinates for the mean anomaly, meaning:
$n = \frac{2\pi}{P}$
so $n = \sqrt{ \frac{\mu}{a^3} }$
Now that's done, we have another way to express the mean anomaly. proof later. action now.
$M(t) = E(t) - e\sin E$
This links us to the eccentric anomaly, which means we can find the true anomaly given the mean anomaly - really, really useful!
$\cos \nu = \frac{\cos E-e}{(1-e\cos E)}$
## Kepler's Equation
# Example Questions
Unify the concepts used for all three anomalies here!
>[!Example]- Example Problem - Bombom's POTD Day 40
>
# Glossary
Tentative glossary.
- *True Anomaly:* The angle between the focus and the position of the particle. Given the variable '$\nu