# The Orbital Velocity Equation *For a recap on Escape Velocities check out it's dedicated section* [[Gravitational Potential Energy#Escape Velocities|here!]] ![[Pasted image 20230914215413.png]] Along with modelling the velocity it takes to escape a gravity well we can also see how fast it'll travel around the object! This is the *orbital velocity* and is defined with the equation: $v_{orb} = \sqrt{ \frac{GM}{r} }$ >[!Success]- Deriving the Equation >We're now in a stable orbit! Congratulations, now you can drop your rods of god on innocent civilians and high-profile targets alike. However, we still need to do some calculations in order to do this - which means finding the orbital acceleration! >As the orbit's now stable, we can imagine that there's a centripetal acceleration vector $\frac{v^2}{r}$ pointing towards the centre of the Earth that we so lovingly orbit. However, there's still a gravitational acceleration pointing downwards - meaning that we can come up with an equation: >$\frac{GM}{r^2} = \frac{v^2}{r}$ >Now we solve for $v$! Algebra below: >$\frac{GM}{r} = v^2$ >$v = \sqrt{ \frac{GM}{r} }$ This only works for circular orbits, which as you may know don't exist perfectly, so we'll have to come up with another equation for elliptical orbits! This gives us the *Vis-Viva Equation*, which we can use to approximate the orbital speed of an object at a certain point in its orbit. Now go forth and bore more people with your knowledge of ellipses! >[!Danger]- Proving it for Ellipses - Orbital Speed >%%NEED DIAGRAM!%% >Thanks to [[Kepler's Laws (with Derivations)#Kepler's Second Law|Kepler's Second Law]] we know that the area an orbiting body sweeps out needs to be equal the entire way. The object's speed is going to be lowest when it's at its furthest from the star and highest when closest to the star. Since this is a part of [[Angular Momentum - Rotational Motion Extended (Mechanics)|conservation of angular momentum]], we can just make the identity: >$m(a-c)v_{max} = m(a+c)v_{min}$ >Where $c$ is the distance between focus and the midpoint of the elliptical orbit. By [[Power, Work Done and Energy (Mechanics, %%marked for rewrite + NEEDS LOADS OF DIAGRAMS, ASK EXTERNAL HELP!%%)#Conservation of Energy (Principle of)|conservation of energy]] the mechanical energy of the system is going to stay the same (even though it doesn't, but more on that later). This gives us a second identity: >$\frac{1}{2}mv_{max}^2 - \frac{GMm}{a-c} = \frac{1}{2}mv_{min}^2 - \frac{GMm} >{a+c}$ >$\frac{1}{2}v_{max}^2 - \frac{1}{2}v_{min}^2 = \frac{GM}{a-c} - \frac{GM}{a+c}$ >Now note the expression we have for the conservation of angular momentum. This will allow us to get an expression for $v_{max}$ in terms of $v_{min}$: >$v_{max} = v_{min} \frac{a+c}{a-c}$ >Let's write $a-c$ as $r_p$ to signal the distance at perihelion and $a+c$ as $r_a$ to signal the distance of the object to the star at aphelion. This way, when we substitute this back into the expression we get: >$\frac{1}{2}v_{min} \left( \frac{r_{a}}{r_{p}} \right)^2 - \frac{1}{2} v_{min}^2 = \frac{GM}{r_{p}} - \frac{GM}{r_{a}}$ >Expand the $(\frac{r_a}{r_p})^2$ term and factorise out the $v_{min}$ term: >$\frac{1}{2}v_{min}^2 ( (\frac{r_{a}}{r_{p}})^2 - 1) = \frac{GM}{r_{p}} - \frac{GM}{r_{a}}$ >$\frac{1}{2}v_{min}^2 = \frac{GM}{r_{p}} - \frac{GM}{r_{a}} / \frac{r_{a}^2 - r_{p}^2}{r_{p}^2}$ >Expand the numerator on the right hand site and simplify by cancelling terms out: >$\frac{1}{2}v_{min}^2 = \frac{r_{a}GM - r_{p}GM}{r_{p}r_{a}} \frac{r_{p}^2}{r_{a}^2 - r_{p}^2}$ >$\frac{1}{2}v_{min}^2 = \frac{GM}{r_{a}} \frac{r_{p}}{r_{a} + r_{p}}$ >The denominator is a difference of two squares that will cancel out. Note that $r_p + r_a = a+c + a-c = 2a$, meaning that we can now write the equation in terms of $r_a$ and $a$ once we make the necessary substitutions: >$\frac{1}{2}v_{min}^2 = \frac{(2a - r_{a})GM}{2a r_{a}} $ >$\frac{1}{2}v_{min}^2 = \frac{GM}{r_{a}} - \frac{GM}{2a}$ >Now all we have to do is solve for $v_{min}$: >$v_{min}^2 = 2GM \left( \frac{1}{r_{a}} - \frac{1}{2a}\right)$ >Now, notice something. The 'minimum' velocity is at a specific $r$ from the barycenter of the system, at apoapsis. In this equation, everything else is a constant **save for the distance to the object**, so the velocity squared can be written as linearly related to the radius. Therefore, the equation will work for all distances from the barycenter along the elliptical orbit $r$. > [^1]: Check out our notes on [[Conics - Circles, Parabolas & More! (Maths)|conic sections]] for more info! # Energies of an Orbit, Continued ![[Pasted image 20240324204607.png]] Remember from [[Gravitational Potential Energy#Continuing with GPE|the orbital energies']] document that the potential energy of an orbiting object is: $U = - \frac{GMm}{r}$ Using $F = ma$, taking $a$ as the centripetal acceleration of the object we can get an expression for the orbital kinetic energy for a circular object: $\frac{GMm}{r^2} = m \frac{v^2}{r}$ Isolating the $v$ gives us an expression for the kinetic energy of the object: $K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$ We can therefore get the total mechanical energy of the system by adding the kinetic and potential energies up: $E_{mec} = K+U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$ Notice that this means the kinetic energy of the object is equal to the negative of its mechanical energy. In other words: $E_{mec} = -K$ Note that from [[Newton's Law of Gravitation - The Intro|the basics' doc]] that there is significant symmetry between ellipsoids and spheres. This parity can be extended into the 2-D plane for circles and ellipses, giving us the formula for ellipsoidal orbit mechanical energies: $E_{mec} = - \frac{GMm}{2a}$ Where $a$ is the [[Conics - Circles, Parabolas & More! (Maths)#Ellipses|semi-major axis]] of the ellipse. >[!Abstract]- Case Study - The Earth's Orbit >[!Danger]- Proving energy for Ellipses - Using Vis-Viva >*Before you take a look at this section, please take a look at our derivation for the velocity in an ellipse - located in the chapter down under!* > >Alright, let's look at the velocity in an elliptical orbit. More information later! > >$v^2 = GM\left( \frac{2}{r} - \frac{1}{a} \right)$ >To get this equation (known as the **vis-viva equation**), we need to use conservation of energy to set up our equations. Here, the kinetic energy of the object as it orbits is equal to its mechanical energy minus its potential energy, giving us: >$\frac{1}{2}mv^2 = \frac{GMm}{r}-\frac{GMm}{2a}$ >$v^2 = GM\left( \frac{2}{r} - \frac{1}{a} \right)$ >Note that the total mechanical energy is actually negative - hence the negative sign in the fraction! # Example Problems A quick guide to using conservation of energy in gravitational systems to approximate speeds, locations and more! (TBA) >[!Example]- # What's Next? This is the first page in a dedicated folder. As a result, here's the rest of the orbital mechanics files! 1. [[The Orbit Equation]] 2. [[Orbital Manoeuvres (Mechanics)]] 3. [[The Roche Limit (Mechanics)]] 4. [[The Two-Body Problem (Mechanics)]] 5. [[Unbound Orbits (Mechanics)]] Good luck! To get back to the main physics page, click: [[The (Incomplete) Physics Almanac]]