October 31, 2017

Transfer Orbit

Transfer orbits are inherently cool

The idea of using gravity as a slingshot to give a boost to a rocket is (probably) enough to get anyone excited, even if you hate physics. However, as is often the case with undergraduate-level courses, there is hardly enough time to skim the contents of orbital mechanics, let alone delve into the details of how something specific with an application in the real world actually works. In this post I’m going to try to do just that, deriving how this would ideally work from (reasonable) start to finish. Topics covered here include centripetal acceleration, the orbit equation, elliptical orbits, and a bunch of geometry throughout. Hopefully it’ll be rewarding and you and I will get to glean a bit of the bigger picture of what is going on here, in a descriptive and numerically-motivated, qualitative way. Let’s get started.


Result:

Before we begin, why don’t you try running the simulation above? The objective of the game is to input the correct addition to the velocity of the orbit around Earth such that the rocket makes it to the moon in an elliptical orbit. The rocket begins in a circular orbit around Earth with radius $2\times10^{7} \ \text{m}$. You can guess and test for now, but don’t worry, we’ll go through the solution in detail in this post. In order to obtain the correct answer you will probably need to look up some constants (like the mass of the Earth), or just continue reading and I’ll provide them all.

Meet the ellipse. This guy is pervasive throughout the universe: just as a square is a subclass of a more general shape - the rectangle - so too is the circle a special kind of ellipse. We can define some necessary quantities which aid in the mathematical description of these kinds of orbits straight away:

Firstly, here is the equation of an ellipse about the origin in plain old Cartesian coordinates, and then in polar coordinates:

$$\begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\ r = \frac{R}{1+e\text{cos}\theta} \end{align*}$$

Geometrically, $R$ and $e$ are just constants which help define the ellipse, the latter being especially important is coined ‘eccentricity’. The value of this parameter determines the overall shape of the orbit, and an eccentricity between 0 and 1 means that the orbit is elliptical. $R$ is also a constant. The definitions of both of these are given below, after a brief explanation of where they come from.

The second equation can be obtained geometrically, but in order to get useful physical properties out of the geometrical definition we will have to equate these values to constants in an equation of motion known as the orbit equation. This equation governs any kind of orbital motion due to a $\frac{k}{r^2}$ force and is given as a differential equation in terms of $u=\frac{1}{r}$.

$$\begin{align*} \frac{du}{d\theta} + u = \frac{km}{L^2} \implies u = Acos\theta + B, \ B = \frac{km}{L^2} \end{align*}$$

Now we can recover our geometrical formula for an ellipse by relabelling

$$\begin{align*} u = \frac{1}{r} = \frac{1}{R}\left(1+ecos\theta\right) \end{align*}$$

so comparing the two we note that the substitution used is just $B= \frac{km}{L^2} = \frac{1}{R}$, which will be very important later on for finding the maximum velocity of the orbit. The definition of eccentricity along with the newly justified definition of $R$ are given below.

$$\begin{align*} e^2 &= 1 - \frac{b^2}{a^2} \\ R &= \frac{b^2}{a} = \frac{L^2}{km} \\ \end{align*}$$

Okay, now on to the physics.

Initially, the rocket is travelling at a certin velocity and radius which allow it to maintain a circular orbit. The condition for this is that $v^2 = \frac{GM}{r}$ because the circular centripetal force from the planet is equal to the force of gravity. Since we are looking at what magnitude of a boost will cause the orbit to become appropriately elliptical, we should find a point in the circular orbit that matches the magnitude of $r$ in an elliptical orbit with the Earth at one of the foci and the moon at the other. This occurs when the rocket is furthest away from the moon i.e. if the moon and Earth lie on the x-axis, the rocket is travelling perpendicular to this axis at the time of the boost. This location, where the rocket is closest to the Earth in its elliptical orbit, is called the perihelion. Looking at the geometry of an ellipse (figure 1), the perihelion occurs when $r = a - ae = a(1-e)$.

Now that we have determined what the orbital radius is in terms of the eccentricity, we can solve for the velocity at this point if the rocket were in the correct elliptical orbit using our definition of $R$ above. We know that $kmR = L^2$ and substituting the definition of angular momentum, we can solve for $v$, the speed, in terms of $R, \ k, \ m$ and $r$. We also know this last value, $r$, in terms of $a$ and $e$ when the rocket is at the perhelion, meaning we can solve for the rocket’s velocity at the perihelion.

$$\begin{align*} \implies m^2r^2v^2 &= kmR \\ v^2 &= \frac{kR}{mr^2} \\ r_{\text{min}} &= a(1-e) \\ \implies v^2|_{r_{\text{min}}} &= \frac{k}{ma}\frac{1+e}{1-e} \end{align*}$$

In other words, the rocket’s velocity needs to change from $v^2 = \frac{Gm}{r}$ to $v^2 = \frac{k}{ma}\frac{1+e}{1-e}$ at exactly the right moment, the point furthest away from the moon. We know $k$ is $GMm$ when we’re dealing with gravity, so

$$\begin{align*} v^2|_{r_{\text{min}}} = \frac{G M}{a}\frac{1+e}{1-e} \end{align*}$$

Now all we need is the value of eccentricity because $a$ is just $\frac{r_{\text{min}}}{1-e}$. Woo!

For that, we can appeal to the geometrical definition (just jump back and forth between geometry and physics as it suits you) and looking at the diagram of the ellipse, we have

$$\begin{align*} 2r_{\text{min}} + d = 2a = \frac{2r_{\text{min}}}{1-e}\\ \text{solving for e}\\ \implies e = 1 - \frac{2r_{\text{min}}}{2r_{\text{min}} + d}\\ &=\frac{d}{2r_{\text{min}} + d} \end{align*}$$

where I’ve labelled the distance to the moon from Earth as $d$.

Finally, all that’s left is to plug in the values which I just Googled. Here they are:

  • distance to the moon $= 384,400 \ \text{km}$
  • mass of the Earth $= 5.972\times 10^24 \ \text{kg}$
  • $G = 6.67 \times 10^{-11} \ \text{m}^3\text{kg}^{-1}\text{s}^{-2}$

where I’ve ignored the mass of the moon for this simulation. Also, I picked the initial radius of the orbit $r_{\text{min}}$ to be $2\times 10^7 \ \text{m}$. Now, solving for $e, \ a, \ v^2|_{r_{\text{min}}}$ and the initial velocity $v^2 = \frac{GM}{r}$, we get

$$\begin{align*} e \approx 0.906 \\ a \approx 2.13 \times 10^8 \ \text{m} \\ v_i \approx 4462.8 \ \text{m/s}\\ v|_{r_{\text{min}}} \approx 6157.8 \ \text{m/s}\\ \implies \text{boost} = \Delta v = v|_{r_{\text{min}}} - v_i = 1695 \ \text{m/s} \end{align*}$$

© Angus Lowe 2018

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