# Mechanics

2016-07-24

Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it. – Isaac Newton, discoverer of the laws of motion and gravitation

Mechanics is the study of movement. In this chapter, we will describe how physicists study the movement of physical objects using mathematical laws. With this knowledge, we will then see how this is relevant to rocket science.

# Referential

## Origin

The first thing we want to do is to find a way to describe where objects are. For this, we will elect an object (an actual physical object or just an arbitrary virtual point in space) to be fixed, and consider all movements relative to this origin.

## Dimensions

If all the objects are on a same line (for instance a race track), then we can describe the position of each object with a single measure: the distance from the origin. Since a single measure is sufficient, a line is said to have one dimension.

If all the objects are on a same surface (for instance, on a checkboard), then we need two numbers to fully describe the position of any of these objects. For example, we may arrange items on a grid a tell their positions as the row and column numbers; we can locate a location on the surface of the Earth using longitude and latitude. Since two measures are needed, a surface is said to have two dimensions.

There are many ways to choose how to describe the position of an object. For instance, rather than using the row and column number, we could use the distance from the center and the angle formed with a fixed line. When we say “there are two dimensions”, the “two dimensions” do not refer to any particular measures, but to the necessity of two different ones. There are many choices of two measures to describe two dimensions.

Finally, we can consider an arbitrary location in space, which needs three dimensions. For example, if we add altitude to longitude and latitude, we can easily locate any object.

A referential is the object we use as a landmark (origin) to keep track of interesting points. A system of coordinates if the kind of data you use to store the position of these points relatively to the origin.

The most common set of measures used in two dimensions are the Cartesian coordinate and the polar coordinate systems. The Cartesian coordinate system extend to three dimensions naturally; there are two ways to extend the polar coordinates system to a third dimension: the cylindrical coordinate and the spherical coordinate systems.

## Cartesian coordinates

Cartesian coordinates are the most common. We basically choose two (or three, in space) directions, and the measures describes how much you have to go each way to get from the origin to the point.

The directions used in Cartesian coordinates (the axes) are usually perpendicular and conventionally named $x$ and $y$, (and $z$)).

### Derivatives

A nice property of the Cartesian coordinate system is that, if we are interested in the variation of some point $P$ with Cartesian coordinates $\left(x,y\right)$, we can simply consider the vector (x, y).

While a point denotes an offset from the origin, a vector denotes an offset from an unspecified point.

## Polar coordinates

Polar coordinates work in two dimensions. One value is simply the distance to the origin while a second is the angle of $P$ with a fixed porientation.

### Derivative

Consider a point $P$ given in polar coordinates $\left(\angle \mathit{\theta },r\right)$. We can use the corresponding Cartesian coordinates: $P=\left(r\mathrm{cos}\left(\angle \mathit{\theta }\right),r\mathrm{sin}\left(\angle \mathit{\theta }\right)\right)$. Then we can derive on each Cartesian coordinate using the derivation rules for function composition.

If we define $\stackrel{^}{r}==\left(\mathrm{cos}\mathit{\theta },\mathrm{sin}\mathit{\theta }\right)$ and $\stackrel{^}{\mathit{\theta }}=\left(-\mathrm{sin}\mathit{\theta },\mathrm{cos}\mathit{\theta }\right)\right)$, then:

$\left\{\begin{array}{cc}\hfill \frac{d}{dt}\stackrel{^}{r}& =\left(-\stackrel{\bullet }{\mathit{\theta }}\mathrm{sin}\mathit{\theta },\stackrel{\bullet }{\mathit{\theta }}\mathrm{cos}\mathit{\theta }\right)=\stackrel{\bullet }{\mathit{\theta }}\stackrel{^}{\mathit{\theta }}\hfill \\ \hfill \frac{d}{dt}\stackrel{^}{\mathit{\theta }}& =\left(-\stackrel{\bullet }{\mathit{\theta }}\mathrm{cos}\mathit{\theta },-\stackrel{\bullet }{\mathit{\theta }}\mathrm{sin}\mathit{\theta }\right)=-\stackrel{\bullet }{\mathit{\theta }}\stackrel{^}{r}\hfill \end{array}$

Now, $P$ is at an offset $\stackrel{\to }{r}$ to the origin, where:

$\stackrel{\to }{r}=\left(r\mathrm{cos}\left(\angle \mathit{\theta }\right),r\mathrm{sin}\left(\angle \mathit{\theta }\right)\right)=r\stackrel{^}{r}$

thus:

$\stackrel{\bullet }{\stackrel{\to }{r}}=\frac{d}{dt}\left(r\stackrel{^}{r}\right)=\stackrel{^}{r}\frac{d}{dt}r+r\frac{d}{dt}\stackrel{^}{r}=\stackrel{\bullet }{r}\stackrel{^}{r}+r\stackrel{\bullet }{\mathit{\theta }}\stackrel{^}{\mathit{\theta }}$

By deriving again, we get the acceleration:

$\stackrel{\bullet \bullet }{\stackrel{\to }{r}}=\frac{d}{dt}\left(\stackrel{\bullet }{r}\stackrel{^}{r}+r\stackrel{\bullet }{\mathit{\theta }}\stackrel{^}{\mathit{\theta }}\right)=\left(\stackrel{\bullet \bullet }{r}-r{\stackrel{\bullet }{\mathit{\theta }}}^{2}\right)\stackrel{^}{r}+\left(2\stackrel{\bullet }{r}\stackrel{\bullet }{\mathit{\theta }}+r\stackrel{\bullet \bullet }{\mathit{\theta }}\right)\stackrel{^}{\mathit{\theta }}$

## Spherical coordinates

Spherical coordinates are simply the generalization of polar coordinates to three dimensions. To the usual coordinates $r$ and $\mathit{\theta }$, we add a third one, $\mathit{\phi }$, which is the angle with the polar plane:

# Newtonian mechanics

## Center of mass

For the sake of simplicity, we will pretend that objects are simple points (e.g. $P$) associated to their mass (e.g. $m$); such points (e.g. $\left(P,m\right)$) are called mass-points. For a given object, the point we will use is called the center of mass (CoM). It is easy to find for regular and uniform objects (e.g. a sphere).

We will usually refer to the position $P$ by using the vector $\stackrel{\to }{r}=\stackrel{\to }{\mathit{OP}}$. The velocity is simply the derivative of the position: $\stackrel{\to }{v}=\stackrel{\bullet }{\stackrel{\to }{r}}$; the acceleration is the derivative of the velocity: $a=\stackrel{\bullet }{\stackrel{\to }{v}}=\stackrel{\bullet \bullet }{\stackrel{\to }{r}}$.

## Newton’s second law

Consider a mass-point $\left(P,m\right)$ subjected to forces ${\stackrel{\to }{F}}_{1},{\stackrel{\to }{F}}_{2},\dots$. We need not consider every force independentely and can sum them as the resultant force $\stackrel{\to }{F}=\sum {\stackrel{\to }{F}}_{i}$.

Then, according to Newton’s second law:

$\stackrel{\to }{a}=\frac{1}{m}\stackrel{\to }{F}$

This equation means several things. First, it means that an object on which is exerted a non-zero resultant force will be accelerated (acceleration encompasses the slowing down of an object). Second, the more mass it has, the less it will accelerate; since this is perceived as a resistance to movement, we call this inertia.

When $\stackrel{\to }{F}=0$, the acceleration is null as well. Thus, the speed is constant; this is Newton’s first law (chapter quote).

## Gravitation

The gravitational force is one of the four fundamental forces that physicist have identified, along with the electromagnetic, weak and strong forces.

Gravitation says that any two objects with a mass attract each other, at any distance. Since it is extremely feeble, we do not observe it within common objects. It becomes perceptible when huge masses are grouped together; for instance, we observe an apple falling because of the vast mass of the Earth effecting the apple.

The amount of force exerted by an object of mass $M$ over an object of mass $m$ is:

$F=G\frac{Mm}{{r}^{2}}$

where $r$ is the distance between the two objects.

The object of mass $m$ attracts the object of $M$ with the exact same mount of force. The reason the Earth does not seem to move towards a falling apple is because of inertia (see Newton’s second law).

# Shell theorem

## Problem

We are interested in knowing the gravitation exerted by a celestial body $C$ (i.e. the gravity). We will assume the celestial body is a ball of radius $R$ and uniform density $\mathit{\mu }$ and consider a mass-point $\left(m,P\right)$ at distance $r$ from $C$:

## Elementary force

The usual way to solve a large problem is to split it in smaller problems and solve those instead. Here, to find the total force exerted, we will split the spherical body in many mass points $\left(Q,dm\right)$.

The force exerted by $Q$ over $P$ is oriented along $\stackrel{\to }{\mathit{QP}}$ and has magnitude:

$F=\stackrel{\to }{dg}m=G\frac{mdm}{{\mathit{PQ}}^{2}}$

Another way to write this is:

$\stackrel{\to }{dg}=G\frac{\mathit{\mu }dV}{{s}^{2}}$

where $s$ is just $\mathit{PQ}$

## Symmetry

Now, to simplify things, we can notice the axial symmetry of the whole situation around $\left(\mathit{CP}\right)$. This means that the resultant force $\stackrel{\to }{g}$ can only be along the axis $\left(\mathit{CP}\right)$: no other could be justified without breaking the symmetry.

Thus, we can only choose to only consider the effect of $Q$ along $\left(\mathit{CP}\right)$; the others effect will cancel out. The effect of gravitation from $Q$ exerted on $P$ along $\left(\mathit{CP}\right)$ is:

$\stackrel{\to }{dg}\bullet \frac{\stackrel{\to }{\mathit{PC}}}{\mathit{PC}}=G\frac{\mathit{\mu }dV}{{s}^{2}}×\mathrm{cos}\mathit{\psi }$

where $\mathit{\psi }$ is the angle $\stackrel{^}{\mathit{CPQ}}$.

## Integrating

Now, it is only a matter of integrating. We first extend the elementary volume $dV$ in spherical coordinates as $\left(\mathit{\rho }\mathrm{sin}\mathit{\psi }d\mathit{\theta }\right)\left(\mathit{\rho }d\mathit{\psi }\right)$. Then we just need to do a substitution and to cancel out terms.

g = _S g
= G V
= G _ =_-^^+ _ =0^ _ =0^2 ( ) ( ) ( )
= 2 G _ =_-^^+ _ =0^
= 2 G _ =0^ 2
= 4 G _u=R^0 u -u u
= 4 G 1 _0^R u^2 u
= G _= V 1

## Result

In the end, $g=G\frac{M}{{s}^{2}}$. It means that, the resultant gravitation force exerted by a planet is the same as the mass-point in the center with the same mass, even when close to it.

This will make all considerations involving the gravitation of a planet relatively easy, as long as we assume it is spherical and uniform.

## Details on the substitution

To get from (*) to (*), we substitute $u=\sqrt{{R}^{2}-{r}^{2}{\mathrm{sin}}^{2}\mathit{\psi }}$ for $\mathit{\psi }$. This means that $\mathit{\psi }=\mathrm{arcsin}\frac{\sqrt{{r}^{2}-{u}^{2}}}{r}$ and subsequently:

$d\mathit{\psi }=\frac{1}{\sqrt{1-\frac{{R}^{2}-{u}^{2}}{{r}^{2}}}}×\frac{-2udu}{2r\sqrt{{R}^{2}-{u}^{2}}}=-\frac{u}{\sqrt{{r}^{2}-{R}^{2}+{u}^{2}}\sqrt{{R}^{2}-{u}^{2}}}du$

We then use the relations $\mathrm{sin}\left(\mathrm{arcsin}x\right)=x$ and $\mathrm{cos}\left(\mathrm{arcsin}x\right)=\sqrt{1-{x}^{2}}$ for $0\le x\le \frac{\mathit{\pi }}{2}$.

# Sphere of influence

## Patched conics

In reality, an object travelling through space is influenced by all moons, planets and stars of the universe. Even when restricting to the closest moon, closest planet and closest star, the combined influence of several bodies is hard to take into account simultaneously.

A simple approximation is to only consider the body with the most influence. The region where a celestial body has the most influence is called the sphere of influence. When leaving the sphere of influence of one body, we transition to that of another. The trajectory is then made of two parts (two conics), hence the name patched conics approximation.

## Influences

We will consider two mass-points $\left({P}_{1},{M}_{1}\right)$ and $\left({P}_{2},{M}_{2}\right)$ and look at their influences on a mass-point $\left(P,m\right)$ between them.

The intensity of forces ${F}_{1}$ and ${F}_{2}$ respectively exerted by ${P}_{1}$ and ${P}_{2}$ are:

For short, we define $r={P}_{1}P$ and $R={P}_{1}{P}_{2}$. Thus, ${P}_{2}P={P}_{1}{P}_{2}-{P}_{1}P=R-r$. We also define ${\mathit{\mu }}_{1}=G{M}_{1}$ and ${\mathit{\mu }}_{2}=G{M}_{2}$ (the gravitational parameters). The accelerations due to each are:

The acceleration exerted by ${P}_{1}$ over ${P}_{2}$, and by ${P}_{2}$ over ${P}_{1}$ are respectively:

## Perturbations

Now, if we look at the acceleration of $P$ relatively to ${P}_{1}$, it is perturbed by the fact that ${P}_{2}$ does not exert the same acceleration over $P$ and over ${P}_{1}$. The absolute perturbation is ${g}_{2}-{g}_{2,1}$ (the difference in acceleration from ${P}_{2}$).

In practice, this value only makes sense when compared to the main acceleration ${g}_{1}$. Here, we will assume that $r<; it makes sense when ${P}_{1}$ is an object significantly smaller than ${P}_{2}$. We define the relative perturbation ${Q}_{1}$ as:

${Q}_{1}=\frac{{g}_{2}-{g}_{2,1}}{{g}_{1}}=\frac{\frac{{\mathit{\mu }}_{2}}{\left(R-r{\right)}^{2}}-\frac{{\mathit{\mu }}_{2}}{{R}^{2}}}{\frac{{\mathit{\mu }}_{1}}{{r}^{2}}}=\frac{{r}^{2}}{{R}^{2}}\frac{{\mathit{\mu }}_{2}}{{\mathit{\mu }}_{1}}\left(\frac{1}{{\left(1-\frac{r}{R}\right)}^{2}}-1\right)$

Now, it can be shown that, when $x$ is small, $\frac{1}{\left(1-x{\right)}^{2}}=1+2x+3{x}^{2}+\dots \approx 1+2x$. With $x=\frac{r}{R}$ we get that:

${Q}_{1}=\frac{{r}^{2}}{{R}^{2}}\frac{{\mathit{\mu }}_{2}}{{\mathit{\mu }}_{1}}×2\frac{r}{R}=2\frac{{r}^{3}}{{R}^{3}}\frac{{\mathit{\mu }}_{2}}{{\mathit{\mu }}_{1}}$

Similarly, the relative perturbation due to ${P}_{1}$ when looking at the acceleration of $P$ relatively to ${P}_{2}$ is:

${Q}_{2}=\frac{{g}_{1}-{g}_{1,2}}{{g}_{2}}=\frac{\frac{{\mathit{\mu }}_{1}}{{r}^{2}}\overline{)-\frac{{\mathit{\mu }}_{1}}{{R}^{2}}}}{\frac{{\mathit{\mu }}_{2}}{\left(R\overline{)-r}{\right)}^{2}}}=\frac{{\mathit{\mu }}_{1}}{{\mathit{\mu }}_{2}}\frac{{R}^{2}}{{r}^{2}}$

Again, we use the fact that $x< to simplify the expression.

Now, we are interested in the point $P$ between ${P}_{1}$ and ${P}_{2}$ where both relative perturbations are of the same magnitude. We thus solve:

Q_1 = Q_2 2 r^3 _2 = _1 R^2
r = 1 R (_1)^2 = 2^- 1 R (M_1)^2

The ${2}^{-\frac{1}{5}}$ factor seems to be forgotten often. Since its value is about 0.87, it is still the right order of magnitude.

# Thrust

## Propulsion

A car can push along the road thanks to friction; a plane can generate lift from air and speed; a balloon can use buoyancy in the atmosphere. However, a rocket needs to be able to operate in the vacuum of outer space, which implies no physical object to push against.

Articulating an object into space will not move its center of mass. It means that a rocket can only gain speed by interacting with the exterior. Thus, the only way a rocket can accelerate is by throwing parts of its mass out. This is exactly what their exhaust do: it ejects lots of mass (propellant) down, so that the rocket will lift up.

Technically, outer space is not empty: light from the Sun is theoretically enough for the low-thrust propulsion of solar sails.

## Conserved momentum

We consider a rocket with center $R$ of mass $m$. From $t$ to $t+dt$, it ejects $dm$ of its propellant ($P$) with relative (exhaust) velocity ${v}_{e}$.

Let us notate $G$ the center of mass of the system $\left\{R,P\right\}$. Since the system did not interact with its environment, the speed of $G$ remains unchanged by the ejection. However, since $P$ is moving downwards, it means the rocket have gained speed.

## Exerted force

We are interested in the changes of speed $d{v}_{R}$ and $d{v}_{P}$ from $t$ to $t+\mathit{dt}$. If we do not take into account external forces, then $d{v}_{g}=0$. Thus:

0 = v_G
= v_R m + v_P m
= v_R (m + m) + v_e m  because  v_P = v_R - v_e

Dividing by $\mathit{dt}$ and with $m+dm\approx m$, it follows that:

$m\frac{d{v}_{r}}{dt}=-{v}_{e}\frac{dm}{dt}$

We finally get the expression of the force of thrust:

${F}_{t}=m{a}_{t}=-{v}_{e}\stackrel{\bullet }{m}\left(1\right)$

## Specific impulse

In KSP, the engines are defined by their maximum thrust ${F}_{t}$ and their ${I}_{\mathrm{sp}}$ (“atmosphereCurve” in configuration files). The specific impulse is just the force exerted per unit (in weight) of fuel used, i.e. ${I}_{\mathrm{sp}}=\frac{{F}_{t}}{\stackrel{\bullet }{m}g}$. Thus:

## Tsiolkovsky rocket equation

We can now integrate equation (1) to get the velocity change of a rocket due to thrust, over a long period of time:

v_t = _0^t t
= _0^t - v_e m t
= _0^t
= v_e m(0)

Remark that we did not make any particular asumption about $m$: the ejection rate $\stackrel{\bullet }{m}$ might grow or shrink in any (continuous) way. This makes it easy to plan complex trajectories with several maneuvers: rockets with different thrusts and staging arrangements will follow similar paths, as long as $\mathit{\Delta }v$ are the same.

Conversely, we can compute the amount of propellant $\mathit{\Delta }m$ to eject to bring a mass $m$ to a given speed:

$\mathit{\Delta }m=m\left(1-{e}^{-\frac{\mathit{\Delta }{v}_{t}}{{v}_{e}}}\right)$