# Launch to orbit

2016-07-24

Can it be that you have come from outer space?

— As a matter of fact, I have! – Yuri Gagarin, first human being in space, to people near his landing site

# Vertical ascent

## Thrust

From the Tsiolkovsky equation (*), we know that:

$$\mathit{\Delta}{v}_{t}=-{v}_{e}\mathrm{ln}(1-\frac{\stackrel{\bullet}{m}}{m(0)}t)$$

If we now assume that $\stackrel{\bullet}{m}$ (and thus the thrust) is constant:

x_t
= - v_e 1
(
(1 - m t)
(1 - m t)
+ m t
)

= v_e
(m(0) - t)
(1 - m t)
+ v_e t

This gives us the displacement done by the thrust of the engines.

## Gravity

Let us notate $\mathit{\mu}=GM$ the gravitational parameter of the body our rocket is landed at, $R$ the radius of that body, and $z$ the current altitude of the rocket. As long as $z<<R$, the acceleration due to gravity will be:

$${a}_{g}=\frac{\mathit{\mu}}{(R+z{)}^{2}}\approx \frac{\mathit{\mu}}{{R}^{2}}$$

Then, integrating twice:

$$\mathit{\Delta}{x}_{g}=\frac{1}{2}{t}^{2}\times {a}_{g}$$

This is the displacement due to gravity.

## General expression

We now compare the graphs for the different approaches (ignoring gravity, constant gravity, or numerical approximation). To get actual values, we simulate a basic rocket made of the following parts:

- Command Pod Mk1
- FL-T800 Fuel Tank
- LV-T30 Liquid Fuel Engine

# Atmospheric drag

## Definition

An object moving through a fluid (gaz or liquid) pushes matter around; by the principle of action-reaction, the object will be subjected to an opposing force. We are only interested in two effects of this: **drag**, which is the **prograde** component (along the velocity vector), and **lift**, a **normal** component.

For now, we will assume an axial symmetry around the velocity vector. In other words, we make it so as not to generate lift. It has been measured that the force of drag was of the form:

$${F}_{d}=\frac{1}{2}\mathit{\rho}\times {C}_{d}A\times {v}^{2}$$

where $\mathit{\rho}$ is the density of air and ${C}_{d}$ and $A$ are aerodynamics factors dependent on the shape of the ship.

## Terminal velocity

Let us consider a falling object subjected only to gravitation and atmospheric drag. Gravitation is a constant force oriented downwards; here, drag is a force oriented upwards and growing with speed. There will hthus be a point where gravitation and drag balances and speed becomes constant. This speed is named **terminal velocity**.

When terminal velocity is reached, $\stackrel{\bullet}{z}$ is constant so $\stackrel{\bullet \bullet}{z}$ is null and:

z = 0
a_g = a_d

F_g = F_d

1 C_d A v_term^2 = m g

v_term =

In Kerbal Space Program pre-1.0, A used to be computed as $\frac{1}{125}{m}^{2}/\mathit{kg}\times m$. This means that the terminal velocity only depended on the air density $\mathit{\rho}$ and the coefficient of drag ${C}_{d}$ (usually about 0.2).

$\frac{m}{125\mathit{kg}/{m}^{2}}$. This means that the terminal

This is not realistic since, according to the square-cube law, mass tends to grow cubically while area tends to grow quadratically: larger ships should experience *proportionally* less drag.

## Ascent optimal speed

Since drag grows with speed, we face the problem of aiming for an ascent speed that minimizes the time of ascent (implying a higher speed) as well as the drag (implying a lower speed).

Ultimately, we are trying to minimize the cost in propellant, measured in $\mathit{\Delta}v$ to reach an altitude $\mathit{\Delta}z$. If we assume $v$ is constant, then so is ${F}_{d}$ (as well as ${F}_{g}$ from above), and:

v
= (a_g + a_d) t

= (F_d + g) t

= (1 C_d A v^2 + m g ) 1 z

= _f(v) z

To find the minimum of $\mathit{\Delta}v$, we try to minimize $f(v)$:

f(v) min
f'(v) = 0

1 C_d A - m g 1 = 0

1 C_d A - m g 1 = 0

v =

v = v_term

From this, we see that the optimal vertical ascent speed is the terminal velocity.

## Air pressure

Assume the atmosphere is in a stable state and consider an infenitesimal volume $dV$. The force exerted on it are the gravity and the pressure surrounding it. By symmetry, the effects of horizontal pressure cancels out; we note $dP$ the difference in pressure below and above $dV$.

The resultant force of pressure on $dV$ is $dAdP$ where $dA$ is the bottom area of $dV$. We have $dV=dAdz$. To fight gravity, we must have $dAdP=-gdm=-g\mathit{\rho}dV$. We can divide by $dV$ on both hands of the equations to get:

$$\frac{dP}{dz}=-\mathit{\rho}g$$

The value $g$ can assumed to be a constant but $\mathit{\rho}$ depends on $P$. According to the ideal gas law:

$$PV=n\mathit{RT}\Rightarrow \mathit{\rho}=\frac{nM}{V}=\frac{PM}{\mathit{RT}}$$

where $V$ is the considered volume, $n$ the number of molecules of gas in it, $R$ a convenient constant, $T$ the temperature and $M$ the mean mass of a molecule of the gas. From this, we can deduce the following differential equation:

$$\frac{dP}{dz}=-\frac{gM}{\mathit{RT}}P$$

To solve it, we need to find a primitive. We will assume that the gravitation is constant and that the temprature decrease linearly with altitude (rate $L$).

$${\int}_{0}^{z}\frac{gM}{\mathit{RT}}dz=\frac{gM}{R}{\int}_{0}^{z}\frac{1}{T-Lz}dz=-\frac{gM}{\mathit{LR}}{[\mathrm{ln}(T-Lz)]}_{0}^{z}=-\frac{gM}{\mathit{LR}}\mathrm{ln}(1-\frac{Lz}{T})$$

Thus:

$$P(z)={P}_{0}{e}^{\frac{gM}{\mathit{LR}}\mathrm{ln}(1-\frac{Lz}{T})}={P}_{0}{(1-\frac{Lz}{T})}^{-\frac{gM}{\mathit{LR}}}$$

## Evolution of terminal velocity

Using the closed expression for the pressure, we can now derive the density and then the terminal velocity depending on the altitude:

# Insertion burn

## Principle

Once the spacecraft has left the atmosphere, it is not subject to air drag anymore and can set its orbit. For this, it simply need to raise its orbit from the center of the body to above the atmosphere. It is more convenient for further maneuvers to have a circular orbit.

## Classical burn

Using the *vis-viva* equation (*), we can compute the requisite speed for orbiting:

$${v}^{2}=GM(\frac{2}{{r}_{a}}-\frac{2}{{r}_{a}+{r}_{a}})=\frac{GM}{{r}_{a}}$$

For example, on Kerbin, you need to reach an horizontal velocity of:

$${v}_{\text{orbit}}=\sqrt{\frac{6.67\times {10}^{11}{m}^{3}/\mathit{kg}/{s}^{2}\times 5.29\times {10}^{22}\mathit{kg}}{600\mathit{km}+80\mathit{km}}}=2279m/s$$

To save $\mathit{\Delta}v$, launches are done close to the equator to go with the movement of the surface due to the body’s rotation. On Kerbin, we save ($T$ is the orbital period):

$${v}_{\text{surf}}=\frac{2\mathit{\pi}R}{T}=174.5m/s$$