Launch to orbit


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


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

Δ vt =- ve ln(1- m m(0) t)

If we now assume that 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.


Let us notate μ=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:

ag = μ (R+z )2 μ R2

The approximation z << R is verified on large bodies since we will usually aim for a low orbit topping the atmosphere. On small bodies, the gravity as well as the atmosphere will usually be negligible.

Then, integrating twice:

Δ xg = 1 2 t2 × ag

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
We could see more clearly the contrast between the different situations by using a large pod (Mk1-2 Command Pod) instead. However, such a rocket will not go as high.

Here, F is the model used for gravity: we either ignore it ( 0), assume it is a constant ( GMm R2 ), or use the exact formula ( GMm (R+z )2 ). From the graphs, we see that assuming that the gravity is constant ( z << R) makes for a good approximation. Moreover, on small bodies (c) (d), gravity has a very limited impact on such short periods.

Atmospheric drag


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:

Fd = 1 2 ρ× Cd A× v2

where ρ is the density of air and Cd 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, z is constant so 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 1 125   m2 /kg×m. This means that the terminal velocity only depended on the air density ρ and the coefficient of drag Cd (usually about 0.2).

m 125kg/ m2 . 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.

The new aerodynamic model (“drag cubes”) avoids this issue.

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 Δv to reach an altitude Δz. If we assume v is constant, then so is Fd (as well as Fg 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 Δv, 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 pressure below dV must be stronger to fight gravity. dP will be a negative value, so that we can keep the vertical axis orientation up

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=-gρdV. We can divide by dV on both hands of the equations to get:

dP dz =-ρg

The value g can assumed to be a constant but ρ depends on P. According to the ideal gas law:

PV=nRTρ= nM V = PM 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:

dP dz =- gM 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).

0 z gM RT dz= gM R 0 z 1 T-Lz dz=- gM LR [ln(T-Lz)]0 z =- gM LR ln(1- Lz T )


P(z)= P0 e gM LR ln(1- Lz T ) = P0 (1- Lz T )- gM 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:

Terminal velocity increases as air density decreases; notice that the terminal velocity at the surface is about 100 m/s in this example

Insertion burn


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.

When P leaves the atmosphere and burns in B, it can change from the red “orbit” to the blue orbit.

Classical burn

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

v2 =GM( 2 ra - 2 ra + ra )= GM ra

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

v orbit = 6.67× 1011   m3 /kg/ s2 ×5.29× 1022  kg 600 km+80 km =2279 m/s

To save Δv, 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 surf = 2πR T =174.5 m/s

Gravity turn