# Optimal speed per altitude for orbit launch

On Kerbin, rockets have an optimal speed depending on altitude for maximum fuel efficiency, as

You can save fuel by being close to your terminal velocity during ascent. Lower velocity wastes delta-V on gravity, higher is wasted on air resistance See http://wiki.kerbalspaceprogram.com/wiki/Basic_maneuvers

I searched for a similar chart on earth but couldn't find any. Can someone give me an approximation of ideal speed for real rockets on earth, or maybe a few other planets ?

• As in Kerbal, the magic speed is terminal velocity (which varies with coefficient of drag), so you might try looking for that. In practice, real world rocket engines don't have the wide range of throttle capability that KSP engines have, so real rockets spend a fair amount of time above the fuel-optimal speed. – Russell Borogove Nov 20 '14 at 4:31
• I've never played the game, but it looks like Kerbin is a rather strange place. It's radius is 1/10 that of the Earth, it's mass is 1/100 of that of the Earth, and it's atmosphere (at least the first 70 km) is fairly Earth-like. On Earth, gravity drag is by far a bigger problem that atmospheric drag. Keeping speed to terminal velocity on Earth is not nearly as big a concern as is minimizing gravity losses. The situation is reversed on Kerbin. – David Hammen Nov 20 '14 at 9:17
• Is this question searching for a better approximation for the optimal velocity with altitude? Or are you just looking for charts of terminal velocity with altitude? The latter will depend on the parameters of the rocket, so it's not so simple to give a general chart. – AlanSE Nov 20 '14 at 16:53
• @AlanSE Optimal velocity with altitude. But it would probably also depends on the rocket's aerodynamics – Antzi Nov 21 '14 at 2:37

If the optimal speed is terminal velocity, this is the formula you need:

$V_t= \sqrt{\frac{2mg}{\rho A C_d }}$

where
- $V_t$ is terminal velocity,
- $m$ is the mass of the falling object,
- $g$ is the Earth's gravity|acceleration due to gravity,
- $C_d$ is the drag coefficient,
- $\rho$ is the density of the fluid through which the object is falling, and
- $A$ is the projected area of the object.

Air density $\rho$ depends on altitude, so you need a table like the International Standard Atmosphere to calculate Vt versus altitude.
m, Cd and A depend on your rocket.

Here's a sample graph, based on a rocket with 3.66 m diameter, 100 ton weight and a constant Cd of 0.75:

Altitude in m on the X axis, terminal velocity in m/s on the Y axis.

As Russell Borogove pointed out, Cd also changes depending on your speed. Here's a graph that shows how it changes:

Unfortunately this means we have a feedback loop: if you want to travel at Vt, Vt depends on Cd which depends on Vt. So instead of a simple formula, you have to do iterative calculations for each altitude.

But, the solution is even more complicated than this. So far, we've ignored gravity losses and looked only at drag. "Goddard's Problem" asks basically the same question: which strategy to use to get the maximum altitude out of a fixed amount of rocket fuel. I've found several papers that tackle this problem, all contain solutions far too complicated for the scope of this answer (and for me to understand without spending days on it).
Many of the solutions for Goddard's problem contain simplifications as well (e.g. the assumption you can do instantaneous thrust, i.e. dump X amount of mass all at once, instead of having to run the engines for Y minutes).

And as a final observation, most rockets don't launch in an optimal-velocity profile: every launch I've seen reaches Mach 2 far below 10 km.
This lecture contains a decent overview of a practical approach to launch profiles.

• Coefficient of drag for a given shape changes sharply when going from subsonic to supersonic, so the terminal velocity curve isn't that simple. For example: braeunig.us/apollo/pics/cd2.gif – Russell Borogove Sep 18 '15 at 3:28
• Above certain threshold of atmosphere density (and not very distant one too) we will inevitably run below the terminal velocity - it simply goes up faster (with atmosphere pressure drop) than our acceleration can follow. And that means losses to gravity drag, which are already huge (order of 1km/s per minute). So it may be beneficial to exceed the optimal ascent speed and sacrifice some fuel to losses in the thick atmosphere, just to emerge at higher speed and reduce the gap between optimal and actual velocity during the later climb - lose more initially, to lose less later. – SF. Jul 12 '16 at 9:18
• @Hobbes - I changed the link in "This lecture contains a decent overview ..." to a wayback machine version. Apparently the course's structure has changed significantly since you wrote this answer. – David Hammen Jan 1 '18 at 16:24
• It's hard to notice that several papers that links to more than one link. This is compounded by the first two words being linked to the same item. Possibly one got dropped? Consider an alternate method perhaps? (e.g. 1, 2, 3) – uhoh Jan 2 '18 at 6:29
• @uhoh -- For better or worse, this is a widely used technique across the SE network. Whenever you see what appears to be a single link with the word "several" in it, look for each word being a separate link. Hover your mouse over each word to see where the link points. (It's always a good idea to use the hover technique even for what is obviously a single link. For computer security reasons, it's best to see where a link would take you before clicking.) – David Hammen Jan 2 '18 at 16:10

The pre-1.0 version of Kerbal Space Program had a very bad aerodynamic model, that the correct rule of thumb was to keep things near the terminal velocity, otherwise you ran the risk of being significantly decreased. The later versions increased the fidelity to something more akin to what is seen. Keeping your speed under control while in the atmosphere is always a good idea, however, it's now possible to be going at a very high speed inside the atmosphere, as in fact it is possible on Earth. See Hobbes answer for the more specifics in real life how drag works.

• I don't know how it worked before 1.0, these days I've found the optimal trajectories can end up quite firey. I've tried using the same rocket with a variety of numbers in MechJeb's ascent profile and once you're in the upper atmosphere you pour on the horizontal velocity. Climbing out of the atmosphere isn't worth it. Also, the bigger the rocket the more of a factor this is, you want to turn even more. – Loren Pechtel Sep 25 '17 at 2:58
• This is really the correct answer. If you look at old delta-V maps they used to have low-kerbin orbit at around 4550 m/s: i.imgur.com/6lStPEh.png, while now they are around 3200 m/s: i.imgur.com/Zx5Lw6L.png. That 1350 m/s difference was due to the pre-1.0 soupy atmosphere. The drag model was also terrible and didn't consider mach numbers or density and only speed. For post-1.0 rockets bang-bang control (100% throttle up) is best for orbital rockets. – lamont Feb 9 at 5:36