Playing Kerbal Space Program, I wonder why they do several burns to get into orbit. Why can't a vehicle get smoothly to it, just pointing slightly off its course?

I saw a Soyuz launch at night from 1000 kms from Baikonur, faint exhaust was clearly visible, and it kept burning continuously, and got into an orbit. (At the cutoff moment, you could see a satellite-like moving star.) So, Soyuz get into in one single burn.

I studied the control theory (thanks for correcting my English), which solves problems like getting to some state by means of some parameters, and taking into account the trajectory. A problem like this is solved with a system of differential equations.

So, is such an optimal path smooth, or not? Is this difference just because of low precision of human flying by hand?


3 Answers 3


No, gradual burning is not delta-v efficient. Basically if you want to change your apopsis it's most efficient to do it by burning at periapsis and vice-versa.

If there was no air and we had infinite thrust then the most efficient launch would have two burns, both horizontal. One on the ground to set the apopsis and then one at apopsis to circularise. Basically a hohmman transfer.

However we do air and we do we have limited thrust. So a real launch profile looks more like.

  1. Vertical ascent, to get out of the worst of the atmosphere quickly while reducing losses to atmospheric drag.
  2. A gradual transition to horizontal burning to efficiently raise the apopsis. If we had infinite thrust this could be purely horizontal but since we have finite thurst we need a vertical component to avoid falling back to earth while we build up horizontal velocity.
  3. A coast phase until we reach the desired orbit height.
  4. A burn to insert into orbit.

There may also be additional burns to fine-tune the orbit or to control where along the orbit the sattelite ends up.

In many cases (including nearly all GEO launches) the final burn is performed by an engine on the sattelite itself rather than the launch vehicle. It's a much smaller engine and may come hours after the launch, so you are unlikely to notice it.

  • 2
    $\begingroup$ Thanks! I think nowadays I'd write about the same, when the whole process is more or less clear to me. But back in 2013 it wasn't. It's quite impressive how valuable is modelling/playing experience. $\endgroup$
    – culebrón
    Jan 8, 2018 at 18:03

Similar to a Hohmann transfer orbit, short high-thrust bursts are more energy-efficient than a continuous low-thrust burn.

So the ideal trajectory has two burns: first the launch, which gets the vehicle on an elliptical orbit whose apogee just touches the final destination orbit, and an orbit-insertion burn, that puts the vehicle on the final orbit.

  • 1
    $\begingroup$ This is close, but not quite... It would be true for taking off from the moon, but from Earth, you have to worry about the atmosphere, which complicates the equation somewhat. $\endgroup$
    – PearsonArtPhoto
    Aug 24, 2013 at 22:57
  • 4
    $\begingroup$ That's why I wrote similar $\endgroup$
    – oefe
    Aug 25, 2013 at 12:32

The typical way that people get to LEO in real life is somewhat similar to the 1ish burn KSS insertion.

  1. Get your rocket so that it's effective apogee is at the appropriate elevation.
  2. Change your momentum so it is horizontal, to get the perigee at the correct altitude.

This is the way that the space shuttle works, for instance. It is very close to the optimal orbit for very low LEO orbits, due to the fact that you have to get out of the atmosphere in order for it to work at all. It also makes things a lot simpler.

In reality, there is a trade off between a lot of factors to get the optimal orbit. These include:

  1. Overall capacity of rocket.
  2. Range safety concerns
  3. Desired orbit, including inclination, apogee, perigee, etc.
  4. Coverage of downrange objects.

Depending on all of these, and more, the optimal mission might be 1 or 2 burns. In general, fewer burns is desired, due to less complex rocketry, missions, etc.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.