# Does terminal velocity play a significant role when launching from Earth?

As mentioned in the question Climbing at terminal velocity minimizes losses? But why? And "of what"? terminal velocity is important when launching space craft from Kerbal (Game Kerbal Space Program). The question has an answer suggesting "Yes it does matter" but the math is over my head and it does not reference any launches from Earth.

Do launches from Earth make any attempt to accelerate at or near terminal velocity? What kind of real life impact is there for fuel consumption?

No, launches don't specifically target terminal velocity. They target an optimized trajectory that minimizes the total propellant required to counter both atmospheric drag and gravity losses on the way to orbit.

If your only objective were to gain altitude, then you would go straight up and, while drag matters, try to track your terminal velocity at the current density. However if you're trying to get to orbit, the altitude gain is a relatively small part of your problem. A small rocket will get a given payload to orbital altitude (and then fall back down).

You will need a much, much bigger rocket to get that same payload to orbital velocity as well. A launch vehicle will start to pitch over almost immediately after clearing the tower in order to start gaining horizontal velocity, at which point the terminal velocity optimization is irrelevant.

• Orbital mechanics made a lot more sense when I realized it was about going fast and not about going high. Apr 24 '15 at 17:38
• @corsiKa Love this "what-if" xkcd: what-if.xkcd.com/58 Apr 24 '15 at 19:33
• Thanks for the link. I am now obligated to include a cartoon. Apr 24 '15 at 19:44
• Is the proportion of propellant that goes to overcoming atmospheric drag a typical number? Apr 24 '15 at 20:11
• I'm sure that such a number could be derived, but I don't know what it is. It would depend on the thrust-to-weight of the launch vehicle and its ballistic coefficient. You need about 9.1 to 9.5 km/s of equivalent free-space $\Delta V$ to get to a 7.8 km/s orbit. So the 1.3 to 1.7 km/s additional is to cover the gravity loss, atmosphere drag, and altitude gain. Apr 24 '15 at 21:32

The most efficient trajectory (which includes a velocity profile) depends on a lot of factors. Spending as little time as possible fighting gravity drag is generally a good thing, but there are limits. Drag is one of those limits.

Drag increases with the square of velocity. So at some point, going faster will be less efficient than going slower. This depends on a lot of factors and is probably not terminal velocity.