# Non-orbital takeoff and landing?

I know it is a naive question, but somehow I couldn't find where it is answered here earlier.

There have been a number of questions and answers detailing how spacecraft need to achieve high orbital speed. It seems to be summed up by a comparatively clear XKCD what-if: https://what-if.xkcd.com/58/

For example, it states:

Getting to space is easy. It's not, like, something you could do in your car, but it's not a huge challenge. You could get a person to space with a small sounding rocket the size of a telephone pole. The X-15 aircraft reached space just by going fast and then steering up.

This concurs with Wikipedia entry on escape velocity:

A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.

XKCD then concludes:

But getting to space is easy. The problem is staying there. Gravity in low Earth orbit is almost as strong as gravity on the surface [...] To avoid falling back into the atmosphere, you have to go sideways really, really fast.

## So, what is the thing that prevents you from continuing the "getting to space is easy" mode of "going fast and then steering up", instead of trying to enter orbit?

Assuming, of course, you didn't really want to enter the orbit, but wanted to e.g. go to the Moon or deep space probing. Especially in conjunction with Wikipedia's note about not having to attain escape velocity to leave gravity well.

The same about landings. Going to a celestial body, entering orbit, then having to shed all that orbital velocity. If you were travelling slow enough as it is, and slowly decelerating all the way, counteracting gravity, it would take you loads of time probably, but surely there must be another reason why it can't be done, else why wouldn't we do it.

– user20636
Jan 22, 2020 at 23:28
• "What prevents you?" - is fuel running out not enough of a reason?
– vsz
Jan 23, 2020 at 5:07
• The value of "escape velocity" decreases as you get further from the planet. If turning off your engine for good does not cause you to eventually fall back to the planet, then you have achieved escape velocity. If you had a magic engine with an unlimited fuel supply, then one way do do that would be to move away from the planet at some constant speed S until you cross the line where escape velocity equals S. Jan 23, 2020 at 18:28
• I don't have a source (so no answer!) but I am very sure that direct polar ascent to the moon was used by the Luna program and I know for a fact that direct descent was used by the Ranger and Surveyor program. When you do not have reliable restartable engines, completing your transfer burns at the same time as your orbital insertion burns can be required. Jan 27, 2020 at 5:50
• I don't see how this is a duplicate; those answers do not answer this question. However, these five answers do!
– uhoh
Jan 29, 2020 at 3:35

TL;DR: it is inefficient. You should play some Kerbal Space Program and see for yourself the effects of travel in this way.

Assuming, of course, you didn't really want to enter the orbit, but wanted to e.g. go to the Moon or deep space probing. Especially in conjunction with Wikipedia's note about not having to attain escape velocity to leave gravity well.

"Escaping", in the sense of what you'll do if you're travelling at escape velocity (and outside of the atmosphere) means leaving the gravitational sphere of influence (SOI) of a body. Obviously, gravity has effectively infinite range, but when you travel far enough from Earth you'll reach a point where the gravitational influence of the Sun far outweighs that of the Earth. You'll have escaped Earth, and entered into a heliocentric orbit.

The requirement for escape is therefore distance, not velocity. If you were travelling at escape velocity (and were outside of the atmosphere), the influence of the body's gravity will be insufficient to slow you to a relative stop and pull you back down, so that's one way to do this.

If you can imagine a pair of very widely separated worlds, joined together by a ladder. If you climb far enough along that ladder, eventually you'll reach a point in space where the gravitational pull of the world you started on has been exceeded by the gravitational pull of the world you're climbing towards. You'll have "escaped" regardless of how fast you can pull yourself along.

(note that if you're travelling to the moon, you don't need to be on an escape trajectory, because the Moon is still in the Earth's gravity well too. that means you'll be going at less than escape velocity, which will save you a bit of fuel)

So, what is the thing that prevents you from continuing the "getting to space is easy" mode of "going fast and then steering up", instead of trying to enter orbit?

The thing that makes "point up and blast away" a bad way to escape a planet is efficiency. Your rocket will be continuously subject to gravity drag... at every point of your escape, you'll need to run the engines hard enough to counter planetary gravity at that altitude, plus a bit more in order to move away from the planet. The longer you take to reach the edge of the planet's sphere of influence, the longer you'll spend having to oppose gravity, and the more fuel you'll waste doing so.

If you have a really powerful rocket so you can travel quickly to the edge of the SOI, great... but you could have just as easily pointed perpendicular to the gravitational field of the planet and run your rocket then, so the full acceleration of your rocket goes into increasing your velocity instead of a load of it going to opposing gravity instead. You wouldn't do this in an atmosphere of course, because then you'd spend a lot longer fighting atmospheric drag. So, you shoot more-or-less upwards to clear the atmosphere, and then you boost more or less sideways to build up lots of speed to reach the edge of the SOI as quickly as possible.

...and suddenly, you'll have found yourself emulating a fairly conventional rocket launch profile, albeit in a "direct injection" rather than the more conventional "enter Earth orbit then transfer to wherever" choice (direct injections are ever so slightly more efficient, but not really enough to offset their inconvenience otherwise).

(there's also something called the Oberth effect that means that using your rocket deep in a gravity well as close to your periapsis as possible is more efficient than using it much further out in an orbit, but one problem at a time).

The same about landings. Going to a celestial body, entering orbit, then having to shed all that orbital velocity. If you were travelling slow enough as it is, and slowly decelerating all the way, counteracting gravity

Obviously, if you're travelling really slowly relative to your destination, it'll take you a really, really long time to get there.

That aside, the issue is once again efficiency. The probably very non-technical term Suicide Burn is used to describe a landing trajectory where all your deceleration is done as late as possible, to minimise the amount of fuel wasted.

ETA: as neph observed, a direct descent trajectory is of course a little more efficient than injecting yourself into an orbit and then deorbiting, but presents timing issues if you wanted to land somewhere in particular and indeed safety issues as it makes bailing out back into orbit somewhat harder.

• TL;DR: there's lots of gravitational potential energy you need to shed when planning a landing. Jan 22, 2020 at 13:24
• @CarlWitthoft that, too. Jan 22, 2020 at 13:47
• What can the KSP game actually teach about spaceflight and orbital mechanics, and what are its limitations? See: ISP (how good the engines are), Delta-V (how much fuel you need), and Max q (how fast you can go before you'll blow up from drag). Jan 23, 2020 at 1:55
• @HeadhunterKev there are even multiple shoutouts on XKCD, so clearly it is the answer to All The Things. Jan 24, 2020 at 10:23
• I would like to clarify that direct descent is more efficient than entering orbit and then descending from that orbit. However, it was not done during the Apollo mission because you cannot abort during a direct descent back to orbit. Jan 27, 2020 at 5:51

Your question is, as I understand it, pointing out that there are two ways to get from the surface of the Earth to the surface of the Moon.

Way one: Burn upwards until through the thickest part of the atmosphere to avoid aero drag. Burn sideways to attain orbital velocity and raise apogee and perigee into space. From Earth orbit, burn prograde to attain translunar trajectory. At perilune, burn retrograde to attain lunar orbit. Burn retrograde further to lower orbit to intersect the surface, and then burn surface-retrograde such that we get a soft landing.

Way two: Figure out how long the journey from Earth to Moon will take if you run the engines continuously. Point rocket at where the Moon will be then. Accelerate constantly in that direction until roughly half way; rotate 180 degrees at the halfway point and accelerate constantly in the other direction, such that you'll reach zero velocity upon reaching the surface of the Moon.

The second method is called a brachistocrone trajectory and it is the fastest way to get from point A to point B. However, as is often the case with the fastest way, it is not at all fuel efficient. It only makes sense if you have an extremely light spacecraft which burns fuel very slowly and very efficiently. The Dawn spacecraft, for example, would be a candidate for this sort of trajectory (though as a commenter notes, it does not have enough thrust to get off the ground!), but we don't have a good reason to want to get from point A to point B as fast as possible.

Rather, we want to get from point A to point B with a reasonable combination of cost and time; the brachistocrone trajectory is extraordinarily expensive because it is so fuel inefficient, so it is not a good balance of cost vs time. The conventional "take off, circularize, transfer, circularize, land" plan takes longer but is many orders of magnitude more fuel efficient, and therefore cheaper.

• Thank you, this is a great summary of what I wanted to know. I am still going to accept Starfish Prime's answer because it has a bit more detail (eg why is going to orbital speed more fuel efficient - going perpendicular to gravity force requires less energy), but I upvoted and if I could I would accept both. :) Jan 23, 2020 at 9:02
• FTR, ion engines like Dawn's are by far not powerful enough to pull off a brachistocrone between Earth and Moon, but could conceivably do this in a binary system between two lightweight asteroids. Of course, in that case it would be even weirder to want to get there as fast as possible, but... Jan 23, 2020 at 10:13
• @leftaroundabout: Right, you would never want to use something like Dawn to ascend or descend from a massive body. Jan 23, 2020 at 17:41

For getting to the moon specifically, there's an extra problem, on top of what's already been mentioned: presumably, your objective when you get to the moon is to either orbit it or do a nice gentle landing on it (if your objective is to turn yourself into a cloud of shrapnel spread over a large area of regolith, feel free to take your approach). That is: you need to get yourself stationary relative to the moon. Since the moon is in orbit around the earth, that means that you need to enter orbit around the earth eventually, whether you like it or not.

For going into deep space, you can take your approach, providing you don't mind horrible efficiency (and techniques like this can work for launching off things that are much smaller than the earth, where the efficiency losses are much smaller), but if you're aiming to rendezvous with something that's in an orbit around the thing that you're launching from, you've absolutely got to get into an orbit around the thing that you're launching from eventually.

Going to a celestial body, entering orbit, then having to shed all that orbital velocity. If you were travelling slow enough as it is, and slowly decelerating all the way, counteracting gravity, it would take you loads of time probably, but surely there must be another reason why it can't be done, else why wouldn't we do it.

Since others (specially Starfish Prime) have already answered with the principles of orbital motion, I will now approach this through a practical angle.

I have the PC version of Kerbal Space Program and I love playing with it. It is a space program simulation game that allows you to see the practical implications of different approaches to space travel.

What you suggest there is possible - I've done it plenty of times in the game. It's just that the approach of parking into an orbit, then deorbiting, is cheaper by a few orders of magnitude.

As you take off into space, your trajectory - the shape of your orbit - is defined by the gravity of whatever you are taking off from and your speed at the moment. You already know that in order to orbit the Earth, your trajectory has to be circular and that requires a lot of horizontal speed.

You may wish to go to the Moon without having to circularize an orbit around the Earth first, and without having to do orbital parking on the Moon. To do that, simply point the rocket up at the right moment and go. That works too.

Let's get into the basics of orbital motion: when you accelerate in the direction you are going (prograde), the shape of your orbit stretches so that the point most distant to you gets higher. Also notice that when your orbit is thus elliptical, the higher you go, the slower your relative velocity to whatever you are orbiting will be.

To get to the Moon by doing a Hoffman transfer, you have to get to Low Earth Orbit (LEO), and then you park at 7.8 km/s. Then you accelerate to 10.4 km/s at a precise point, and that will get you in an elliptical trajectory that intersects with the Moon's own orbit, so that you and the Moon meet at the precise same time and you can land. Your speed will remain 10.4 km/s relative to Earth at the point where you accelerated, but it will be reduced to around 1 km/s at the farther end of your orbit (not considering acceleration from the Moon).

When you get close to the Moon your relative speed to the Moon itself will be smaller - at 60,000 km from it your relative speed should be close to 1.1 km/s. You will be accelerated by the Moon towards it, and then you have to shed that speed. Apollo 11 came to 118 km of the surface of the Moon at 1.8 km/s (due to gravity acceleration by the Moon), then decelerated to 1.6 km/s to park in its orbit. The lander had to shed that much speed to touch down.

In contrast, if you wish to take a trajectory as straight to the Moon as you can, you can do like the New Horizons did. Once you are past the atmosphere, reach 16.2 km/s. You will reach the Moon in less than nine hours (compare with the ~3 days it took Apollo). You will have much more speed too - in this case your apogee is WAY beyond the Moon, which means you will lose very little speed along the way. You will practically reach the Moon with that same speed, so now you have ~16 km/s to shed.

All in all, you are changing velocity ~3x more by going in a less circular path. That does not mean your expenses are simply multiplied by 3, though. Quoting Randall Munroe:

It's what engineers call the tyranny of the rocket equation: As the amount that you want to change your speed ("delta-v") goes up, the fuel required increases exponentially.

I don't have the math in me, but from my experience in KSP you might need an Energia to land a very small probe on the Moon if you were to try and pull this off in real life. Unfortunately Energias are both prohibitively expensive and discontinued.

What is the thing that prevents you from... going fast and then steering up...

Isn't it the atmosphere?

Going up first then speeding up makes spaceflight possible. No rocket could accelerate to mach-20+ at 1 atmosphere and then sustain it all the way to space.

• To steer up and transform your horizontal speed to vertical you need the atmosphere. But the atmosphere prevents you going hypersonic to orbital speed.
– Uwe
Jan 22, 2020 at 11:42
• @Uwe to steer, can't engines simply be gimbaled?
– uhoh
Jan 22, 2020 at 12:39
• To be exact, atmospheric drag makes this choice of flight path extremely energy-intensive, not to mention high structural risk. Jan 22, 2020 at 13:25
• @uhoh if you simply steer in a vacuum by gimballing the engines, you are now pointing in the right direction...and still going in your original direction. To actually change course you need to cancel all your original speed, which is very expensive with engines, but fine for aircraft where you can push off the air to change direction "for free". Jan 22, 2020 at 18:44
• @Skyler oh I see; in order to "transform your horizontal speed to vertical" using the atmosphere, one could use aerodynamic lift and conserve at least some of the momentum attained.
– uhoh
Jan 22, 2020 at 23:50

The answer is entirely gravity loss. When a rocket burns straight up it constantly is incurring gravity losses equivalent to local gravitational acceleration (9.8 m/s on the Earth's surface). A rocket that is burning straight up at the surface of the Earth at 9.8 m/s is exactly counteracting the gravitational force and hovering and going nowhere. The whole output of the rocket is going into gravity loss. Rockets that burn horizontal incur no gravity loss, they aren't "trying" to hover at all.

This is one way of looking at why a Hohmann transfer orbit is most efficient, because the rocket burns horizontal both for injection and capture burns.

For a torch drive, efficiency is useless and you want to get there fast so you just point at your target at go, burn at 1g then flip and decelerate. But since we don't live in The Expanse, we care about getting places like the Moon with the least amount of fuel possible.

And the most efficient way for a rocket to launch to the Moon would actually be burning roughly horizontal to the surface. Rockets burn upwards to launch because the surface of the Earth gets in the way, drag losses due to the atmosphere and the orbital speeds necessary to reach the Moon would burn up the rocket at sea level. But if you removed the surface of the Earth and the atmosphere (minor concerns), rockets would launch horizontally and do something that looked like a Hohmann transfer.

So rockets burn up only enough to get them away from the Earth and the atmosphere, then they all flop over so that they can burn as horizontal as they can.

There's a bit of a complication where something like a direct ascent trajectory won't spend as much time horizontal as using a parking orbit, but it trades that off by dumping more fuel lower in the gravity well and taking advantage of the Oberth effect, but gravity losses are the primary concern. A "direct ascent" trajectory is still very horizontal and not "point at the moon and go" like you'd do if you had a torch drive, and it saves little over a parking orbit, while the parking orbit lets you assess if you made it off the Earth safely and make a go-no-go decision before committing to trans-lunar injection.