# Time gap between leaving the Earth's atmosphere and initiation of orbital motion by space shuttle

It is sometimes difficult to imagine for people like me how the space shuttle initiates its orbital motion (at a speed of 17500 km/h).

So when the space shuttle goes out of the Earth's atmosphere, how does it then initiate going into the orbit or vice versa, when it has to enter the Earth's atmosphere? And what is the time gap between the two operations that is, time at which it comes out of the atmosphere and that when it initiates the orbital motion. And what is the initial speed from which it reaches 17500 km/h (if this number is correct)?

• All motion is orbital. Even standing still you are in orbit, but the earths surface prevents motion. If you were driving a car and the earth became passible somehow, you'd fall in a very thin ellipse, past the center, up the other side, and down agann and back to where you started. The faster horizontally you go, the more this elliptical path becomes circular. Start high over the atmosphere and fast enough for a near circular path hnd you have nothing stopping the motion, no air, no surface, you'd stay going around forever – Innovine Jun 11 at 21:51

When the space shuttle goes out of the Earth's atmosphere, how does it then initiate going into the orbit?

You're thinking of the climb out of the atmosphere and the entry into orbit as two separate operations, but it's really one continuous thing.

It takes several minutes for even powerful rocket engines to accelerate to orbital speeds. Most of the horizontal acceleration has to happen at high altitude, above the densest part of the atmosphere, otherwise too much power would be lost to air resistance. So the strategy is to start going straight up, then gradually tip over toward the horizontal, with the rocket engines firing the whole time. As more of the acceleration goes toward the horizontal, the vertical rate of ascent slows, and ideally you hit the target altitude just as your horizontal speed reaches that needed to maintain circular orbit, and then turn off the engines.

The space shuttle did it a little bit differently; the main engines would shut down while it was on an elliptical path that would take it to its desired altitude, but not going quite fast enough for circular orbit. Once at the desired altitude, the secondary Orbital Maneuvering System engines would fire, providing the last increment of speed to circularize the orbit.

Here's a plot of the the space shuttle's altitude over time, for the main part of the ascent, from a simulation of the STS-129 mission from FlightClub.io:

The lavender arc represents the trajectory of the solid rocket boosters after they burn out and separate; the pink track is the orbiter. You can see that the vertical speed is increasing continuously until the SRBs burn out. The main engines produce less thrust than the boosters, and the orbiter is pitched over, only about 36 degrees up from horizontal, at that point, so the vertical climb begins to slow as horizontal speed builds up.

The Kármán line of 100km altitude is generally accepted as the border between atmosphere and space, reached at about 4 minutes into the ascent, but as Organic Marble points out, there is no actual dividing line; the atmosphere just gets thinner and thinner as altitude increases.

Here's the pitch angle over the same time period:

For the first fifteen seconds of flight, the shuttle goes straight up (90º pitch), then quickly pitches over; by 85 seconds into the flight it's at a 45º angle, gaining both horizontal and vertical speed rapidly. It continues slowly lowering the angle until main engine cutoff at about 510 seconds. At this point, it's coasting to a higher altitude, and the Earth's surface is curving away below it, so the plot shows the effective pitch angle is increasing as it goes. (At some point, the real orbiter would have to return to a horizontally pitched attitude to do the circularization burn, but I suspect this simulation doesn't cover that.)

Here's the total velocity of the orbiter over time, but it doesn't tell you what direction the velocity is. (The pitch angle is pretty close to the velocity direction, though.)

As the engines and solid rocket boosters burn fuel, the mass of the shuttle decreases rapidly, while thrust remains generally level, so the velocity curve increases as it goes for the first two minutes before the SRBs burn out. Velocity then increases more slowly, with the acceleration rate once again increasing as fuel is burned off.

Reentry is pretty nearly the reverse. The OMS engines would fire to slow the orbit slightly, just enough to bring the low point of the orbit down into the atmosphere. This requires a velocity change ("delta-v") of only about 90 m/s down from the 7700 m/s orbital velocity. Once the orbiter started to meet the thin air of the upper atmosphere, it would slow down further, and reentry would be assured.

• What's the purple line in the 2nd graph? And is "elevation" = "pitch"? Or flight path angle? – Organic Marble Jun 11 at 21:27
• Purple line is the SRBs again; I don't think their post-burnout behavior is authentic. I think Elevation is supposed to be the pitch of the stack relative to local horizontal; there's a separate plot for "angle of velocity vector" which I probably should have used instead. – Russell Borogove Jun 11 at 22:03
• Thanks for the info. – Organic Marble Jun 11 at 22:18
• Great answer, but I feel like the altitude vs. time plot is likely to be misleading, since I suspect that many people skimming the answer will instinctively read it as an altitude vs. ground distance plot (i.e. basically a picture of the launch trajectory) and not notice their mistake. You might want to consider including an actual plot of altitude vs. ground distance downrange (e.g. something like this) for comparison. – Ilmari Karonen Jun 12 at 19:19

Two quibbles with the question:

• 17500 is roughly orbital speed in miles per hour, not km/h
• It's probably best not to discuss it in terms of "leaving the Earth's atmosphere". The shuttle never really did, the atmosphere in some form goes out very far indeed.

However we can look at some times, velocities and burns.

For STS-135 these are the important times (given in minutes and seconds from liftoff)

• Solid rocket booster staging: 2:03
• Main engine cut-off: 8:23.8
• OMS-2 perigee-raising burn: 37:45

And the corresponding velocities

• Main engine cut-off: 25817 ft/s (28328 km/hr using the units you mentioned)
• OMS-2 delta-v: 96.8 ft/s (106 km/hr)

And the orbital parameters (units are nautical miles given as altitudes, all numbers are approximate)

• Main engine cutoff: Apogee 118 nm, Perigee 31 nm

• Post OMS-2: Apogee 124 nm, Perigee 84.9 nm

Sources:

• ...I'd never really noticed that 1 ft/s is approximately 1 km/h. Handy shortcut to remember! – Andrew Jun 11 at 21:30
• @Andrew yes - and 1ft is 1 nano light second! – Tim Jun 12 at 9:13

It's really important to take heed that any launch vehicle pitches continuously (i.e. its path gradually curves from vertical to its "horizontal" direction that will put it into LEO) whilst it is still in the atmosphere.

The reason is that the vehicle is then using aerodynamic lift - the reaction of the atmosphere on its body - to change its direction from vertical to horizontal. You can then get the horizontal component of the delta v needed to achieve LEO "for free", or at least for less fuel cost than if you waited until the vehicle were above the atmosphere to begin pitching. The engines impart the $$\Delta v$$ vertically and fins etc on the launch vehicle can tip it over.

If one waited till one were above the atmosphere to tip the rocket over into a LEO orbit, the horizontal component of the velocity would have to come wholly from the momentum of burnt fuel. At the same time, pushing through the atomosphere is wasteful. So there is an optimal, minimum fuelk use trajectory, that profits from the atmosphere's presence to help with aerodynamic pitching whilst keeping the time that the launch vehicle feels atmospheric drag relatively low.

• Are you describing a "gravity turn"? Is aerodynamic lift necessary for that? en.wikipedia.org/wiki/Gravity_turn – Organic Marble Jun 13 at 14:42
• I believe this is incorrect, and that most spacecraft use gravity turns to avoid the aerodynamic stress that would come from relying on aerodynamic lift to change direction. – notovny Jun 13 at 14:42
• Agreed, normal launchers (not the shuttle) fly at near zero alpha and have little lift. – Organic Marble Jun 13 at 14:44
• @notovny I've been told by people who actually work these things out at our institute (DLR) that gravity, drag, everything is a delicate balancing act. Aerodynamics definitely help in the practical sense and the whole lot is a complicated optimization problem – Selene Routley Jun 14 at 1:29