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A sub-orbital spaceflight is a spaceflight in which the spacecraft reaches outer space, but its trajectory intersects the surface of the gravitating body from which it was launched. Hence, it will not complete one orbital revolution, will not become an artificial satellite nor will it reach escape velocity.

For example, the path of an object launched from Earth that reaches the Kármán line (about 83 km  – 100 km ^[2] above sea level), and then falls back to Earth, is considered a sub-orbital spaceflight. Some sub-orbital flights have been undertaken to test spacecraft and launch vehicles later intended for orbital spaceflight. Other vehicles are specifically designed only for sub-orbital flight; examples include crewed vehicles, such as the X-15 and SpaceShipTwo, and uncrewed ones, such as ICBMs and sounding rockets.

Flights which attain sufficient velocity to go into low Earth orbit, and then de-orbit before completing their first full orbit, are not considered sub-orbital. Examples of this include flights of the Fractional Orbital Bombardment System.

A flight that does not reach space is still sometimes called sub-orbital, but cannot officially be classified as a "sub-orbital spaceflight". Usually a rocket is used, but some experimental sub-orbital spaceflights have also been achieved via the use of space guns.^[3]

Altitude requirement

By definition, a sub-orbital spaceflight reaches an altitude higher than 100 km (62 mi) above sea level. This altitude, known as the Kármán line, was chosen by the Fédération Aéronautique Internationale because it is roughly the point where a vehicle flying fast enough to support itself with aerodynamic lift from the Earth's atmosphere would be flying faster than orbital speed.^[4] The US military and NASA award astronaut wings to those flying above 50 mi (80 km),^[5] although the U.S. State Department does not show a distinct boundary between atmospheric flight and spaceflight.^[6]

Orbit

During freefall the trajectory is part of an elliptic orbit as given by the orbit equation. The perigee distance is less than the radius of the Earth R including atmosphere, hence the ellipse intersects the Earth, and hence the spacecraft will fail to complete an orbit. The major axis is vertical, the semi-major axis a is more than R/2. The specific orbital energy ${\displaystyle \epsilon }$ is given by:

$${\displaystyle \varepsilon =-{\mu \over {2a}}>-{\mu \over {R}}\,\!}$$

where ${\displaystyle \mu \,\!}$ is the standard gravitational parameter.

Almost always a < R, corresponding to a lower ${\displaystyle \epsilon }$ than the minimum for a full orbit, which is ${\displaystyle -{\mu \over {2R}}\,\!}$

Thus the net extra specific energy needed compared to just raising the spacecraft into space is between 0 and ${\displaystyle \mu \over {2R}\,\!}$.

Speed, range, and altitude

To minimize the required delta-v (an astrodynamical measure which strongly determines the required fuel), the high-altitude part of the flight is made with the rockets off (this is technically called free-fall even for the upward part of the trajectory). (Compare with Oberth effect.) The maximum speed in a flight is attained at the lowest altitude of this free-fall trajectory, both at the start and at the end of it.^{[citation needed]}

If one's goal is simply to "reach space", for example in competing for the Ansari X Prize, horizontal motion is not needed. In this case the lowest required delta-v, to reach 100 km altitude, is about 1.4 km/s. Moving slower, with less free-fall, would require more delta-v.^{[citation needed]}

Compare this with orbital spaceflights: a low Earth orbit (LEO), with an altitude of about 300 km, needs a speed around 7.7 km/s, requiring a delta-v of about 9.2 km/s. (If there were no atmospheric drag the theoretical minimum delta-v would be 8.1 km/s to put a craft into a 300-kilometer high orbit starting from a stationary point like the South Pole. The theoretical minimum can be up to 0.46 km/s less if launching eastward from near the equator.)^{[citation needed]}

For sub-orbital spaceflights covering a horizontal distance the maximum speed and required delta-v are in between those of a vertical flight and a LEO. The maximum speed at the lower ends of the trajectory are now composed of a horizontal and a vertical component. The higher the horizontal distance covered, the greater the horizontal speed will be. (The vertical velocity will increase with distance for short distances but will decrease with distance at longer distances.) For the V-2 rocket, just reaching space but with a range of about 330 km, the maximum speed was 1.6 km/s. Scaled Composites SpaceShipTwo which is under development will have a similar free-fall orbit but the announced maximum speed is 1.1 km/s (perhaps because of engine shut-off at a higher altitude).^{[citation needed]}

For larger ranges, due to the elliptic orbit the maximum altitude can be much more than for a LEO. On a 10,000-kilometer intercontinental flight, such as that of an intercontinental ballistic missile or possible future commercial spaceflight, the maximum speed is about 7 km/s, and the maximum altitude may be more than 1300 km. Any spaceflight that returns to the surface, including sub-orbital ones, will undergo atmospheric reentry. The speed at the start of the reentry is basically the maximum speed of the flight. The aerodynamic heating caused will vary accordingly: it is much less for a flight with a maximum speed of only 1 km/s than for one with a maximum speed of 7 or 8 km/s.^{[citation needed]}

The minimum delta-v and the corresponding maximum altitude for a given range can be calculated, d, assuming a spherical Earth of circumference 40000 km and neglecting the Earth's rotation and atmosphere. Let θ be half the angle that the projectile is to go around the Earth, so in degrees it is 45°×d/10000 km. The minimum-delta-v trajectory corresponds to an ellipse with one focus at the centre of the Earth and the other at the point halfway between the launch point and the destination point (somewhere inside the Earth). (This is the orbit that minimizes the semi-major axis, which is equal to the sum of the distances from a point on the orbit to the two foci. Minimizing the semi-major axis minimizes the specific orbital energy and thus the delta-v, which is the speed of launch.) Geometrical arguments lead then to the following (with R being the radius of the Earth, about 6370 km):

$${\displaystyle {\text{major axis}}=(1+\sin \theta )R}$$

$${\displaystyle {\text{minor axis}}=R{\sqrt {2\left(\sin \theta +\sin ^{2}\theta \right)}}={\sqrt {R\sin(\theta ){\text{semi-major axis}}}}}$$

$${\displaystyle {\text{distance of apogee from centre of Earth}}={\frac {R}{2}}(1+\sin \theta +\cos \theta )}$$

$${\displaystyle {\text{altitude of apogee above surface}}=\left({\frac {\sin \theta }{2}}-\sin ^{2}{\frac {\theta }{2}}\right)R=\left({\frac {1}{\sqrt {2}}}\sin \left(\theta +{\frac {\pi }{4}}\right)-{\frac {1}{2}}\right)R}$$

The altitude of apogee is maximized (at about 1320 km) for a trajectory going one quarter of the way around the Earth (10000 km). Longer ranges will have lower apogees in the minimal-delta-v solution.

$${\displaystyle {\text{specific kinetic energy at launch}}={\frac {\mu }{R}}-{\frac {\mu }{\text{major axis}}}={\frac {\mu }{R}}{\frac {\sin \theta }{1+\sin \theta }}}$$

$${\displaystyle \Delta v={\text{speed at launch}}={\sqrt {2{\frac {\mu }{R}}{\frac {\sin \theta }{1+\sin \theta }}}}={\sqrt {2gR{\frac {\sin \theta }{1+\sin \theta }}}}}$$

(where g is the acceleration of gravity at the Earth's surface). The Δv increases with range, leveling off at 7.9 km/s as the range approaches 20000 km (halfway around the world). The minimum-delta-v trajectory for going halfway around the world corresponds to a circular orbit just above the surface (of course in reality it would have to be above the atmosphere). See lower for the time of flight.

An intercontinental ballistic missile is defined as a missile that can hit a target at least 5500 km away, and according to the above formula this requires an initial speed of 6.1 km/s. Increasing the speed to 7.9 km/s to attain any point on Earth requires a considerably larger missile because the amount of fuel needed goes up exponentially with delta-v (see Rocket equation).

The initial direction of a minimum-delta-v trajectory points halfway between straight up and straight toward the destination point (which is below the horizon). Again, this is the case if the Earth's rotation is ignored. It is not exactly true for a rotating planet unless the launch takes place at a pole.^[7]

Flight duration

In a vertical flight of not too high altitudes, the time of the free-fall is both for the upward and for the downward part the maximum speed divided by the acceleration of gravity, so with a maximum speed of 1 km/s together 3 minutes and 20 seconds. The duration of the flight phases before and after the free-fall can vary.^{[citation needed]}

For an intercontinental flight the boost phase takes 3 to 5 minutes, the free-fall (midcourse phase) about 25 minutes. For an ICBM the atmospheric reentry phase takes about 2 minutes; this will be longer for any soft landing, such as for a possible future commercial flight.^{[citation needed]}

Sub-orbital flights can last from just seconds to days. Pioneer 1 was NASA's first space probe, intended to reach the Moon. A partial failure caused it to instead follow a sub-orbital trajectory, reentering the Earth's atmosphere 43 hours after launch.^{[citation needed]}

To calculate the time of flight for a minimum-delta-v trajectory, according to Kepler's third law, the period for the entire orbit (if it did not go through the Earth) would be:

$${\displaystyle {\text{period}}=\left({\frac {\text{semi-major axis}}{R}}\right)^{\frac {3}{2}}\times {\text{period of low Earth orbit}}=\left({\frac {1+\sin \theta }{2}}\right)^{\frac {3}{2}}2\pi {\sqrt {\frac {R}{g}}}}$$

Using Kepler's second law, we multiply this by the portion of the area of the ellipse swept by the line from the centre of the Earth to the projectile: