Editing Sub-orbital spaceflight (section)

==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 [[standard gravity|acceleration of gravity]], so with a maximum speed of 1&nbsp;km/s together 3 minutes and 20 seconds. The duration of the [[flight]] phases before and after the free-fall can vary.{{citation needed|date=May 2024}}

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|date=May 2024}} [[SpaceX Starship integrated flight test 4|Test flight 4 of the SpaceX 'Starship']] performed such a flight with a lift off from Texas and a simulated soft touchdown in the Indian Ocean 66 minutes after liftoff.

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.<ref>{{cite web |url=https://nssdc.gsfc.nasa.gov/nmc/spacecraft/display.action?id=1958-007A |title=Pioneer 1 - NSSDC ID: 1958-007A |publisher=NASA NSSDC}}</ref>

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:

<math display="block">\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}}</math>

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:

<math display="block">\text{area fraction} = \frac{1}{\pi}\arcsin\sqrt{\frac{2\sin\theta}{1 + \sin\theta}} + \frac{2\cos\theta\sin\theta}{\pi\text{(major axis)(minor axis)}}</math>

<math display="block">\begin{align}
  \text{time of flight}
    &= \left(\left(\frac{1 + \sin\theta}2\right)^\frac{3}{2}\arcsin\sqrt{\frac{2\sin\theta}{1 + \sin\theta}} + \frac{1}{2}\cos\theta\sqrt{\sin\theta}\right)2\sqrt\frac{R}{g} \\
    &= \left(\left(\frac{1 + \sin\theta}2\right)^\frac{3}{2}\arccos\frac{\cos\theta}{1 + \sin\theta} + \frac{1}{2}\cos\theta\sqrt{\sin\theta}\right)2\sqrt\frac{R}{g} \\
\end{align}</math>

This gives about 32 minutes for going a quarter of the way around the Earth, and 42 minutes for going halfway around. For short distances, this expression is [[asymptotic]] to <math>\sqrt{2d/g}</math>.

From the form involving arccosine, the derivative of the time of flight with respect to ''d'' (or θ) goes to zero as ''d'' approaches {{val|20000|u=km}} (halfway around the world). The derivative of Δ''v'' also goes to zero here. So if ''d'' = {{val|19000|u=km}}, the length of the minimum-delta-v trajectory will be about {{val|19500|u=km}}, but it will take only a few seconds less time than the trajectory for ''d'' = {{val|20000|u=km}} (for which the trajectory is {{val|20000|u=km}} long).