This might be a beginner's question, but I cannot find a full answer.
I'd like to clarify the theoretical speeds that change the nature of the trajectory of an object launched at some altitude on Earth, horizontally and in the equator plane, assuming no atmosphere, without any propulsion mean after the launch. My incomplete reasoning:
If the object is launched with an initial horizontal move with a low speed it will fall free (parabolic trajectory)
If the speed is increased, the object may start to move in elliptical orbit, such that it will just reach the sea level at the opposite point of the launch (perigee) and return to apogee at the launch point.
Increasing further the speed, the orbit should become circular.
Increasing again, the orbit will be elliptical again, with the perigee being at the launch point, and the apogee at the opposite point.
Increasing again, up to what I believe is the "critical speed", the ellipse change back to a parabolic curve, the trajectory is open, and the object will not return (the null speed being reached only at an infinite distance).
Increasing past this speed, the trajectory continue to be open, but the curve is now hyperbolic.
A common illustration of the experience:
To sum up, what I think true, and what I don't know about launch speeds:
- $0 \leq s < s_1$, free fall / parabolic
- $s_1 \leq s < s_2$, elliptical, apogee at launch point
- $s = s_2$, circular
- $s_2 \leq s < s_3$, elliptical, perigee at launch point
- $s = s_3$, parabolic, open
- $s_3 < s$, hyperbolic, open
Question: What are the names and values (for Earth) of $s_1$, $s_2$ and $s_3$?
From the answers and comments provided (thanks!) and additional searches:
- Orbital speed $s_2$, the speed for injection into a circular orbit, depends on the altitude above the sea level: 7.91 km/s at sea level, 7.73 at 300 km.
- Escape velocity ($s_3$) is the speed at which an object will not orbit the Earth, but escape it along a parabolic trajectory. 11.18 km/s at sea level, 10.93 at 300 km/s.
Below escape velocity, the object will move along an ellipse, if ellipse doesn't intersect the ground surface (free fall in this case).
The circular orbit is a particular case of elliptical orbit. We learned at school that the free fall trajectory is a parabola, this is not valid in real life where the Earth is not flat, and the gravitational force varies with the distance to the Earth center.
For orbits, below $s_2$, the launch point becomes the apogee and above $s_2$, it becomes the perigee.
Above escape velocity the curve is hyperbolic, with no return.
See also: Orbital speed on wikipedia.