Treva31 said:
Yea ok let's go with the first one, minimal delta-v on the "second burn" certainly is the aim.
Alright, then this is the case Janus calculated for you in the other thread in his first response:
Janus said:
However if you just fired it horizontally at 8.03 km/sec, it would arrive at a 400 km altitude moving horizontally at just 117.5 m/sec slower than orbital speed. Meaning you only need 117.5 m/sec to achieve circular orbit. Total delta v ~8.15 km/sec. (assuming all velocities are measured with respect to the Earth's center and not its surface and not accounting for the additional initial speed needed to overcome atmospheric drag)
You don't need to change the direction in this manoeuvre.
The projectile is fired horizontally w/r to the surface, with initial velocity higher than what is needed for a circular orbit at this radius from the centre of the gravity field. I.e. greater than the 'first cosmic velocity':
$$V=\sqrt{\frac{GM}{r}}$$
G - gravitational constant
M - mass of the Earth
r - radius of the circular orbit (here, radius of the Earth)
Launch velocity higher than that makes the orbit elliptical, where the projectile starts horizontally to the Earth's surface at the periapsis. It then climbs higher, exchanging some of its kinetic energy for potential energy (i.e. slowing down), and arrives at the apoapsis of the ellipse - again horizontally to the surface. At this point, it has too low a velocity to stay at this distance, so it starts descending back towards the periapsis.
That's why there's no need to change direction, and a simple insertion burn at apoapsis, making the velocity at that point equal to the velocity needed for a circular orbit, is all that is needed.
You can calculate the velocity at a given height of the apoapsis from the conservation of energy equations.
The total orbital energy is the sum of the potential and kinetic energies, and it stays constant throughout the orbit. It also equals half the potential energy at the distance equal to the semi-major axis.
$$E=const=\frac{mV^2}{2}-G\frac{Mm}{r}=-G\frac{Mm}{2a}$$
E - total energy
M - mass of Earth
m - mass of the smaller body (cancels out anyway, so use 1)
r - distance from the centre of the Earth
a - semi major axis
V - speed of the orbiting body
You can calculate the total energy using the ##-G\frac{Mm}{2a}## part, and deduct from it the potential energy at the desired orbital distance (i.e. apoapsis; assuming here: radius of the Earth+400 km). The difference is the kinetic energy at apoapsis. Knowing kinetic energy it's trivial to get velocity at apoapsis.
It will be lower than the velocity needed for the circular orbit at that same distance (use first equation above). The difference between the two is the required Delta-V for insertion.
In the same way, you can calculate velocity at periapsis, i.e. the launch velocity needed for raising the orbiter to the apoapsis distance. The sum of the two calculated velocities is the total Delta-V needed to put the craft on a circular orbit at a given height.
I'm getting the same numbers Janus got from his calculations in the quoted paragraph, btw.
Treva31 said:
Are there names for those 2 types?
First one is called Hohmann transfer orbit. It is a special case of the second type, which is a more generic two-impulse transfer. I would swear there was some name attached to the non-Hohmann type, or at least to its flight time-minimising case, but for the love of me I can't remember what it was.
Anyway, here's some further reading:
https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec17.pdf
Wikipedia articles on 'orbital mechanics' and 'orbital maneuver' are also well-written, with lots of relevant equations, but arguaby less pedagogical.