The 'Cost' of stopping in orbit on the way to an escape trajectory.
Some argue, correctly, that a direct ascent to escape trajectory is superior to
a ascent to orbit then add an additional maneuver to place a spacecraft on an
The arguments against such a maneuver are generally that you want the extra
time on orbit to check out systems, tweak position and velocity data, add abort
capabilities, and screw with phasing and departure geometry. An additional
parameter might be the use of a reusable/recoverable launch vehicle such as a
The argument for using a direct ascent is generally that it is more efficient, which
is to say that less DeltaV is required due to the gravitational losses. There is also
the possibility of using more of a lower stages propellants.
Since most of these arguments are really mission specific, We can look at the
difference in DeltaV required to hit escape velocity as a function of altitude.
Case #1: (The Jules vern approach) Do all the DeltaV at the launch site.
IE a cannon.
Case #2: Do an idealized DeltaV burn at a parking orbit of about 200km.
The orbit can be circular or elliptical with it's lowest point (Perigee) at the 200km.
(this assumes the ellipticity is due to using all available propellants from the lower
stages. There is not an astrodynamic difference, but practical staging utilization
might make it better.)
Case #3 similar to case #2 except that the altitude would be about 50km. This is
the equivalent of a 'Direct' trajectory to escape.
Case #1; The 'Orbit' of the launch site can be considered to have an apogee of
6378 km with a tangential apogee velocity of 464 m/s.
Escape velocity from Rearth is
Rearth = 6378.141km MuEarth = 3.986E5
Vesc = sqrt(2*MuEarth/R) = sqrt(2*3.986e5/6378) = 11.18 km/sec
DeltaV required (ignoring Air drag) 11.18 - 0.464 = 10.72 km/sec
Case #2, Escape Velocity from 200 km is 11.0 km/s
circular Velocity at 200 km is Vcirc = sqrt(MuEarth/R) = 7.78 km/sec
which is to say that the DeltaVesc from orbit is 3.22 km/sec
To get the initial 200km orbit, the best that can be done (ignoring air drag) is a
Homan transfer from the launch site witch takes two DeltaV maneuvers
totaling 7.5 km/sec (Note: 1)
so minimum escape velocity change is 10.73 km/sec
Case #3 Escape Velocity from 50km is 11.14, minus the initial rotational velocity
of 0.464 Plus the velocity change to get to 50 km.
This analysis is a bit problematic because it requires a knowledge of how to get
to 50 km. The smallest answer has to be between 10.73 and 10.72 km/s.
Conclusion: Since neither Case #1 nor Case #2 really includes the 1.7 km/sec
that real launch vehicles use to overcome air drag and gravitational
(slow acceleration) losses, then
the astrodynamic gain in efficiency of only 0.010 km/s
is really lost in the noise. .
(*1) The minimum orbital launch ignoring air drag and having nearly
instantaneous accelerations is first entering into a elliptical orbit and then
circularizing (Homan- transfer) In this case a 200km x 0km intermediate orbit.
semi-major axis, A = 6478 km Rp = 6378, Ra = 6578
Perigee velocity = Vp = 7.966 km/s
Apogee Velocity = Va = 7.72 km/s
To make this a transfer from the equator to the circular orbit we need
surface DeltaV = 7.966 - 464 = 7.502 km/s
Circularizing kick velocity = 7.78 - 7.72 = 0.06 km/sec
Total theoretical DeltaV from equator to orbit is 7.508 km/sec
reality is more like 9.5 km/sec because of air drag and the limited acceleration
rate (nonzero elapsed time gives 'Gravitational losses)
Used Astrodynamics Formulas
MuEarth = 3.98600 km3/sec2 Rearth = 6378.141 km
R = Alt + Rearth Rp = Perigee radius Ra = Apogee Radius
Vcirc = sqrt(Mu/R) Vesc = sqrt(2*Mu/R) = 1.414*Vcirc
A = (Rp+Ra)/2 e = Ra/a - 1 = 1 - Rp/A = (Ra-Rp)/(Ra+Rp)
Ra = A(1+e) Rp = A(1-e) RpVp = RaVa
Vn = sqrt[Mu (1+2eCos(n) + ee)/(A*(1-ee)]
Vp (n=0) = sqrt[Mu(1+e)/A(1-e)] Vp = sqrt[Mu*Ra/(A*Rp)]
Va = sqrt[Mu(1-e)/A(1+e)] Va = sqrt[Mu*Rp/(A*Ra)]