Bear in mind that I am no expert in orbital mechanics, and I have no computer codes to analyze 3-body problems. But I do understand the basics of 2-body problems, especially elliptical orbits, fairly well. There are simple formulas for those. That is the basis for the discussion topic here.
It is one thing to build and fuel a large craft in low Earth
orbit. It is quite another to propel it
from there onto an interplanetary trajectory.
The velocity requirement is to reach a speed somewhat over escape, from only circular orbit speed.
The math says that escape speed is larger than circular
orbit speed by a factor of the square root of 2, or about 1.414. For Earth,
that is about 11 km/s for escape, and about 8 km/s for circular
orbit, with the velocity requirement
being the difference, at about 3
km/s. To depart, we actually need a bit
more than that 3 km/s difference,
because we actually need to exceed escape speed.
By using a space tug and an extended elliptical orbit about
the Earth, that velocity requirement on
the departing spacecraft can be substantially reduced. The space tug is only reusable if it ends up remaining
in that extended ellipse, from which it
can then return to low circular orbit.
Geometrically, the
situation is symmetrical, so that this
concept also works for arrivals from interplanetary trajectories. But the sequence of events that must
happen is not symmetrical, so the
arrival trajectory plan cannot be the exact mirror reverse of the departure
trajectory plan.
The example here is for Earth, using low Earth orbit as the basis, as this is the most easily reached orbit for
launches from the surface. But the same
arguments and math apply to departures and arrivals at any planetary body! Only the numbers are different.
Let us start with the arrival scenario depicted as
sequential sketches in Figure 1.
The arriving craft is coming in on a hyperbolic path with respect to the
Earth, with its perigee specifically located
at low Earth circular orbit altitude,
and a speed at perigee somewhat larger than Earth escape speed at that same
altitude.
As the arriving craft reaches its hyperbolic perigee, it must impulsively make a modest speed
reduction (delta-vee, or dV) to slow to
a speed just under Earth escape speed at that altitude. That puts it onto an extended elliptical
orbit about the Earth, instead of
continuing on into deep space.
The space tug that is going to retrieve this craft from that
extended elliptical orbit is not already on that ellipse, and in fact cannot be on that ellipse! It must be somewhere on the low circular
orbit with the short period of about 90 minutes. The extended ellipse has a much longer period
near 4 (or even 5) days!
Because the timing of the craft’s arrival into that extended
ellipse, relative to where the tug is
located on that low circular orbit, is
not something that can be controlled,
the tug could be “anywhere” around that circular orbit! It will pass the ellipse perigee point
multiple times, while the arriving craft
makes one circuit about the extended ellipse.
There will be one tug circuit about the circular orbit where
it and the arriving craft are both very near each other at the ellipse
perigee, at the same time! That is where the tug must fire its
propulsion to accelerate impulsively to ellipse perigee speed, thus matching both position and velocity with
the arriving craft. The closer the orbit
period ratio is to an integer, the
closer together this rendezvousing pair of vehicles will be, and the lower the required rendezvous dV
budget.
Actual rendezvous and docking take a finite interval that is
not trivial! The two can dock, but by the time that is accomplished, they will no longer be anywhere near the
ellipse perigee point! The docked pair
must complete another circuit about the ellipse. When they reach ellipse perigee the second
time, the tug can decelerate the docked
pair into low circular orbit, precisely
because the perigee altitude is preserved by making burns there, and not anywhere else along the orbit.
From there, normal
rendezvous procedures can be used to reach any desired orbital station or
facility.
Figure 1 – The Arrival Scenario
The departure scenario sequence of sketches is given
in Figure 2. The difference in the event sequence is that
the departing craft and the tug are docked together when this scenario
starts! They must only wait in
circular orbit until the geometry lines up with the intended departure
path, and then the tug fires to bring
the docked pair to the perigee speed of the extended ellipse.
They undock as that burn terminates, and the departing craft immediately makes its
modest burn to reach the intended departure speed. Meanwhile,
the tug just coasts one circuit about the ellipse. When it reaches perigee, it can decelerate back into low circular
orbit.
From there, standard
rendezvous procedures can return the tug to any desired orbital station or
facility. Starting out docked together
is precisely what provides the proper timing of the departure events
sequence, that is inherently lacking
during the arrival sequence! Undocking
and moving away a short distance, for
departure burn safety, is something that
only takes several seconds, not the
several minutes to an hour or so, for close-in
rendezvous and docking.
Figure 2 – The Departure Scenario
For the sake of argument,
presume the following data:
Vcirc = 7.8 km/s
Vesc = 11.0 km/s
Vper = 10.9 km/s
Vdep = 11.5 km/s = Varr
Those values produce the following
departure results:
Laden tug dV = Vper – Vcirc = 3.1
km/s
Craft departure dV = Vdep – Vper =
0.6 km/s (unassisted this is 3.7 km/s)
Unladen tug back to circular dV = Vper – Vcirc = 3.1 km/s
They produce similar results values for the arrival scenario, reflecting the geometric symmetry:
Craft arrival dV = Varr – Vper =
0.6 km/s (unassisted this is 3.7 km/s)
Unladen tug onto ellipse dV = Vper
– Vcirc = 3.1 km/s
Laden tug back onto circular dV = Vper – Vcirc = 3.1 km/s
To these figures one should add some dV budgets for
rendezvous and docking for the tug,
likely near 0.2 km/s for each such maneuver. For departure, there is likely only one such maneuver, conducted unladen, as the tug returns to the appropriate
facility in low orbit. For the arrival
scenario, there are likely two such
maneuvers: one unladen to rendezvous and
dock with the craft in the extended ellipse,
the other laden to rendezvous (and dock) the docked pair with the
appropriate facility in low orbit. This
reflects the decided asymmetry of the events sequences and circumstances, for arrival versus departure.
For the departure scenario,
the tug sees these dV requirements in the listed order:
Laden onto ellipse dV = 3.1
km/s (this is the total, and it is done first!)
Unladen back to circular dV = 3.1 km/s
Unladen rendezvous and dock dV = 0.2 km/s
Total unladen dV
= 3.3 km/s (second)
For the arrival scenario,
the tug sees these dV requirements in the listed order:
Unladen onto ellipse dV = 3.1 km/s
Unladen rendezvous and dock dV = 0.2 km/s
Total unladen dV
= 3.3 km/s (first)
Laden back onto circular dV = 3.1 km/s
Laden rendezvous and dock dV = 0.2 km/s
Total laden dV
= 3.3 km/s (second)
Either way, the
interplanetary craft sees the same departure and arrival dV requirements: some 0.6 km/s for both of these
scenarios. The weight statements for
these cannot be the same, so you cannot
simply sum the dV’s, as neither scenario
(as described) includes operations, any
staging, or any refueling, at the destination. All of those are set by the overall mission
design.
As for the tug, the
laden vs unladen weight statements are drastically different, so you cannot sum the laden and unladen total
dV’s for one simple rocket equation calculation. Those must be two separate but linked rocket
equation sizing calculations. And
the order in which the events occur controls the linkage between the
calculations! That linkage is via
the weight statements.
The asymmetry of the events sequence also shows up in how
many times the craft and the tug must each go around the ellipse.
For the arrival sequence,
the craft will go around the ellipse either 0 (unlikely) or 1 time, before the tug can enter the ellipse with
it. Then the docked pair go around the
ellipse a second time, for either 1 (unlikely)
or 2 (likely) total ellipse circuits on arrival for the craft, and just 1 for the tug.
In contrast, for the
departure sequence, the craft never goes
around the ellipse at all, and the tug
must go around once.
What if the departure/arrival speed requirement is
higher?
A higher departure/arrival speed requirement than the
nominal 11.5 km/s discussed above, just
means the difference between those arrival/departure speeds from 11.5
km/s, simply adds directly to the
interplanetary craft dV requirement.
There is no change to the tug dV requirements, because the actual departure/arrival speed is
not part of its dV requirement calculations.
Those only figure into the interplanetary craft dV requirements.
What would affect both is electing a different extended
ellipse from the 10.9 km/s perigee speed used for this discussion. For that reason, it is recommended that for any given mission
to be designed, first you pick a low
circular orbit and get its speed and period.
You don’t really need escape speed at that altitude, except as a check value (arrival/departure
must be higher than that).
Then pick an extended ellipse such that its perigee is at the
low circular altitude, and its period is
an integer multiple of the circular period.
Then get its perigee and apogee speeds.
I would recommend apogee altitudes almost out to the moon, in order to get perigee speeds up as close as
you can, to escape at perigee
altitude. That is the way to reduce (as
far as is possible) the dV requirement on the interplanetary craft.
Analyze separately what your near-Earth departure/arrival
speed requirement is, for the interplanetary
mission that you want to fly. The
formulas for the dV’s to be figured, are
given in the text right after Figure 2 above.
Final Remarks
The topic here has been propulsive arrivals and
departures, assisted by a space
tug. For arrivals only, there is also the possibility of the craft
aerobraking in a pass down in the atmosphere,
such that it is on an extended ellipse as it leaves the atmosphere. However,
this is not without propulsive burn requirements on the part of the
arriving craft, so it is no
“freebie”.
First, at the ellipse
apogee, the craft will have to make a
small burn posigrade to lift the ellipse perigee up out of the atmosphere and
yet further to circular orbit altitude.
This is to avoid an unintended entry on the next perigee pass.
Second, the craft
will have to make some sort of modest burn on the next raised perigee
pass, to get the “right” apogee, such that the ellipse period is an integer
multiple of low circular orbit period.
Third, bear in mind
that peak heating starts before peak deceleration gees in any sort of
entry, and the gees required here are
significant, meaning the pass has to
go deep in the atmosphere. In turn, that means the peak heating on this type of a
deceleration pass will be of similar magnitude to that of a direct entry! And that means the vehicle must be fairly
compact, have no parallel-mounted
nacelles or other structures, and must
be protected by what amounts to a fully-capable direct entry heat shield. Those are very restrictive design
requirements!
Once these orbital adjustments are made by the arriving
craft, then the tug rendezvous-and-assist
back to circular can proceed, exactly as
described above. (There is no such
aerobraking thing for departure,
though.)
For more details about aerobraking deceleration and elliptic capture, please see my article “Elliptic Capture” on this site, dated 1 October 2024.
