Orbit Variations
The optimal parking orbit selected is based on a balance between launch cost and Orbit Transfer
Vehicle (OTV) cost. The spacecraft requires the majority of the transfer time and energy to
escape Earth’s gravity; therefore, increasing the parking orbit slightly results in a significant
decrease in transfer costs but an increase in launch cost. In this analysis we compare various
orbit altitudes, eccentricities, and launch vehicles to determine the optimal parking orbit based on
the total cost.
Higher Circular Parking Orbit
The most effective way of reducing the transfer cost involves performing the trans-lunar
injection from a higher circular parking orbit. By varying the parking orbit altitude in
Spiral_EOM_Script.m, we determine the initial mass, thrust and power requirements, and the
transfer cost for the OTV. For this analysis we select a mass flow rate of 7.1 mg/s, a flight time
(TOF) of 150 days, and assume the power cost as $1000 per Watt as estimated in section X.X.X.
As shown below in table #-1, increasing the parking orbit to Medium Earth Orbit (MEO)
altitudes results in a significant cost reduction. A parking orbit at 15,000 km alt requires 50 kg
less and approximately one third the cost for solar arrays.
Table #-1 Transfer Costs for Varying Circular Parking Orbit Altitudes
Earth Parking Orbit Altitude
km
Initial Mass
kg
Thrust Required
mN
Power
kW
Solar Array Mass
kg
Power Cost
Million $
200
650
123
2.2
14.6
2.2
2000
630
102
1.7
11.3
1.7
15 x 104
600
55
0.7
4.7
0.7
36 x 104
580
31
0.4
2.7
0.4
Although the higher parking orbit has a profound effect on the transfer cost, we see in table #-2
that the launch costs still negate the savings. Table #-2 shows the capability for the currently
selected Dnepr Launch Vehicle. Cite where this came from At an altitude of 900 km, the OTV
mass exceeds the launch vehicle capability. From these results we conclude that departing from
a higher circular parking orbit increases the mission cost.
Table #-2 Dnepr Launch Vehicle Capability
Circular Altitude
km
200
300
400
500
600
700
800
900
Deliverable Mass
kg
4400
3700
3400
2750
1900
1200
650
300
Eccentric Orbits
Although a high circular parking orbits does not reduce the mission cost, we consider departing
from an eccentric parking orbit. We discover that performing trans-lunar injection from an orbit
with a 400 km periapsis and 1000 km apoapsis still reduces the transfer cost compared to a 400
km circular orbit. For this analysis, we run Spiral_EOM_Script_Elliptic.m iteratively with
EP_Sizing.m to determine the OTV initial mass and power necessary to achieve requirements.
Assuming an engine mass flow rate of 5.6 mg/s, we size the OTV for six flight times between
351 and 196 days. For each flight time, the parking orbit apoapsis altitude varies between 400 to
10 x 104 km. (Periapsis is fixed at a 400 km altitude). This analysis results in 42 different flight
scenarios. Because two engines with different flow rates are currently considered, we repeat this
entire analysis for an engine mass flow rate of 7.1 mg/s. A sample of the results is provided in
table #-3. Departure Orbit Variations.xlsx contains 12 tables similar to table #-3 for six flight
times and two mass flow rates in the “Curve Fits sheet.
Orbit Energy Plots
The data from table #-2 specifies the launch capability to various circular orbits; however, it does
not indicate how much mass can be launched into an elliptical orbit. Without knowing the
deliverable mass to a particular elliptical orbit, we cannot determine the cost. To alleviate this
problem, we calculate the amount of mass that can be delivered to an orbit of a specific energy.
The orbit energy for a particular orbit can be found by eq. #-1 where μ is the gravitational
parameter for Earth, and a is the semi major axis of the parking orbit. (eq. #-2)
𝜀=−
𝜇
2𝑎
(#-1)
𝑎 = 1⁄2 (𝑟𝑝 + 𝑟𝑎 ) (#-2)
where rp and ra are the periapsis and apoapsis radii, respectively.
As an example, table #-2 shows that 2750 kg can be delivered to a circular orbit at 500 km
altitude. From eq. #-1 and #-2, we find that a 500 km circular orbit has energy of -28.97 km2/s2.
Thus we assume that 2750 kg can be delivered to an orbit with energy of -28.97 km2/s2
regardless of the eccentricity. Manipulating eq. #-1 and #-2, an orbit of 400 km by 603 km has
the same orbit energy. This shows that we can deliver the same mass of 2750 kg 103 km higher
for the same launch cost.
The data compiled in Departure Orbit Variations.xlsx indicates that an elliptical parking orbit
reduces the cost, and the example above shows that it is possible to launch more mass to higher
altitudes with eccentric orbits. For this reason we investigate the problem further. Implementing
eq. #-1 and #-2, we take the data from Departure Orbit Variations.xlsx and relate it to orbit
energy. See table #-3 below for the sample of this data.
Table #-3 Sample of Sizing Results
TOF
days
286
Mass Flow Rate
kg/s
5.6
Apoapsis Altitude
km
400
500
750
1000
2000
5000
10000
Semi-major Axis
km
Orbit Energy
Km2/s2
6778.14
-29.40
6828.14
6953.14
7078.14
7578.14
9078.14
11578.14
-29.19
-28.66
-28.16
-26.30
-21.95
-17.21
Power
Cost
Initial
OTV
Mass
Million
kg $
649.6
649.4
649.0
648.6
646.8
643.1
639.3
Power Required
kW
3.50
3.47
3.41
3.35
3.11
2.60
2.09
The 12 tables compiled in Departure Orbit Variations.xlsx contain 84 different flight
scenarios. By plotting these data points, we develop linear relationships for how the initial OTV
mass and power requirements vary. Figure #-1 displays the relationship of initial OTV mass
with orbit energy. A different curve was created for the six flight times tested. As we see the
initial OTV mass decreases as orbit energy increases. An increase in orbit energy corresponds to
an increase in altitude, so this behavior is expected. We notice that shorter flight times result in a
reduced mass. The engine operates at a constant mass flow rate, so the propellant mass can be
found by eq #-3. Clearly shorter flight times result in reduced OTV mass.
𝑚𝑝𝑟𝑜𝑝 = 𝑚̇𝑡 (#-3)
We also note that the amount of mass savings for different orbit energies varies with flight time.
For a TOF of 351 days, the mass only decreases by about 8 kg between a 400 km circular and a
400 by 10 x 104 km orbit. The mass reduces approximately 21 kg for the same energy change
when the flight time is 196 days. Thus as flight time decreases, we want to increase the
departure altitude as much as possible.
690
680
670
660
TOF=351
650
TOF=316
640
TOF=286
630
TOF=256
620
TOF=226
610
TOF=196
600
590
-31
-29
-27
-25
-23
-21
-19
-17
-15
Orbit Energy (km^2/s^2)
Fig. #-1 Initial OTV Mass as Function of Orbit Energy for mdot=5.6 mg/s
Similar to fig. #-1, we create fig #-2 to display the relationship with the power required. The
power required decreases as the orbit energy increases due to the reduction in thrust necessary to
reach the Moon in the same amount of time. As the TOF decreases, the power requirements
increase due to additional thrust needed. Similar to mass, the savings becomes more pronounced
with shorter flight times.
We create two similar plots for mass flow rates of 7.1 mg/s as well (See Departure Orbit
Variations.xlsx). A mass flow rate of 7.1 mg/s results in larger propellant mass but reduced
power requirements compard to 5.6 mg/s; however, the it is consistent with the relationships
between orbit energy and TOF.
7
6
TOF=351
TOF=316
TOF=286
TOF=256
TOF=226
TOF=196
5
4
3
2
1
-31
-29
-27
-25
-23
-21
Orbit Energy (km^2/s^2)
-19
-17
-15
Fig. #-2 Power Required as Function of Orbit Energy for mdot=5.6 mg/s
We also create relationships for launch capability with orbit energy with LV Performance. Xlsx
in the “Orbit Variations” sheet. In addition to the Dnepr, we include four other launch vehicles.
See fig. #-3 below.
25000
20000
Dnepr
15000
Falcon 9
Delta IV
10000
Taurus
Rockot
5000
0
-35
-33
-31
-29
-27
-25
Orbit Energy (km^2/s^2)
-23
-21
Fig. #-3 Launch Capability as Function of Orbit Energy
-19
-17
Although some of the launch vehicles cost more than the Dnepr (Falcon 9, Delta IV), they can
provide significantly more mass to orbit. By including larger launch vehicles, we determine
whether the increased capability, which corresponds to a reduction in cost per kilogram, offsets
the additional purchase cost.
Cost Comparisons
With the results from the analysis above, we investigate the scenario that yields the lowest cost.
Eq. #-4 displays a model for estimating mission cost. Although this model neglects several
factors, it includes the major contributions to the cost. The total cost is found by adding the
launch cost, power cost, and propellant cost. CLaunch_Vehicle represents the total launch vehicle
cost. The ratio
𝐶𝐿𝑎𝑢𝑛𝑐ℎ_𝑉𝑒ℎ𝑖𝑐𝑙𝑒
𝑚𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑎𝑏𝑙𝑒
represents the cost per kilogram to a particular altitude. The
product of the cost per kg and spacecraft mass (msc) yields the launch cost. Psc represents the
power required for the spacecraft with Prate being the estimated $1000 per Watt. Lastly mprop
represents the propellant required for the OTV with Xrate being the rate of Xenon ($1200 per kg).
Cite this source
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝑚𝑠𝑐 ∗
𝐶𝐿𝑎𝑢𝑛𝑐ℎ_𝑉𝑒ℎ𝑖𝑐𝑙𝑒
𝑚𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑎𝑏𝑙𝑒
+ 𝑃𝑠𝑐 ∗ 𝑃𝑟𝑎𝑡𝑒 + 𝑚𝑝𝑟𝑜𝑝 ∗ 𝑋𝑟𝑎𝑡𝑒 (#-4)
We perform this analysis with Orbit_Altitude_Cost_Comparisons.m. As seen in table #-3, the
orbit energies range between -29.4 and -17.2 km2/s2, so we find the total cost for hundreds of
energy values in this range. Applying the curve fit equations from fig. #-3, we find the mdeliverable
for each energy value. Similarly msc and Psc are found using the curve fit equations from figures
#-1 and #-2. We indicate mprop from the data in Departure Orbit Variations.xlsx and
CLaunch_Vehicle from cite this source. With all the conditions known, we calculate the total cost
from eq. #-4. We repeat this analysis for all flight times and mass flow rates considered in
Departure Orbit Variations.xlsx.
With five launch vehicles and six flight times, we create 30 plots similar to figures #-4 and #-5,
below.
9.5
mdot=5.6
mdot=7.1
9
Total cost (Million $)
8.5
8
7.5
7
6.5
6
5.5
5
0
2000
4000
6000
8000
Apoapsis Altitude (400 x ra) km
10000
12000
Fig. #-4 Total Cost with Varying Altitude for Falcon 9 at 351 day TOF
10
mdot=5.6
mdot=7.1
9.8
9.6
Total cost (Million $)
9.4
9.2
9
8.8
8.6
8.4
8.2
8
0
2000
4000
6000
8000
Apoapsis Altitude (400 x ra) km
10000
12000
Fig. #-5 Total Cost with Varying Altitude for Falcon 9 at 196 day TOF
We see in fig. #-4 that the cheapest mission cost occurs at the lowest altitude of 400 km. The
cost with a mass flow rate of 5.6 mg/s is consistently less than with 7.1 mg/s. Figure #-5 reveals
that departing from a higher parking orbit does reduce the cost as compared to departing from a
400 km circular orbit. As mentioned above, the savings become more pronounced with shorter
flight times, so the cost savings for the OTV mass and power requirements offset the increased
cost of launching to a higher parking orbit. Because the power requirement savings become so
prounced at shorter TOF, the engine operating at 7.1 mg/s is consistently cheaper than at 5.6
mg/s.
The 30 plots created similar to fig. #-4 and #-5 show the results for a particular launch vehicle at
a certain TOF. Running Orbit_Altitude_Cost_Comparisons.m, we determine the minimum
cost for all vehicles analyzed and the TOF and altitude for which that occurs.
14
Delta 4
Falcon 9
Dnepr
Rockot
Taurus
13
Total cost (Million $)
12
11
10
9
8
7
6
5
399
399.5
400
400.5
Apoapsis Altitude (400 x ra) (km)
401
401.5
Fig. #-6 Minimum Total Cost for Each Launch Vehicle by Altitude
We see in fig. #-6 that the overall minimum cost occurs for the Falcon 9 at approximately 5.3
million dollars. Dnepr has the next lowest cost at approximately 5.8 million dollars. For all
launch vehicles, the minimum cost occurs when departing from a 400 km circular orbit. We
seeing fig. #-7 that the minimum cost occurs at a flight time of 351 days.
14
Delta 4
Falcon 9
Dnepr
Rockot
Taurus
13
Total cost (Million $)
12
11
10
9
8
7
6
5
350
350.2 350.4 350.6 350.8 351 351.2 351.4 351.6 351.8
Time of Flight (days)
352
Fig. #-7 Minimum Total Cost for Each Launch Vehicle by Flight Time
Forming Conclusions
We see in this analysis that we can achieve the minimum cost by launching with the Falcon 9.
At this time, the Falcon 9 has not yet launched; therefore, we cannot select it as our launch
vehicle due to reliability issues. Based on cost we confirm our choice of launch vehicle to be the
Dnepr.
This analysis shows that in some cases launching to an eccentric orbit does reduce the total
mission cost, particularly when flight times are shorter. As we see in fig. #-7, the minimum cost
still occurs for longer flight times. Reaching the Moon in less time requires more thrust, which
consequently requires more power. Even departing from higher parking orbits does not offset
the cost of this additional power. If flight time was more of a limitation, this method would be
practical. We have no constraints against a 351 day flight time; therefore we choose the longest
flight time possible. The results show that for a 351 day flight time, the lowest parking orbit
possible yields the minimum cost.
Parking Orbit Selection
Because the analysis in section X.X.X reveals that we must reach 400 km to negate drag effects,
we choose to launch into a 400 km circular orbit with the Dnepr launch vehicle. Barring
propulsion system constraints, we choose the flight time to be as long as possible.