Deltav (velocity increment) budget
The deltav budget provides information on the types of maneuvres to be performed, their number and their size (in terms of velocity change to be accomplished). An example deltav budget is given in the next table.
A deltav budget usually stems from oribital analysis. However, the following data subdivided over three broad categories may help you to generate a preliminary Dv budget for the mission at hand without the need for timeconsuming calculations.
Given the preliminary character of the data, it is advised to add a proper margin to this data. Note that in case you require more accurate data (for instance to be able to reduce the design margin), the values given can be verified through orbit analysis in the later stages of the design. Fundamentals of orbit analysis and design are treated in amongs others TUDelfts' course AE1801.
Launch into Low Earth Orbit (LEO)
Table: Required Delta V for launch into LEO
Manoeuvre  Delta V, km/s 
Launch into LEO (including drag and gravity loss)  9.5 
Impulsive shot manoeuvres
Table: Typical Delta V value(s) for impulsive shot space manoeuvres
Manoeuvre  Delta V, km/s 
Orbit transfer: 

Orbit control:


Orbit control: Drag compensation


Attitude control: 3axis control  26 m/s per year 
Auxiliary tasks:


Constant low thrust space manoeuvres
Because of gravity loss, low thrusttoweight (T/W) propulsion systems suffer a loss in performance equivalent to increasing the effective mission DV. For example, the impulsive DV for a high T/W transfer from LEO to GEO is 4.2 km/s; for a low T/W transfer, the effective DV is about 5.9 km/s. However, even with gravity losses, low T/W propulsion systems can still outperform high T/W impulsive systems, because the very high specific impulse of some low T/W systems (greater than 1000 s) more than compensates for the increase in effective Dv. Table: Typical DV value(s) for constant low thrust (acceleration < 0.001 m/s^{2}) orbit transfer (propellant mass is negligible)
Manoeuvre  Delta V, km/s  Transfer time 
LEO (200 km altitude) to GEO (no plane change)  4.71  a is 0.001 m/s^{2}: 
LEO (200 km altitude) to GEO (including 28 deg. plane change)  5.97  a is 0.001 m/s^{2}: 
LEO to MEO (19150 km altitude; no plane change)  3.83  a is 0.001 m/s^{2}: 
LEO to Earth escape for different values of initial accelerationtolocal gravitational acceleration:



LEO to Lunar orbit  ~8  monthsyear 
LEO to Mars orbit  ~15  ~2.2 years 
^{ Transfer or trip time for constant thrust spiral is is calculated by dividing total propellant mass by mass flow. Total propellant mass is calculated using the rocket equation also known as Tsiolkowsky's equation. In case of negligible propellant mass (constant acceleration), transfer time can be calculated by dividing the velocity change by the acceleration.}
^{ DV for LEO to GEO transfer orbit calculated using T.N. Edelbaum's equation: DV = SQRT(V1 2  2 V1 V2 cos (pi/2 deltai ) + V2 2 ) where V1 is circular velocity initial orbit, V2 is circular velocity final orbit, and deltai is plane change in degrees.}
^{ Values for LEO to Earth escape taken from Rocket Propulsion and Spaceflight dynamics, by Cornelisse, Schoyer & Wakker, for jet exhaust to initial circular velocity ratio equal to 10.}
^{ Value for GTO to Lunar orbit taken from SMART1, by D. Racca}
^{ Value for LEO to Low Lunar Orbit taken from Optimized LowThrust Orbit Transfer for Space Tugs, by Pukniel}
^{ Value for LEO to Mars orbit taken from NASAJPL.}
Figure: Low thrust LEOtoGEO orbit transfer (oneway only) including 28.5 degrees plane change (launch from Eastern Test Range)
Note: A thrusttoweight ratio of 0.1 corresponds to an initial acceleration of about 1 m/s^{2}.
Travel time calculation example:
Taking a(n) initial thrusttoweight (T/W) ratio of 0.001 (initial acceleration of 0.01m/s^{2} ) to achieve a velocity increment of 5.9 km/s and assuming a constant acceleration leads to a transfer or travel time of about 6.8 days. For a spacecraft with an initial weight of 20000 N (~2000 kg mass), we find for the required thrust level a value of 20 N. Assuming a specific impulse of 2000 s this gives then a mass flow of ~ 1g/s. Multiplying by the travel time, this gives a propellant mass of 616 kg. On the other hand, using the rocket equation, we find an initialtoempty mass ratio of 1.34 or an empty mass of 1489 kg. This leads to a propellant mass of 511 kg, which is ~ 100 kg below the propellant mass estimated assuming a constant acceleration. Since mass flow of propellant is constant, we find for the travel time 511kg / 1 g/s = 511000 s = 5.91 days.