Delta-v – wikipedia gas vs electric oven for baking cakes


If for example 20% of the launch mass is fuel giving a constant v exh {\displaystyle v_{\text{exh}}\,} of 2100 m/s (a typical value for a hydrazine thruster) the capacity of the reaction control system is Δ v = 2100 ln ⁡ ( 1 0.8 ) m/s = 460 m/s . {\displaystyle \Delta {v}=2100\ \ln \left({\frac {1}{0.8}}\right)\,{\text{m/s}}=460\,{\text{m/s}}.}

The acceleration ( 2) caused by the thruster force is just an additional acceleration to be added to the other accelerations (force per unit mass) affecting the spacecraft and the orbit can easily be propagated with a numerical algorithm including also this thruster force. [2] But for many purposes, typically for studies or for maneuver optimization, they are approximated by impulsive maneuvers as illustrated in figure 1 with a Δ v {\displaystyle \Delta {v}\,} as given by ( 4). Like this one can for example use a "patched conics" approach modeling the maneuver as a shift from one Kepler orbit to another by an instantaneous change of the velocity vector.

This approximation with impulsive maneuvers is in most cases very accurate, at least when chemical propulsion is used. For low thrust systems, typically electrical propulsion systems, this approximation is less accurate. But even for geostationary spacecraft using electrical propulsion for out-of-plane control with thruster burn periods extending over several hours around the nodes this approximation is fair. Production [ edit ]

Delta- v is typically provided by the thrust of a rocket engine, but can be created by other engines. The time-rate of change of delta- v is the magnitude of the acceleration caused by the engines, i.e., the thrust per total vehicle mass. The actual acceleration vector would be found by adding thrust per mass on to the gravity vector and the vectors representing any other forces acting on the object.

The rocket equation shows that the required amount of propellant dramatically increases with increasing delta- v. Therefore, in modern spacecraft propulsion systems considerable study is put into reducing the total delta- v needed for a given spaceflight, as well as designing spacecraft that are capable of producing larger delta- v.

This is convenient since it means that delta- v ‘s can be calculated and simply added and the mass ratio calculated only for the overall vehicle for the entire mission. Thus delta- v is commonly quoted rather than mass ratios which would require multiplication. Delta- v budgets [ edit ]

When designing a trajectory, delta- v budget is used as a good indicator of how much propellant will be required. Propellant usage is an exponential function of delta- v in accordance with the rocket equation, it will also depend on the exhaust velocity.

It is not possible to determine delta- v requirements from conservation of energy by considering only the total energy of the vehicle in the initial and final orbits since energy is carried away in the exhaust (see also below). For example, most spacecraft are launched in an orbit with inclination fairly near to the latitude at the launch site, to take advantage of the Earth’s rotational surface speed. If it is necessary, for mission-based reasons, to put the spacecraft in an orbit of different inclination, a substantial delta- v is required, though the specific kinetic and potential energies in the final orbit and the initial orbit are equal.

When rocket thrust is applied in short bursts the other sources of acceleration may be negligible, and the magnitude of the velocity change of one burst may be simply approximated by the delta- v. The total delta- v to be applied can then simply be found by addition of each of the delta- v’s needed at the discrete burns, even though between bursts the magnitude and direction of the velocity changes due to gravity, e.g. in an elliptic orbit.

Delta- v is also required to keep satellites in orbit and is expended in propulsive orbital stationkeeping maneuvers. Since the propellant load on most satellites cannot be replenished, the amount of propellant initially loaded on a satellite may well determine its useful lifetime. Oberth effect [ edit ]

• ^ Can be the case for a "blow-down" system for which the pressure in the tank gets lower when fuel has been used and that not only the fuel rate ρ {\displaystyle \rho \,} but to some lesser extent also the exhaust velocity v exh {\displaystyle v_{\text{exh}}\,} decreases.

• ^ The thrust force per unit mass being f ( t ) m ( t ) = v exh ( t ) m ˙ ( t ) m ( t ) {\displaystyle {\frac {f(t)}{m(t)}}=v_{\text{exh}}(t){\frac {{\dot {m}}(t)}{m(t)}}\,} where f ( t ) {\displaystyle f(t)\,} and m ( t ) {\displaystyle m(t)\,} are given functions of time t {\displaystyle t\,}

• ^ "Delta-V Calculator". Archived from the original on March 12, 2000. Gives figures of 8.6 from Earth’s surface to LEO, 4.1 and 3.8 for LEO to lunar orbit (or L5) and GEO resp., 0.7 for L5 to lunar orbit, and 2.2 for lunar orbit to lunar surface. Figures are said to come from Chapter 2 of Space Settlements: A Design Study Archived 2001-11-28 at the Library of Congress on the NASA website (dead link).

• ^ The sum of LEO to GTO and GTO to GEO should equal LEO to GEO. The precise figures depend on what low earth orbit is used. According to Geostationary transfer orbit, the speed of a GTO at perigee can be just 9.8 km/s. This corresponds to an LEO at about 700 km altitude, where its speed would be 7.5 km/s, giving a delta-v of 2.3 km/s. Starting from a lower LEO would require more delta-v to get to GTO, but then the total for LEO to GEO would have to be higher.

• ^ The speed of the earth going around the sun is 29.78 km/s, equivalent to a specific kinetic energy of 443 km 2/s 2. One must add to this the potential energy depth of LEO, about 61 km 2/s 2, to give a kinetic energy close to Earth of 504 km 2/s 2, corresponding to a speed of 31.8 km/s. Since the LEO speed is 7.8 km/s, the delta-v is only 24 km/s. It would be possible to reach the sun with less delta-v using gravity assists. See Parker Solar Probe.