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84-417: A simple example might be the case of a conventional rocket-propelled spacecraft, which achieves thrust by burning fuel. Such a spacecraft's delta- v , then, would be the change in velocity that spacecraft can achieve by burning its entire fuel load. Delta- v is produced by reaction engines , such as rocket engines , and is proportional to the thrust per unit mass and the burn time. It is used to determine

168-419: A Δ v {\displaystyle \Delta v} of 9.8 m/s each second). If the possible rate is only g {\displaystyle g} or less, the maneuver can not be carried out at all with this engine. The power is given by where F {\displaystyle F} is the thrust and a {\displaystyle a} the acceleration due to it. Thus

252-478: A Δ v {\displaystyle \Delta v} of ca. 9.5 km/s (mostly for the speed to be acquired), but if the engine could produce Δ v {\displaystyle \Delta v} at a rate of only slightly more than g , it would be a slow launch requiring altogether a very large Δ v {\displaystyle \Delta v} (think of hovering without making any progress in speed or altitude, it would cost

336-406: 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}}.} If v exh {\displaystyle v_{\text{exh}}}

420-406: 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}}.} If v exh {\displaystyle v_{\text{exh}}}

504-402: A computer. All reaction engines lose some energy, mostly as heat. Different reaction engines have different efficiencies and losses. For example, rocket engines can be up to 60–70% energy efficient in terms of accelerating the propellant. The rest is lost as heat and thermal radiation, primarily in the exhaust. Reaction engines are more energy efficient when they emit their reaction mass when

588-426: A computer. Some effects such as Oberth effect can only be significantly utilised by high thrust engines such as rockets; i.e., engines that can produce a high g-force (thrust per unit mass, equal to delta-v per unit time). In the ideal case m 1 {\displaystyle m_{1}} is useful payload and m 0 − m 1 {\displaystyle m_{0}-m_{1}}

672-418: A mission, for example, when launching from or landing on a planet, the effects of gravitational attraction and any atmospheric drag must be overcome by using fuel. It is typical to combine the effects of these and other effects into an effective mission delta-v . For example, a launch mission to low Earth orbit requires about 9.3–10 km/s delta-v. These mission delta-vs are typically numerically integrated on

756-418: A mission, for example, when launching from or landing on a planet, the effects of gravitational attraction and any atmospheric drag must be overcome by using fuel. It is typical to combine the effects of these and other effects into an effective mission delta-v . For example, a launch mission to low Earth orbit requires about 9.3–10 km/s delta-v. These mission delta-vs are typically numerically integrated on

840-535: A spacecraft's delta- v , then, would be the change in velocity that spacecraft can achieve by burning its entire fuel load. Delta- v is produced by reaction engines , such as rocket engines , and is proportional to the thrust per unit mass and the burn time. It is used to determine the mass of propellant required for the given maneuver through the Tsiolkovsky rocket equation . For multiple maneuvers, delta- v sums linearly. For interplanetary missions, delta- v

924-413: A specific impulse that is both high and fixed such as Ion thrusters have exhaust velocities that can be enormously higher than this ideal, and thus end up powersource limited and give very low thrust. Where the vehicle performance is power limited, e.g. if solar power or nuclear power is used, then in the case of a large v e {\displaystyle v_{e}} the maximum acceleration

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1008-429: A stationary engine does no useful work. Exhausting the entire usable propellant of a spacecraft through the engines in a straight line in free space would produce a net velocity change to the vehicle; this number is termed delta-v ( Δ v {\displaystyle \Delta v} ). If the exhaust velocity is constant then the total Δ v {\displaystyle \Delta v} of

1092-403: A thrust to weight ratio of more than one. To do this with the ion or more theoretical electrical drives, the engine would have to be supplied with one to several gigawatts of power, equivalent to a major metropolitan generating station . From the table it can be seen that this is clearly impractical with current power sources. Alternative approaches include some forms of laser propulsion , where

1176-417: A vehicle can be calculated using the rocket equation, where M is the mass of propellant, P is the mass of the payload (including the rocket structure), and v e {\displaystyle v_{e}} is the velocity of the rocket exhaust . This is known as the Tsiolkovsky rocket equation : For historical reasons, as discussed above, v e {\displaystyle v_{e}}

1260-476: Is Conclusions: These results apply for a fixed exhaust speed. Due to the Oberth effect and starting from a nonzero speed, the required potential energy needed from the propellant may be less than the increase in energy in the vehicle and payload. This can be the case when the reaction mass has a lower speed after being expelled than before – rockets are able to liberate some or all of the initial kinetic energy of

1344-462: Is a measure of the impulse per unit of spacecraft mass that is needed to perform a maneuver such as launching from or landing on a planet or moon, or an in-space orbital maneuver . It is a scalar that has the units of speed . As used in this context, it is not the same as the physical change in velocity of said spacecraft. A simple example might be the case of a conventional rocket-propelled spacecraft, which achieves thrust by burning fuel. Such

1428-428: Is a non-constant function of the amount of fuel left v exh = v exh ( m ) {\displaystyle v_{\text{exh}}=v_{\text{exh}}(m)} the capacity of the reaction control system is computed by the integral ( 5 ). 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

1512-427: Is a non-constant function of the amount of fuel left v exh = v exh ( m ) {\displaystyle v_{\text{exh}}=v_{\text{exh}}(m)} the capacity of the reaction control system is computed by the integral ( 5 ). 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

1596-627: Is achieved, the exhaust stops in space and has no kinetic energy; and the propulsive efficiency is 100% all the energy ends up in the vehicle (in principle such a drive would be 100% efficient, in practice there would be thermal losses from within the drive system and residual heat in the exhaust). However, in most cases this uses an impractical quantity of propellant, but is a useful theoretical consideration. Some drives (such as VASIMR or electrodeless plasma thruster ) actually can significantly vary their exhaust velocity. This can help reduce propellant usage and improve acceleration at different stages of

1680-422: Is also notable that large thrust can reduce gravity drag . 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. From power considerations, it turns out that when applying delta- v in

1764-422: Is also notable that large thrust can reduce gravity drag . 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. From power considerations, it turns out that when applying delta- v in

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1848-445: 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

1932-445: 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

2016-422: Is approximately 3000 m/s, using a Hohmann transfer orbit . For the sake of argument, assume the following thrusters are options to be used: Observe that the more fuel-efficient engines can use far less fuel; their mass is almost negligible (relative to the mass of the payload and the engine itself) for some of the engines. However, these require a large total amount of energy. For Earth launch, engines require

2100-451: Is commonly paraphrased as: "For every action force there is an equal, but opposite, reaction force." Examples include jet engines , rocket engines , pump-jets , and more uncommon variations such as Hall effect thrusters , ion drives , mass drivers , and nuclear pulse propulsion . The discovery of the reaction engine has been attributed to the Romanian inventor Alexandru Ciurcu and to

2184-428: Is even more so when the planet is a large one with a deep gravity field, such as Jupiter. Due to the relative positions of planets changing over time, different delta-vs are required at different launch dates. A diagram that shows the required delta- v plotted against time is sometimes called a porkchop plot . Such a diagram is useful since it enables calculation of a launch window , since launch should only occur when

2268-428: Is even more so when the planet is a large one with a deep gravity field, such as Jupiter. Due to the relative positions of planets changing over time, different delta-vs are required at different launch dates. A diagram that shows the required delta- v plotted against time is sometimes called a porkchop plot . Such a diagram is useful since it enables calculation of a launch window , since launch should only occur when

2352-418: Is for the ideal case again, with no energy lost on heat, etc. The latter causes a reduction of thrust, so it is a disadvantage even when the objective is to lose energy (deceleration). If the energy is produced by the mass itself, as in a chemical rocket, the fuel value has to be v e 2 / 2 {\displaystyle \scriptstyle {v_{\text{e}}^{2}/2}} , where for

2436-399: Is inversely proportional to it. Hence the time to reach a required delta-v is proportional to v e {\displaystyle v_{e}} . Thus the latter should not be too large. On the other hand, if the exhaust velocity can be made to vary so that at each instant it is equal and opposite to the vehicle velocity then the absolute minimum energy usage is achieved. When this

2520-465: Is just the rocket equation applied to the sum of the two maneuvers. This is convenient since it means that delta- v 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. When designing a trajectory, delta- v budget is used as a good indicator of how much propellant will be required. Propellant usage

2604-465: Is just the rocket equation applied to the sum of the two maneuvers. This is convenient since it means that delta- v 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. When designing a trajectory, delta- v budget is used as a good indicator of how much propellant will be required. Propellant usage

Delta-v - Misplaced Pages Continue

2688-415: Is often plotted on a porkchop plot , which displays the required mission delta- v as a function of launch date. Δ v = ∫ t 0 t 1 | T ( t ) | m ( t ) d t {\displaystyle \Delta {v}=\int _{t_{0}}^{t_{1}}{\frac {|T(t)|}{m(t)}}\,dt} where Change in velocity

2772-422: Is reaction mass (this corresponds to empty tanks having no mass, etc.). The energy required can simply be computed as This corresponds to the kinetic energy the expelled reaction mass would have at a speed equal to the exhaust speed. If the reaction mass had to be accelerated from zero speed to the exhaust speed, all energy produced would go into the reaction mass and nothing would be left for kinetic energy gain by

2856-445: Is roughly linear , and little reaction mass is needed. If Δ v {\displaystyle \Delta v} is comparable to v e , then there needs to be about twice as much fuel as combined payload and structure (which includes engines, fuel tanks, and so on). Beyond this, the growth is exponential; speeds much higher than the exhaust velocity require very high ratios of fuel mass to payload and structural mass. For

2940-439: Is sometimes written as where I sp {\displaystyle I_{\text{sp}}} is the specific impulse of the rocket, measured in seconds, and g 0 {\displaystyle g_{0}} is the gravitational acceleration at sea level. For a high delta-v mission, the majority of the spacecraft's mass needs to be reaction mass. Because a rocket must carry all of its reaction mass, most of

3024-407: Is the coordinate acceleration. When thrust is applied in a constant direction ( ⁠ v / | v | ⁠ is constant) this simplifies to: Δ v = | v 1 − v 0 | {\displaystyle \Delta {v}=|v_{1}-v_{0}|} which is simply the magnitude of the change in velocity . However, this relation does not hold in

3108-407: Is the coordinate acceleration. When thrust is applied in a constant direction ( ⁠ v / | v | ⁠ is constant) this simplifies to: Δ v = | v 1 − v 0 | {\displaystyle \Delta {v}=|v_{1}-v_{0}|} which is simply the magnitude of the change in velocity . However, this relation does not hold in

3192-405: Is the exhaust velocity), which is simply the energy to accelerate the exhaust. Comparing the rocket equation (which shows how much energy ends up in the final vehicle) and the above equation (which shows the total energy required) shows that even with 100% engine efficiency, certainly not all energy supplied ends up in the vehicle – some of it, indeed usually most of it, ends up as kinetic energy of

3276-432: Is the specific energy of the rocket (potential plus kinetic energy) and Δ v {\displaystyle \Delta v} is a separate variable, not just the change in v {\displaystyle v} . In the case of using the rocket for deceleration; i.e., expelling reaction mass in the direction of the velocity, v {\displaystyle v} should be taken negative. The formula

3360-677: Is useful in many cases, such as determining the change in momentum ( impulse ), where: Δ p = m Δ v {\displaystyle \Delta \mathbf {p} =m\Delta \mathbf {v} } , where p {\displaystyle \mathbf {p} } is momentum and m is mass. In the absence of external forces: Δ v = ∫ t 0 t 1 | v ˙ | d t {\displaystyle \Delta {v}=\int _{t_{0}}^{t_{1}}\left|{\dot {v}}\right|\,dt} where v ˙ {\displaystyle {\dot {v}}}

3444-677: Is useful in many cases, such as determining the change in momentum ( impulse ), where: Δ p = m Δ v {\displaystyle \Delta \mathbf {p} =m\Delta \mathbf {v} } , where p {\displaystyle \mathbf {p} } is momentum and m is mass. In the absence of external forces: Δ v = ∫ t 0 t 1 | v ˙ | d t {\displaystyle \Delta {v}=\int _{t_{0}}^{t_{1}}\left|{\dot {v}}\right|\,dt} where v ˙ {\displaystyle {\dot {v}}}

Delta-v - Misplaced Pages Continue

3528-574: The reaction mass does not provide the energy required to accelerate it, with the energy instead being provided from an external laser or other beam-powered propulsion system. Small models of some of these concepts have flown, although the engineering problems are complex and the ground-based power systems are not a solved problem. Delta-v (physics) Delta- v (also known as " change in velocity "), symbolized as Δ v {\textstyle {\Delta v}} and pronounced /dɛltə viː/ , as used in spacecraft flight dynamics ,

3612-508: The vacuum I sp is used for calculating the vehicle's delta- v capacity via the rocket equation . In addition, the costs for atmospheric losses and gravity drag are added into the delta- v budget when dealing with launches from a planetary surface. Orbit maneuvers are made by firing a thruster to produce a reaction force acting on the spacecraft. The size of this force will be where The acceleration v ˙ {\displaystyle {\dot {v}}} of

3696-507: The vacuum I sp is used for calculating the vehicle's delta- v capacity via the rocket equation . In addition, the costs for atmospheric losses and gravity drag are added into the delta- v budget when dealing with launches from a planetary surface. Orbit maneuvers are made by firing a thruster to produce a reaction force acting on the spacecraft. The size of this force will be where The acceleration v ˙ {\displaystyle {\dot {v}}} of

3780-500: The French journalist Just Buisson  [ fr ; ro ] . For all reaction engines that carry on-board propellant (such as rocket engines and electric propulsion drives) some energy must go into accelerating the reaction mass. Every engine wastes some energy, but even assuming 100% efficiency, the engine needs energy amounting to (where M is the mass of propellent expended and V e {\displaystyle V_{e}}

3864-507: The Soyuz spacecraft makes a de-orbit from the ISS in two steps. First, it needs a delta- v of 2.18 m/s for a safe separation from the space station. Then it needs another 128 m/s for reentry . Reaction engine A reaction engine is an engine or motor that produces thrust by expelling reaction mass (reaction propulsion), in accordance with Newton's third law of motion . This law of motion

3948-554: The direction of the velocity the specific orbital energy gained per unit delta- v is equal to the instantaneous speed. This is called the Oberth effect. For example, a satellite in an elliptical orbit is boosted more efficiently at high speed (that is, small altitude) than at low speed (that is, high altitude). Another example is that when a vehicle is making a pass of a planet, burning the propellant at closest approach rather than further out gives significantly higher final speed, and this

4032-505: The direction of the velocity the specific orbital energy gained per unit delta- v is equal to the instantaneous speed. This is called the Oberth effect. For example, a satellite in an elliptical orbit is boosted more efficiently at high speed (that is, small altitude) than at low speed (that is, high altitude). Another example is that when a vehicle is making a pass of a planet, burning the propellant at closest approach rather than further out gives significantly higher final speed, and this

4116-481: The energy produced by the fuel minus the energy gain of the reaction mass. The larger the speed of the rocket, the smaller the energy gain of the reaction mass; if the rocket speed is more than half of the exhaust speed the reaction mass even loses energy on being expelled, to the benefit of the energy gain of the rocket; the larger the speed of the rocket, the larger the energy loss of the reaction mass. We have where ϵ {\displaystyle \epsilon }

4200-407: The energy required is 50 MJ per kg reaction mass, only 20 MJ is used for the increase in speed of the reaction mass. The remaining 30 MJ is the increase of the kinetic energy of the rocket and payload. In general: Thus the specific energy gain of the rocket in any small time interval is the energy gain of the rocket including the remaining fuel, divided by its mass, where the energy gain is equal to

4284-423: The exhaust. If the specific impulse ( I s p {\displaystyle I_{sp}} ) is fixed, for a mission delta-v, there is a particular I s p {\displaystyle I_{sp}} that minimises the overall energy used by the rocket. This comes to an exhaust velocity of about ⅔ of the mission delta-v (see the energy computed from the rocket equation ). Drives with

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4368-400: The flight. However the best energetic performance and acceleration is still obtained when the exhaust velocity is close to the vehicle speed. Proposed ion and plasma drives usually have exhaust velocities enormously higher than that ideal (in the case of VASIMR the lowest quoted speed is around 15 km/s compared to a mission delta-v from high Earth orbit to Mars of about 4 km/s ). For

4452-424: The fuel value also the mass of the oxidizer has to be taken into account. A typical value is v e {\displaystyle v_{\text{e}}} = 4.5 km/s, corresponding to a fuel value of 10.1   MJ/kg. The actual fuel value is higher, but much of the energy is lost as waste heat in the exhaust that the nozzle was unable to extract. The required energy E {\displaystyle E}

4536-403: The general case: if, for instance, a constant, unidirectional acceleration is reversed after ( t 1 − t 0 )/2 then the velocity difference is 0, but delta- v is the same as for the non-reversed thrust. For rockets, "absence of external forces" is taken to mean the absence of gravity drag and atmospheric drag, as well as the absence of aerostatic back pressure on the nozzle, and hence

4620-403: The general case: if, for instance, a constant, unidirectional acceleration is reversed after ( t 1 − t 0 )/2 then the velocity difference is 0, but delta- v is the same as for the non-reversed thrust. For rockets, "absence of external forces" is taken to mean the absence of gravity drag and atmospheric drag, as well as the absence of aerostatic back pressure on the nozzle, and hence

4704-599: The initially-expended reaction mass goes towards accelerating reaction mass rather than payload. If the rocket has a payload of mass P , the spacecraft needs to change its velocity by Δ v {\displaystyle \Delta v} , and the rocket engine has exhaust velocity v e , then the reaction mass M which is needed can be calculated using the rocket equation and the formula for I sp {\displaystyle I_{\text{sp}}} : For Δ v {\displaystyle \Delta v} much smaller than v e , this equation

4788-482: The integration variable from time t to the spacecraft mass m one gets Assuming v exh {\displaystyle v_{\text{exh}}\,} to be a constant not depending on the amount of fuel left this relation is integrated to which is the Tsiolkovsky rocket equation . 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

4872-482: The integration variable from time t to the spacecraft mass m one gets Assuming v exh {\displaystyle v_{\text{exh}}\,} to be a constant not depending on the amount of fuel left this relation is integrated to which is the Tsiolkovsky rocket equation . 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

4956-495: 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

5040-447: 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

5124-487: 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 . For examples of calculating delta- v , see Hohmann transfer orbit , gravitational slingshot , and Interplanetary Transport Network . It

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5208-487: 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 . For examples of calculating delta- v , see Hohmann transfer orbit , gravitational slingshot , and Interplanetary Transport Network . It

5292-488: 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

5376-488: 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

5460-607: The mass of propellant required for the given maneuver through the Tsiolkovsky rocket equation . For multiple maneuvers, delta- v sums linearly. For interplanetary missions, delta- v is often plotted on a porkchop plot , which displays the required mission delta- v as a function of launch date. Δ v = ∫ t 0 t 1 | T ( t ) | m ( t ) d t {\displaystyle \Delta {v}=\int _{t_{0}}^{t_{1}}{\frac {|T(t)|}{m(t)}}\,dt} where Change in velocity

5544-721: The mass ratios of the maneuvers, and v 1 , v 2 are the delta- v of the first and second maneuvers m 1 m 2 = e V 1 / V e e V 2 / V e = e V 1 + V 2 V e = e V / V e = M {\displaystyle {\begin{aligned}m_{1}m_{2}&=e^{V_{1}/V_{e}}e^{V_{2}/V_{e}}\\&=e^{\frac {V_{1}+V_{2}}{V_{e}}}\\&=e^{V/V_{e}}=M\end{aligned}}} where V = v 1 + v 2 and M = m 1 m 2 . This

5628-721: The mass ratios of the maneuvers, and v 1 , v 2 are the delta- v of the first and second maneuvers m 1 m 2 = e V 1 / V e e V 2 / V e = e V 1 + V 2 V e = e V / V e = M {\displaystyle {\begin{aligned}m_{1}m_{2}&=e^{V_{1}/V_{e}}e^{V_{2}/V_{e}}\\&=e^{\frac {V_{1}+V_{2}}{V_{e}}}\\&=e^{V/V_{e}}=M\end{aligned}}} where V = v 1 + v 2 and M = m 1 m 2 . This

5712-533: The mission is within the capabilities of the vehicle to be employed. Delta- v needed for various orbital manoeuvers using conventional rockets; red arrows show where optional aerobraking can be performed in that particular direction, black numbers give delta- v in km/s that apply in either direction. Lower-delta- v transfers than shown can often be achieved, but involve rare transfer windows or take significantly longer, see: Orbital mechanics § Interplanetary Transport Network and fuzzy orbits . For example

5796-530: The mission is within the capabilities of the vehicle to be employed. Delta- v needed for various orbital manoeuvers using conventional rockets; red arrows show where optional aerobraking can be performed in that particular direction, black numbers give delta- v in km/s that apply in either direction. Lower-delta- v transfers than shown can often be achieved, but involve rare transfer windows or take significantly longer, see: Orbital mechanics § Interplanetary Transport Network and fuzzy orbits . For example

5880-439: The nodes this approximation is fair. 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

5964-439: The nodes this approximation is fair. 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

6048-399: The object. The total delta- v needed is a good starting point for early design decisions since consideration of the added complexities are deferred to later times in the design process. 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

6132-399: The object. The total delta- v needed is a good starting point for early design decisions since consideration of the added complexities are deferred to later times in the design process. 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

6216-424: The propellant. Also, for a given objective such as moving from one orbit to another, the required Δ v {\displaystyle \Delta v} may depend greatly on the rate at which the engine can produce Δ v {\displaystyle \Delta v} and maneuvers may even be impossible if that rate is too low. For example, a launch to Low Earth orbit (LEO) normally requires

6300-512: The rocket and payload. However, if the rocket already moves and accelerates (the reaction mass is expelled in the direction opposite to the direction in which the rocket moves) less kinetic energy is added to the reaction mass. To see this, if, for example, v e {\displaystyle v_{e}} =10 km/s and the speed of the rocket is 3 km/s, then the speed of a small amount of expended reaction mass changes from 3 km/s forwards to 7 km/s rearwards. Thus, although

6384-435: The spacecraft and the orbit can easily be propagated with a numerical algorithm including also this thruster force. 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

6468-434: The spacecraft and the orbit can easily be propagated with a numerical algorithm including also this thruster force. 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

6552-492: The spacecraft caused by this force will be where m is the mass of the spacecraft During the burn the mass of the spacecraft will decrease due to use of fuel, the time derivative of the mass being If now the direction of the force, i.e. the direction of the nozzle , is fixed during the burn one gets the velocity increase from the thruster force of a burn starting at time t 0 {\displaystyle t_{0}\,} and ending at t 1 as Changing

6636-492: The spacecraft caused by this force will be where m is the mass of the spacecraft During the burn the mass of the spacecraft will decrease due to use of fuel, the time derivative of the mass being If now the direction of the force, i.e. the direction of the nozzle , is fixed during the burn one gets the velocity increase from the thruster force of a burn starting at time t 0 {\displaystyle t_{0}\,} and ending at t 1 as Changing

6720-416: The theoretically possible thrust per unit power is 2 divided by the specific impulse in m/s. The thrust efficiency is the actual thrust as percentage of this. If, e.g., solar power is used, this restricts a {\displaystyle a} ; in the case of a large v e {\displaystyle v_{\text{e}}} the possible acceleration is inversely proportional to it, hence

6804-520: The time to reach a required delta-v is proportional to v e {\displaystyle v_{\text{e}}} ; with 100% efficiency: Examples: Thus v e {\displaystyle v_{\text{e}}} should not be too large. The power to thrust ratio is simply: Thus for any vehicle power P, the thrust that may be provided is: Suppose a 10,000 kg space probe will be sent to Mars. The required Δ v {\displaystyle \Delta v} from LEO

6888-483: The total delta- v needed for a given spaceflight, as well as designing spacecraft that are capable of producing larger delta- v . Increasing the delta- v provided by a propulsion system can be achieved by: Because the mass ratios apply to any given burn, when multiple maneuvers are performed in sequence, the mass ratios multiply. Thus it can be shown that, provided the exhaust velocity is fixed, this means that delta- v can be summed: When m 1 , m 2 are

6972-483: The total delta- v needed for a given spaceflight, as well as designing spacecraft that are capable of producing larger delta- v . Increasing the delta- v provided by a propulsion system can be achieved by: Because the mass ratios apply to any given burn, when multiple maneuvers are performed in sequence, the mass ratios multiply. Thus it can be shown that, provided the exhaust velocity is fixed, this means that delta- v can be summed: When m 1 , m 2 are

7056-401: The vehicle is travelling at high speed. This is because the useful mechanical energy generated is simply force times distance, and when a thrust force is generated while the vehicle moves, then: where F is the force and d is the distance moved. Dividing by length of time of motion we get: Hence: where P is the useful power and v is the speed. Hence, v should be as high as possible, and

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