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28

Besides limiting aerodynamic stress and drag losses as you and Antzi mention, using the core engine only at high altitude means the engine can be optimized for low-pressure use by putting a larger nozzle on it. This optimizes expansion of the exhaust, and in the case of GSLV MkIII contributes to a ~6% increase in specific impulse over the sea-level version ...

18

What is needed is not thrust (above a certain basic amount), but delta-v, a function of the type of engine and the ratio between fueled mass and dry mass, according to the Tsiolkovsky equation. The required delta-v from low earth orbit to low lunar orbit is about 4040 meters per second. If I remember rightly, the plan for LK was to have the Soyuz 7K-LOK ...

16

The Tsiolkovsky rocket equation tells you how much delta-V you get for a given exhaust velocity and full/empty mass ratio per stage. Typically you'll want to divide the total 9400m/s requirement into two (or more) stages and work backward from the uppermost stage. Select an appropriate engine for the stage, decide how much dry tankage/structural mass you ...

16

Your assumption that we need max thrust at takeoff is partially wrong. Although right at takeoff you do want max thrust, it might be counterproductive short after: Your rocket and payload are Max G rated. You can't have an arbitrary high acceleration Atmospheric drag is higher at lower altitude and increase with the square of speed. If you go too fast too ...

16

While Russell Borogove mentioned it I think it deserves more focus: Thrust is the wrong thing to measure here. Thrust is how hard you can push--akin horsepower in your car. Your question is thus akin to asking how much horsepower does it take to drive from New York to Los Angeles. Rather, you are interested in whether it has the fuel to get there. With ...

15

The implication of the rocket equation is that linear increases in ∆v require exponential increases in mass ratio for a single stage. There's not strictly a maximum delta-v -- if you redo your plot on a log scale, you'll see that it doesn't go vertical. Getting very high mass ratios (much above 10:1) is difficult to do on a single stage, so there is a ...

13

Your question is about the behavior of the Tsiolkovsky rocket equation itself, in the limit of very small final mass (dry mass). Roughly: "is there any limit to delta-v in theory?" Using MathJax: $$\Delta v=v_e \ln\frac{m_0}{m_f}.$$ If you just look at the velocity ratio and the mass ratio: $$\frac{\Delta v}{v_e}=\ln\frac{m_0}{m_f}=-\ln\frac{m_f}{m_0}, ... 12 For each phase of flight (stage or throttle variation or what have you) you weight the specific impulse (= exhaust velocity) by the propellant mass flow rates (i.e. consumption rates) of the different engines operating. So if, for example the boosters are consuming 5000kg/s at 240s specific impulse and the core is consuming 1000kg/s at 310s, it works out ... 12 I guess that it goes approximately like this: assume that the enthalpy change (I'll denote it \Delta H) is fully converted to the kinetic energy of the exhaust and the exhaust moves with velocity v relative to the engine. Then we have:$$ mv^2/2 = m \Delta H; v = \sqrt{2\Delta H};I_{sp} = \sqrt{2\Delta H}/g.$$But that is assuming that all ... 12 I_{sp} is inversely related to log of mass ratio if delta v is held constant, yes, but that's not how the rocket equation is usually applied. The way the rocket equation is usually applied is that you have a delta-v requirement given by a particular mission -- for example, the 4100 m/s needed to get from low Earth orbit to lunar orbit. Your v_e will be ... 12 Considering fuel consumption or energy expenditure may be misleading, because of the huge change in mass over the flight as fuel is expended. 2/3 of the fuel is expended by the first stage, which only produces 1/3 of the total velocity, for example. Another way to look at the question is through delta-v expenditure; according to Bob Braeunig's simulation ... 12 Any multi-stage rocket design has to obey three rules to achieve good performance: The fuel should have a high specific impulse. A real rocket will achieve a fixed percentage thereof, depending on structural weight and payload. Each stage's payload (which can be another stage) should outweigh the stage's structure, otherwise most of the energy is wasted ... 7 Any excess energy, potential and kinetic, the boosters carry after separation is wasted. Therefore, you want to separate them at the lowest possible altitude and velocity, saving the core rocket's fuel for when it must no longer drag the boosters with it. This holds true for any stage separation and it's the reason we use staged rockets at all. It ... 6 No, that is not quite right. Let's first state and describe the Tsiolkovsky Rocket Equation: \displaystyle \Delta v = V_e \times \ln(\frac{m_i}{m_f}) \Delta v is delta v, the change in velocity in km/s V_e is the effective exhaust velocity in km/s (it's another way of measuring specific impulse) \ln() is just the natural logarithm, or log base e (... 6 If it reaches Earth and Mercury, then it already sounds like the delta-V requirements for Venus capture are beyond hope. If it were to make close approaches to Venus, you might think of using Venus assist to "circularize" its solar orbit at same altitude as Venus around the Sun, putting it in a very elongated orbit, with apoapsis somewhere towards Sun-Venus ... 6 Welcome to the site ThaNoob! I believe the answer to the question you are asking is yes, just pay attention to your choice of coordinate system. You can eliminate the fictitious coriolis and centrifugal force terms through your choice of an inertial reference frame no matter the coordinate system. However, there may still be a "coriolis" and "centrifugal" ... 5 The derivation is based on the rate of change of total momentum$$\frac{dp_{tot}}{dt}$$. For each engine,$$\frac{dp}{dt}=v_{ex}\frac{dm}{dt},$$the exhaust velocity times the rate that the propellants' mass is ejected (kg/sec). You calculate that for each engine, then add them up. So where you would normally have$$m \frac{dV}{dt} = -v_{ex}\frac{...

5

You have to consider each "phase" of flight. For each phase of flight you need to work out the initial mass, the final mass and the overall specific impulse (total thrust/total propellant flow rate). You can then apply the rocket equation to each phase of the flight and add up the results.

5

Check out the diagram at the top of the page that you got the equation from. Let's define our terms. $\dot{m}_e V_e$ is the momentum thrust term $\dot{m}_0 V_0$ is the incoming momentum term $(p_e - p_0) A_e$ is the pressure thrust term The incoming momentum term is important for jet engines because the engine swallows the incoming stream and then ...

5

Rocket equation starts with conservation of momentum: $$\frac{dp}{dt} = m\frac{\partial v}{\partial t} + v\frac{\partial m}{\partial t}$$ But at such a high energy, the rest mass of the proton can be ignored - it's about 1E-11 (as $m_0c^2$) as big as the energy. So drop the second term. $$\frac{dp}{dt} \approx m\frac{\partial v}{\partial t}$$ $$\frac{dv}... 5 These graphs show different visualizations of the Tsiolkovsky rocket equation: (source: wikipedia article) The equation relates change in velocity to engine efficiency and the propellant mass consumed. For example the first graph shows that to achieve a delta-v of 30,000 m/s with an engine that has an I_{sp} of 1000, you will need a mass ratio of 21. ... 4 If you're in vacuum, the power represented by the kinetic energy of the thrust is the thrust force times the exhaust velocity (g times Isp) over two:$$P={g I_{sp}T\over 2} One SSME in vacuum delivers almost seven million horsepower. (I think they might have forgotten to divide by two on this Aerojet Rocketdyne page, or maybe they were calculating ...

4

Ideally if you could design the nozzle to match the exhaust pressure in a vacuum (i.e. nearly zero), the third term drops automatically. If $p_0$ is zero, then $p_e$ would have to go to zero as well because an ideally designed nozzle results in no pressure drag (i.e. ambient freestream pressure and exhaust pressure are the same). In reality, such a nozzle ...

4

That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it. You must specify the flowrate in $\frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $\frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So ...

3

I'm not sure I understand the question completely, but I'll work off of the comment Of course the faster you go, the faster you reach max-q. So you must be able to say "by going this fast, you will reach max-q by this altitude" surely. and take it to suggest that there may be a way to include altitude and therefore density implicitly in an expression ...

3

The graph posted was for a single stage rocket launching vertically, and showed that there comes a point where adding fuel stops increasing the final speed. The author concludes, that for an engine of 440 MN, the limit is "somewhere around $4 \cdot 10^6$ kg of fuel mass" This point corresponds to the point at which the takeoff mass exceeds the thrust of ...

3

When you burn that oxidiser, you are losing $\Delta v$ due to using engines with lower $I_{sp}$, but gaining $\Delta v$ because you are getting rid of unnecessary mass. Then the right question to ask is "How much oxidiser should be burnt to maximise the $\Delta v$?". In some cases the best solution might even be "burn some oxidiser, but not all of it." To ...

3

Is this theoretically correct? Is this practically feasible? I get the same result using the rocket equation, yes. The tug described can move a 47 ton payload through that Hohmann transfer; it will then be completely out of fuel and unable to maneuver anywhere else.

3

$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$. $P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from. The rest of the unit cancellation is confusing because of the ...

3

Wikipedia gives sea level pressure as 101325 Pa, that would make the answer zero, otherwise OK, what you have is close enough. Looks good to me Looks good to me Looks good to me Looks good to me Looks good to me I don't like that equation. Let's check it with some real world data from here. The burn took 156.92 seconds. We can calculate mass flow from Isp ...

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