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24

Let's take, for example, the Saturn V rocket. Payload can be 140 tonnes to LEO. The third stage has a dry weight of about 10 tonnes, 119 tonnes fuelled. The second stage has a dry weight of about 36 tonnes, 480 tonnes fuelled. The first stage has a dry weight of about 131 tonnes, 2300 tonnes fuelled. Each stage doesn't care if it is lifting payload, ...


20

According to Wikipedia, Saturn V could launch 48600 kg to translunar injection. From there, you need about 2410m/s of ∆v to soft-land on the moon. Let's take a little additional fuel for safety margin and call it 2700m/s. Per the rocket equation, assuming you're using a rocket that uses storable hypergolic propellants with a specific impulse of 312s (...


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 ...


14

It typically takes a total expenditure of 9400-10000 meters per second of delta-v to reach LEO. Per the rocket equation, delta-v is proportional to the log of the propellant mass ratio, but also proportional to the exhaust velocity of the rocket engines or their specific impulse. Solid rocket boosters have relatively low specific impulse: 275 sec for ...


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}, $...


13

I'm not up to a complete exhaustive survey of every possible contestant, so I've focused on relatively recent orbiters. I found a few that beat Cassini. The figures I've found so far are occasionally a bit contradictory or squirrelly -- masses are rounded differently here and there, some sources give the design tankage while others give the actual flown ...


11

There are three components to a rocket (in some sense). Payload, vehicle and fuel (where vehicle includes fuel tanks, engines, etc. and fuel includes oxidiser). What you want, of course, is payload to make up as much of the mass as possible. However, given the orbit you want to reach and the type of fuel you are using, the ratio of fuel to everything else is ...


10

In almost all cases, the overall mass versus time curve for a multistage launcher will approximate an exponential decay curve; this is the nature of the rocket equation. Within each stage, if all the stage's engines fire at a continuous throttle setting, the mass decline over that portion of the flight will be linear. Many launchers throttle down over the ...


8

Paul's comment is correct. As always, when you try to figure out what is "better", you have to define what you are talking about. In this case, the author means better as in Higher delta V The higher your propellant mass fraction, the higher your delta V.


8

It's always difficult to make apples-to-apples comparisons between the space shuttle and other launchers, because the orbiter is ambiguously part launcher and part payload. This is compounded by the broadness of the term "LEO"; shuttle payloads went to a variety of altitudes and inclinations. However, since the title of the question specifies "mass to LEO" ...


7

You'll lose more energy hauling that hose up than you could possibly save by pumping fuel in. A hose full of fuel is going to be heavy, not to mention you'll have quite a time keeping it out of the rocket exhaust.


7

Even better than refilling a rocket through hoses as it rises is to put the rocket on a flying platform. It is quite possible to lift a rocket up to 50 miles or higher with the right platform. This platform is of course, known as "the first stage".


7

There exist one way for this to work As pointed out by others before, a hose is heavy to carry along. However, if the propellant station had the same altitude and velocity as the rocket, it may be pretty simple engineering. And the obvious way of propelling the tank is by the means of a rocket engine. This is known as "propellant cross feed" or "asparagus ...


6

How is payload to LEO calculated? These days, they run simulations on everything. The payload to LEO is either an analytical assessment of the payload using worst case assumptions, or else some threshold of simulation. Basically, the rocket is designed to meet a minimum LEO threshold, using a nominal orbit. These values are broken down into what every piece ...


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

And here is the answer from the Apollo era perspective: "Popular Science" text from 1966: "What We'll Do on the Moon", written by dr. Wernher von Braun himself! - Google Books link Unmanned cargo landers, launched by Saturn V rockets, could soft-land 30,000 pounds apiece on moon - opening way to large stationary and mobile lunar labs and, ultimately, ...


6

All good answers and comments so far. The question summary, "Conceptually, the lower the propellant mass fraction the better, right?" is correct if you abandon the assumption of a specified propellant and the propulsion system achieves the mission's required ∆V; but see the concluding paragraph below. In addition to the mass ratio the rocket equation has ...


5

Should be pretty close to the payload to LEO: 22.8 tons. Maybe a bit more if you design a new second stage with larger tanks and a nose cone.


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. ...


5

The simple solution is to launch from much higher up. Either an air launched rocket or launch from a raised platform. (Both have been described, and one has been used.) Or just begin the launch with a supergun. Really, there are so many engineering problems associated with hoses... just put the fuel in detachable boosters. (If you think about it, ...


5

Additional problems I see are having a pump that can move the required amount of fuel at the rate required to the height required while overcoming shock losses as the hose unrolls and travels upwards. Also, a trailing hose would be problematic with rocket stability and streamlining the rocket as it moves through the air.


5

The high price of payload to orbit is precisely the point: Pressure stabilization means that the entire transport of the tanks has to be done with the greatest care. Any dropped tool can easily punch a hole through the paper-thin tank walls. They also have to be constantly wetted with an oil film to prevent corrosion. The additional labour adds greatly to ...


5

Yes, the Wikipedia answer applies to your problem and is correct. I get $v_e=0.6275\Delta V$. (Your "ve/v of about 1.6" is backwards -- it should be v/Ve, as shown correctly on the Wikipedia plot.) This gives a mass ratio of about $4.9$. There is no closed form for that number, $x\approx 0.6275$. You need to iteratively solve ${1\over 2}=x\left(1-e^{-1/...


4

In addition to the other answers, larger rockets are more efficient: Larger tanks have a better volume-to-surface ratio, so there's less structural weight per kg of content. Some parts on a rocket don't scale up when the rocket gets larger. For instance the guidance system on a Saturn V isn't 14 times larger than that of a Falcon 9. And speaking of the ...


4

I believe Russell Borogove's answer is correct but maybe could be stated a little more directly: The reason your 2 factors of payload mass and liftoff mass don't correlate is that you are ignoring the other factor in the equation. The 3 main factors in the rocket equation are mass ratio, delta-v, and specific impulse. Since your delta-v is basically fixed ...


4

What the equations I used completely ignore is initial thrust to gross launch mass which'd surely affect gravity drag? ... What is the assumption behind the 9.7 km/s delta-v on the Wikipedia page as to gravity drag fraction and initial launch acceleration? 9.7km/s is towards the high end of delta-v to orbit requirements. It varies with both the aerodynamics ...


3

We need a fair number of approximations here... Let's ignore atmospheric losses, since according to Russell Borogov's comment these are dwarfed by gravity losses. Let's also assume that the rocket instantly gets to orbital altitude and velocity, so that there is no influence of the trajectory on how efficient we can be (more on that later). The difference ...


3

Sorry, but this comment is rather long. Regarding the Saturn V LEO payload questions... Check out this source: http://forum.nasaspaceflight.com/index.php?topic=12519.20 The Saturn V launch vehicle was upgraded throughout its lifetime. Its engines were uprated and numerous measured were taken to reduce the weight of the rocket. Thus, by the time Apollo 17 ...


2

I don't know if that adds anything useful, but lets see: $$ \Delta v = c_e* ln(m_0/m_b) = c_e* ln(\sigma/\mu_L) $$ with $\Delta v$ : characteristic velocity (constant for fixed orbits) $c_e$ : velocity of the rocket exhaust $m_0$ : initial total mass $m_b$ : burnout mass $\sigma$ : structural mass ratio [σ=(mM+mS)/m0] $m_M$ : mass of rocket motor $m_S$ : ...


2

How far a rocket can go is measured in $\Delta v$ (delta-v). In orbital mechanics, you reach places by changing your velocity and thus your orbit. $\Delta v$ is given by the Tsiolkovsky rocket equation $ {\displaystyle \Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}} $ where $m_0$ is the initial total mass, including propellant ("wet mass"). $m_f$ is the ...


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