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If you could miniaturize each component uniformly, you're correct that the rocket equation terms would all balance out and the rocket would be capable of the same delta-v performance. However, the impact of atmospheric drag would be much worse; drag force is proportional to cross sectional area, not to volume. A Saturn V loses less than 1% of its delta v ...

39

You've got a few problems: When a model is scaled to 1:2 it's size, its area drops by 1/4 but its volume by 1/8. This is known as the square-cube scaling problem. So you will have one quarter the drag but one eighth the impulse. Russell expands on this, so I'll address the other issues. The Reynolds number of the air does not scale with the model. So you ...

32

Page 331 in the Shuttle Crew Operations Manual, an official NASA astronaut training document, confirms that The deorbit burn usually decreases the vehicle's orbital velocity anywhere from 200 to 550 fps, depending on orbital altitude. The deorbit burn was not intended to reduce the Orbiter's velocity to a small value, but rather to change its orbital ...

31

Addressing is the sun's mass and other quantities known well enough for this to be absolutely accurate? Well, the key to this is the vis-viva equation in your question. It's not actually important for us to know the mass of the Sun precisely, so long as we know the product $GM$ (another answer makes mention of this). And that product is, of course, the ...

29

If you're just looking for an intuitive handle on it, try this: In circular LEO, your orbital period is about 90 minutes. If you apply a velocity change of 90 m/s, then wait half an orbit -- 45 minutes -- you should expect to be out of position by 90 m/s * 45 min * 60 s/min = 243,000m, or 243km. The distorting effect of Earth's gravity means that the ...

24

The amount by which a spacecraft is able to change its velocity is called it's Δv (delta-velocity) budget. You can calculate the Δv-budget of each stage of a rocket using the Tsiolkovsky rocket equation which reads: $$\Delta v = I_{sp} * 9.81 * \ln \frac {Mass_{full}} {Mass_{dry}}$$ where Isp is the specific impulse ("fuel-efficiency") of the engine. The ...

24

You are correct that Voyager did not change from above escape velocity to below escape velocity shortly after launch. The plot is misleading in that it is just not very accurate right there at 1 AU. The plot lines are kind of thick and a smidge off. Now that I look at it more closely, the escape velocity line in that plot is wrong in other places as well. ...

24

To flyby or impact Venus varies from 3.45 to 3.6 km/s from LEO for the optimal time every 19 months. Mars varies from 3.55 to 3.9 km/s for the optimal time every 26 months. So on average, getting to Venus is a little less energy than getting Mars. But not by much. It could even be a tiny bit more in some years. If you also want to get barely into orbit ...

24

The Vis-viva equation is $$v = \sqrt{ GM \left(\frac{2}{r} - \frac{1}{a} \right) },$$ The $GM$ product for the Sun is 1.327E+20 m^3/s^2. If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s. If you then change to an ellipse with aphelion still at 1 AU and ...

23

In addition to the other answers, you'd also have to take temperature into the equation. Your cryogenics tanks have much thinner thermal insulation, primarily LH2 (LOX was "insulated" by the tank hull + ice, yes, really) but the cryogenic is still at the same low temperature. Much more ice build-up on the outside in relation to the Saturn's total weight. ...

21

This is a large question, but we can certainly boil it down. You need several levels of requisite knowledge. I'll break it down as so: Relevance of Delta v for propellent budget Conversion between gravitational potential and its corresponding velocity The basic physics of Hohmann transfers Non-ideal factors going from surface to orbit Not all of these ...

20

With current technologies, this is unfortunately well outside our reach. However, there is promise on the horizon! Chemical Engines The Tsiolkovsky equation is always your friend when calculating Δv for conventional engines (or your enemy, depending how you look at it!): $$\Delta v = I_{sp} \times g \times \ln \frac {Mass_{full}} {Mass_{dry}}$$ ...

17

Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO. In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to ...

16

Yes, those are the three factors. Your third factor shows up as the $v_\infty$ of the spacecraft relative to the object. The first two are the GM of the object, $\mu$ and the closest approach distance $r$. The $\Delta V$ you can get is: $$2\,v_\infty\over 1+{r\,v_\infty^2\over\mu}$$ As you surmised, lower $v_\infty$ is good since you spend more time ...

16

Jack did a great job describing how to do it using propulsive engines. I have a different answer: We already (almost) planned to do it (inadvertently). The original plan for the recently launched Parker Solar Probe was to do a gravity assist at Jupiter for a subsequent fly-by of the Sun at a relative speed of more than 300 km/s. So, to get to a speed of ...

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 ...

15

There's no conflict here. Because delta-v is a scalar figure, the result of applying velocity changes in different directions to a single object in 3 dimensional space over a period of time doesn't necessarily add up linearly. If a spacecraft applies a delta v of 100 m/s in one direction, then another maneuver of 100 m/s in a perpendicular direction, it ...

14

One thing you're missing seems to be Oberth's effect. To go from LEO to solar system escape velocity, you have to counter Earth's escape velocity, but after that, you get an additional multiplier by doing the burn at a higher initial velocity (at LEO). Your method here also has a problem: Hence an additional 11.2 km/s - 9.5 km/s = 1.7 km/s is needed to ...

14

The image (original at Wikimedia Commons) is only an approximation, as evidenced by the noticeable change in shape of the solar system escape velocity line at 14AU. The line is only defined with three points, and my guess would be that the creator of the graph tried to shape the curve manually. According to Wikipedia the solar system escape velocity at the ...

14

I think that something visual may be of help This is a bit more to scale than most peoples pictures, but the shuttle only orbits at 200 miles, while the earth itself is almost 8000 miles across, so the orbit its more like a thick skin on an orange.. we fly very close to it. In the picture the red dot is the ship, the thick line is earth, thin lines show ...

14

The second table here essentially answers your question. Venus transfer from Low Earth Orbit is 3.5 km/s, Mars transfer is 3.6. This will allow you to impact either body (on Venus you will need to make sure your vehicle is tough enough to actually impact, rather than dissolving in the atmosphere, but that's not really the point). In either case, you can ...

13

Voyager 1: separated from Centaur stage with velocity 18.3 km/s (relative to Earth, Dave Doody, Deep Space Craft: An Overview of Interplanetary Flight, 2010, page 120). Then 76.5N Injection Propulsion Unit of V1 did burn its solid fuel in 43 seconds giving additional 1,7 km/s (?), injecting the spacecraft into Hohmann transfer orbit. IPU was separated, and ...

13

Here's how, using an approximate patched conic technique. The $\Delta V$ using instantaneous (e.g. chemical propulsion) maneuvers can be determined by repeated application of this equation that simply says that the total energy is the sum of the kinetic energy and the potential energy: $\mathcal{E}=\frac{v^2}{2}-\frac{\mu}{r}$ where $\mathcal{E}$ is the ...

13

The minimum $\Delta V$ is effectively a Hohmann transfer. You would de-orbit just enough to barely touch the surface at periapsis, and right at the surface you would do an instantaneous impulsive burn (also known as a "suicide burn") to exactly cancel your velocity relative to the surface. Done! A perfect landing. In the real world however you don't have ...

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

Here's the original source of the diagram: https://www.reddit.com/r/space/comments/29cxi6/i_made_a_deltav_subway_map_of_the_solar_system/ Your concerns are also discussed there. 27000m/s is a very theoretical value, that takes into account the losses of a chemical rocket escaping the thick atmosphere of Venus. This makes it clear that launching a rocket ...

12

The requirements for this can be found at Wikipedia, and here's the general budget. These are measured in terms of Delta-V, which is the only thing that really matters. Let's take this 1 at a time. The total requirement to land on Mars from the Moon is the sum of the lunar escape velocity and the Mars Insertion Orbit. That is 2.8+0.6 km/s, or 3.4 km/s. The ...

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That's a brachistochrone or constant-acceleration trajectory; Heinlein was fond of it and called the turnover the "skew-flip maneuver".

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The delta-v needed from low Earth orbit to a Hohmann transfer orbit with a periapse inside the Sun is actually "just" 21,300 m/s. But there is a better option. A bi-elliptic transfer to just hit a central body is better when the ratio between the orbital radius and the radius of the central body is larger than 4.82. The orbital radius of the Earth divided ...

12

Even setting aside the other problems, rocket engines themselves do not scale indefinitely. You decrease the scale of a rocket motor, you decrease the thickness of the rocket chamber walls, which in turn decreases the bursting pressure of said rocket chamber. If you scale the rocket down enough, this will become a limiting factor. At that point you will ...

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