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108

Because the earth goes very fast around the sun. If you want to get to the sun, you need to slow down almost completely so that your speed relative to the sun becomes almost zero. If you don't slow down (almost) completely, your probe will miss the sun when you 'drop' it, so it will eventually come back and you'll end up in an elliptical orbit. Kind of like ...


65

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 ...


44

Wouldn't i inevitably spiral to sun surface even if i was faster than 0km/s ? No. On reasonable timescales, an orbit will have a fixed distance of closest approach, called "periapsis." (These timescales shorten if you're close enough to what you're orbiting that an atmosphere can drag you down). You don't really need to "drop in straight line" (which ...


41

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 ...


34

Changing orbits requires delta-v. To reach the Sun, you need to subtract delta-v such that your velocity relative to the Sun is near zero, which allows you to "fall straight down" into the Sun - your required delta-v is nearly equal to your orbital speed. To escape the solar system, you need to add sufficient delta-v in order to reach escape ...


33

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

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. ...


31

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 ...


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

The amount of delta-V needed to get from the Moon to low Moon orbit is only 1.87 km/s or about 1/5 of that needed to get from Earth to LEO. That amount is easily attainable even if your ascent vehicle has a high payload fraction (i.e. only a small part of your vehicle is fuel). Other sections of an Earth-to-Moon-and-back mission require much more delta-V. ...


25

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. ...


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

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 ...


22

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!): $$ \Delta v = I_{sp} \times g \times \ln \frac {Mass_{full}} {Mass_{dry}} $$ ...


20

Escaping the solar system requires adding orbital velocity to the spacecraft. Similarly, getting closer in the solar system requires removing orbital velocity. It turns out Earth is more out of the Sun's gravity well than it's in it. In other words, the simple answer is that Mercury is "farther away" in terms of the change of velocity that's ...


18

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 ...


18

Given the mass costs in terms of consumables and the risk and support costs of keeping humans in space for longer, it seems unlikely that the multiple Earth-Venus flybys used by a lot of robot probes to get out to Jupiter or in to Mercury will ever be a sensible choice for humans. A Jupiter flyby on the way to Saturn is probably a no-brainer apart from maybe ...


18

You need below 2866 m/s of orbital velocity at 1 AU to crash into the Sun. You technically don't need to slow down exactly to 0 m/s relative to the Sun in order to crash into it. Let's calculate the approximate velocity required to graze the "surface" of the Sun. This is an excellent answer on how to calculate apoapsis and periapsis of an orbit. So first, ...


18

There have been a few purely solid-fuel orbital rockets over the years. The first was the Scout from 1961; the only ones in current use are the Long March 11 and the Minotaur/Minotaur-C family. There are more "nearly-pure" solid-fuel rockets such as the Shavit-2. These use three or more solid-fuel stages to get into an orbit, and then a liquid-...


17

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 ...


17

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 ...


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

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 ...


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 ...


16

And note that if you want to hit the sun the cheaper (but slow!) way to do it is to head out. 12.32km/sec will take you to infinity, at infinity a burn of 0m/sec will kill your orbital velocity and you'll come straight in. Of course this will take infinite time, but even going only as far as Jupiter's orbit means you use less energy to drop your periapsis ...


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

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}, $...


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 ...


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