Hot answers tagged

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 ...


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 ...


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 ...


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 ...


10

It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope. As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched ...


7

It assumes no gravity and no air resistance. For ascent from Earth's surface to LEO, it's typical to expend around 9.4 km/s of ∆v, to reach a final orbital speed of around 7.8 km/s. Most of the loss, around 1400-1500 m/s, is in fighting gravity, with around 100-200 m/s lost to air drag -- the exact numbers vary with the characteristics of the launch ...


6

It doesn't assume anything about your environment, but rather simply indicates what your engines can do. If gravity is a factor some of your delta-v is used to counter gravity rather than actually accelerating your rocket.


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

Good question, I’m also interested if someone has a more specific answer to share! On the following table, you can find the required delta-v per year for different orbits. And about formula, I'm not sure but maybe this page can give you an idea: https://en.wikipedia.org/wiki/Delta-v


5

For any spherical body with a density $\rho$ and radius $R$ and no atmosphere, we can calculate this easily. Let's assume you launch to a very low orbit that just skims the surface of the sphere. You can add 10% or 20% later for a planet like Earth with its atmosphere. Venus would be a lot harder! (so I've asked separately Launch to orbit delta-v penalty ...


5

If you can accelerate to a speed you can slow down from that speed, if you accelerate a ship to .6c and then back to 0c the net acceleration is 0, you aren't breaking any laws of physics. If all things are equal (mass of the ship for example) the delta-v to decelerate the ship to 0c is the same it takes to speed it up to .6c in the first place. As you state ...


5

The amount of propellant required to achieve a certain delta-V is dependent on the ratio between the starting and ending mass of the spacecraft, according to the Tsiolkovsky rocket equation; a given thruster and fuel supply will get you more delta-V on a smaller spacecraft and less delta-V on a larger one. That is, 0.058 km/s per kg is not an inherent ...


4

Specific impulse is a measure of the fuel efficiency of a rocket engine. It’s established by the design of the engine, not by the size of the rocket. Specific impulse of multiple stages is definitely never added together. For a very rough analogy, imagine a car that gets 30 miles to a gallon of gas. Ten of them driving in a convoy do not achieve 300 miles ...


4

These data points are all variations from the optimal Hohmann Transfer, a specific orbital maneuver to go from circular orbit to circular orbit minimizing DeltaV. A certain alignment of the planets results from a certain departure and arrival date. From the geometry of the situation, different planet alignments have different DeltaV requirements and travel ...


4

Going by delta-v, the Moon is pretty close. The reason we needed such a large rocket for the Apollo missions was that we wanted to land a really heavy spacecraft on the Moon. It needed to be that heavy to be able to take off again, and get back to Earth. The CSM+LM weighed about 45 tons at the start of the mission. In contrast, the Voyagers weighed only ...


3

note: This is a very helpful extended comment that may be of use to the OP but it can't currently be posted as a comment until this user reaches 50 reputation points. Oh this is a fantastic question. It is common to fall into the following trap when making these types of calculations. Check carefully your reference frames. Celta V numbers are all relative,...


3

I too am going to start with the orbital speed as a first approximation, but we can do slightly better than that. $$\Delta v = v_{orbit} = \sqrt{\frac{\mu}{r}}$$ If you are using this approximation alone, you would want to use the objects radius for $r$ and not the radius of the low orbit, as that gives a slightly higher cost that better accounts for the ...


3

Some good practices I'm aware of are: As you mentioned, maneuvers are simulated before they are commanded and their effect is evaluated on ground so that thruster parameters and tank filling are updated, so if anything funny is happening during maneuvers this can be identified. If propulsion is electric (which is still not so common), then thrusting is ...


3

I will give my current best shot at this problem, and others should feel free to strengthen the argument with additional mathematics. (Or poke holes!) You ask two questions, I will answer the first as the second has been partially answered by the update. Are there other inclination change strategies that are more efficient for some values of $\alpha$? ...


3

Despite watching a few times it looks like I linked the use of the term "capture" at this point in the video with the use of the term "capture" at 08:20 in the video. But that's about the mechanical capture of Snoopy during a docking maneuver and this is about the gravitational capture of the combined pair in the Earth's gravitational field. What Manley is ...


2

Energy is always conserved, but different oberservers will disagree about how much energy there is, and what forms it takes. Also you have to be sure to include the whole system. Let's return to the Phobos example from the linked question, but be a bit more careful. Suppose what we actually do is use our teraton nukes to split Phobos into two equal halves ...


2

Supplemental answer: Here are the relevant equations and discussion from Sutton, 4th edition.


2

Once relativity becomes a factor, you always have to phrase your statements so as to make it clear what point of view you are taking. From the perspective of the rocket, and assuming you have somehow got unlimited fuel, you can keep accelerating as long as you like. From your perspective, doing so will shorten your journey to a distant star, which can ...


2

It depends on if you are doing a direct to leave the solar system or doing a flyby, but probably. If you are directly leaving the solar system, then the close you are to the Earth's inclination around the Sun, the more your velocity will count. If you go completely perpendicular to that, it will take quite a bit more fuel, as you have to do an inclination ...


2

Here is a rough estimate. tl;dr: raising 445 to 550 km 58 m/s keeping it there 20 m/s bringing it down 112 m/s Total 190 m/s using about two kilograms of krypton, which is about 5 liters at 100 atmospheres. Raising them up to 550 km I looked at the current TLEs and plotted eccentricity versus altitude ...


1

$\rm L_1$ is a similarly circular orbit, and according to this source, the Earth-Sun $\rm L_1$ is $\approx$ 1.5million km from the Earth. Between circular orbits, the cheapest transfer is the Hohman transfer. According to the Wiki page, the required $\Delta v$ for Hohman-transfer orbits is $\Delta v = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2 r_2}{r_1+...


1

In short: These maneuvers would require 1.2c $\Delta v$. Using 0.6c $\Delta v$ for linear acceleration would only result $\frac{1}{\sqrt{1+\frac{v^2}{c^2}}} \approx 0.51c$ . The $c$ limit exists only for the possible speed of objects in reference frames. If you sum speeds, or not related speeds, you can get any big speed num, but it won't have physical ...


1

Escape velocity for 2 body systems is not a super clean issue, and I would avoid using it as it clouds the question a little. However, to answer yours: The dv to change between an Earth-Moon transfer orbit and a 100km lunar orbit is of the order of $700\text{m}\text{s}^{-1}$ (directly, in one orbit). There is lots of info on the options and compromises of ...


Only top voted, non community-wiki answers of a minimum length are eligible