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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-...


12

I've got a set of Keplerian orbital elements $e_0$, $a_0$, $i_0$, $\omega_0$, $\Omega_0$, and $\theta_0$, and I'd like to get to a different orbit with orbital elements $e$, $a$, $i$, $\omega$, $\Omega$, and $\theta$. How do I calculate (a) the amount of delta-v I'll need for this maneuver or set of maneuvers, and (b) which maneuver or set of maneuvers I ...


10

The SS-520 is a solid fuel rocket and is described in detail in my answer to Do launchers using only solid propellant exist? SS-520-5 has put the Tricom cubesat into orbit, therefore it also qualifies as a "completely solid fuelled orbital rocket". The second attempt at becoming the smallest orbital launching rocket was made on 3 February 2018. ...


9

What you're looking for is Lambert's problem, which is used both for trajectory design and orbit determination, and to produce porkchop plots. Your hunch that this is not a simple problem is correct. pykep has a solver for Lambert's problem that supports multiple revolutions as well as solvers for various related problems such as low-thrust trajectories.


8

I can't say how all-solid launchers compare cost-wise with liquid or mixed launchers, but they're viable. The Minotaur I and Minotaur IV rockets are all-solid, four-stage orbital launchers. The R&D and production costs for the Minotaurs were offset to some degree by the use of converted ICBMs. The Vega rocket is a 4-stage orbital launcher with solid ...


8

Almost certainly the vehicle with the most Delta-v post-booster separation was the Dawn spacecraft, with an incredible 11 km/s! Put another way, that is the same amount as the rocket that launched it roughly. This is because of the unique nature of an ion drive, being vastly more efficient than chemical rockets. If it were a chemical rocket, it would have to ...


7

Another way of saying delta V to LEO = 10 km/s is this: To be in orbit, a thing needs to move horizontally at a speed of at least 7.8 km/s To get to orbit, the rocket delivering the thing will have to get up to that speed, and get out of the atmosphere While it does that, the drag from gravity and air resistance make it have to exert as much force as if it ...


6

This is only a partial answer, but there are two major factors that make the performance of the three vehicles not comparable at all. There are two reasons why Starship is so much worse for missions with a large C3, and both are design features: First, it's built to land (repeatedly) on planets. This makes it a lot more sturdy than other second stages. The ...


6

ISAS, now a part of JAXA, has a long history of using solid fuel to launch scientific satellites. Eight different vehicles, not including the SS-520.


5

The rocket equation is meant to work with constant specific impulse. If you want to stay with the Rocket Equation, you can 'split' any stage into more 'virtual' stages (where the initial mass of the next stage is equal arbitrarily chosen dry mass of the previous one), find what delta-V you need to reach roughly 10km altitude and generate a 'virtual' stage ...


5

The simple theoretical delta V to achieve a particular orbit is constant, but in practice (or on more detailed analysis) Delta V is not constant for a number of reasons. For launches from the surface of a moon or planet delta V will be greater than the theoretical value because: A rocket will not be able to achieve orbit instantly, it will need to ...


5

As $\Delta v$ is just change in velocity, we can just integrate the norm of the acceleration function over time: $$\Delta v = \int|\mathbf{a}(t)| dt$$ You're out of luck getting a closed form of that integral though. As far as analytical solutions goes, we can note that at $t = \frac{\pi}{2}$, all of $a_x$, $a_y$ and $a_z$ are maxed out, and hence $\Delta v &...


5

First of all: great observation! This is indeed the reason why pressure fed rocket engines are limited in possible chamber pressure, the added weight from the tanks isn't worth it at a certain point. Which is why we have pump fed rocket engines. Question 1: Some equations from Ideal Rocket Theory: Specific Impulse is the equivalent velocity divided by ...


5

Do you really want to compute a new TLE, or just a new orbit? The TLE format itself is a significant problem, so it's best to avoid if possible. If you just need to look for changes in the orbit state, you should use SGP4 to convert into position and velocity, propagate the state with and without the maneuver using something other than SGP4, and convert ...


4

To enjoy the greatest advantage from the Earth's rotation, you want two things: Launch from as close to the equator as possible. Launch as close to directly east as possible. The first part is fairly simple. The rotational velocity is proportional to $\cos(latitude)$, so the Guiana Space Centre gets over 99% of the effect, Kennedy Space Centre around 88%, ...


3

A dynamic system, with at least 3 massive bodies, will have chaos that can, in theory, be exploited to reach (almost) arbitrary positions within said system at close to zero $\Delta v$ over very long time spans. This is the "Interplanetary Transport Network". This sounds very alluring, but it's easy to be misled into believing this has much ...


3

Barely, but it involving shenanigans even less practical than a Jupiter flyby. The lowest delta-v cost trajectory without any planetary flybys from Earth to a 1.35AU-5.4AU 79.11° orbit is the following: Do a burn in LEO, reaching a solar system escape trajectory. Cost: 8750m/s At the edge of the solar system, do an inclination change and set the periapsis ...


3

While it's a very loose lower bound, it does perhaps have some value to present one most trivial such bound. With basis in the fact that under no circumstances there exists any more efficient way to increase apoapsis than a prograde burn at periapsis, the following bound exists: $$\Delta v \geq \sqrt{\frac{2}{r_{P1}} - \frac{2}{r_{P1} + r_{A2}}} - \sqrt{\...


2

The answer is much depending on how realistic you try to model "your world" and how small your error margin (are you trying to hit a "POINT" or an "AREA"?) is. Do you need the solution for minimum delta V or "a possible" delta V? Most simple: reduce it to a 2D scenario with no atmospheric drag - simple hohmann transfer ...


2

But if we examine it closely, what do we see? Hopefully answers will address the following: Does the command contain a set point for engine cut-off based on integrating one accelerometer, or three, or something else? Does the command include a direction in an inertial frame, or the spacecraft's frame? How would this technique be compared and contrasted to ...


2

The calculation uses the following model for "total propulsive delta-v": $$\Delta v_{total} = \Delta v_{spacecraft} + \Delta v_{launcher}$$ Here, $\Delta v_{spacecraft}$ is what propulsive capabilities the probe has by itself after leaving the Earth system entirely, and is presumed to be a known value that can be looked up. $\Delta v_{launcher}$ is ...


2

TL;DR It works for a very large amount of orbit changes First, let's take your equation and rearrange it to get rid of inconvenient values like force, time and mass. $$\frac{\Delta v_{\text{orb}}}{\Delta t} = \frac{F_{\text{retro}}}{m}$$ $$\Delta v_{\text{orb}} = \frac{F_{\text{retro}} \Delta t}{m} = - \Delta v_{\text{maneuver}}$$ $\frac{F\cdot t}{m}$ is ...


2

For the purpose of $\Delta v$ planning, an orbit at Earth-Moon L4 or L5 is about identical to just any orbit with a radius of 380,000 km. At L4/5 distance to Earth and Moon are identical and as the gravitational force changes with $r^2$, the pull of the Moon is just 0.01% of that of Earth at this point and can be neglected. It's sufficient to keep the orbit ...


2

I was going to leave this is a comment because I don't have a great answer, but I ran into a character limit with hyperlinks. Sadly I can't give a specific answer for non-NRHO halos. But I did find this source, which mentions that ISEE-3 (the first mission to go to a Halo orbit) used about 8 m/s per year of delta-V for stationkeeping. This mission was in a ...


2

I see two main things throwing off your calculations: You can not simply subtract the velocity of one planet from another to get interplanetary transfer costs. An optimal transfer consists of an elliptic orbit touching the orbit of the inner planet at perihelion, and the outer planet at aphelion. Thus, the numbers you should try to obtain are the ...


2

The ISRO had two fully solid fuelled launch systems called the Satellite Launch Vehicle ( SLV ) and Augmented Satellite Launch Vehicle ( ASLV ). SLV was a four stage launch vehicle with a payload capacity of 40kg to LEO and ASLV was 5 stage with 150kg to LEO. Both launch vehicles were considered as experimental and never intended to have a long service life. ...


1

Partial answer, waiting for a rocket scientist to chime in. This is a cool question! Celta-v calculated from exhaust velocity using the Tsiolkovsky rocket equation for each stage would be a huge overestimate because it doesn't account for things like atmospheric drag or gravity. So you'd have to numerically integrate over a specific trajectory for a final ...


1

What you're asking here is how to take into account waste variables such as gravity loss, aero-forces, and ISP loss at sea level. These all depend on your rocket's specific flight profile. For example, rockets with high thrust to weight ratios will experience less loss due to gravity, but much greater loss due to air resistance. Your thrust to weight ratio ...


1

This is to some extent equivalent to Lambert's problem. Namely, if you pick any point along the two orbits you can draw transfer orbits between. This can be constrained to one orbit if you for example also specify the true anomaly when departing or arriving (if the two points and the celestial body do lie on one line). You could also specify the transfer ...


1

This is indeed a difficult question. On this site, you can for instance see that even my quest to resolve the co-planar case hasn't been particularly resolved, and yet even more restricted forms like optimal inclination change between circular orbits still quickly grow complicated. However, that is for optimal transfers. There exists a simple strategy that ...


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