# Tag Info

18

Techically a Hohmann transfer is a single instant, but as you mentioned, there is a window about the Hohmann that is nearly optimal. Just to give you an idea, here is the launch window sizes for some upcoming/ recent missions InSight- 33 days Curiosity- 26 days Maven- 20 days MRO- 21 days Of some note is this statement in the MRO link: The launch ...

14

First off, Hohmann transfers are a fiction. A very useful fiction, but a fiction nonetheless. Hohmann transfers are for transferring from one perfectly circular Keplerian orbit about a planet (or star) with a perfectly spherical mass distribution to another perfectly circular Keplerian orbit. There is no such thing as a perfectly circular Keplerian orbit. ...

11

For reference, that number is: $$5+4\sqrt 7 \cos\left({1\over 3}\tan^{-1}{\sqrt 3\over37}\right)$$ It is the positive root of: $$x^3−15x^2−9x−1=0$$ If we take the equation for the total $\Delta V$ of a Hohmann transfer between two circular orbits, and express it in terms of the ratio of the radius of the larger to the radius of the smaller orbit, $x$, ...

9

TL; DR: Trajectory optimization for continuous thrust is difficult and this field is very active in research. Edit: Concerning the rotation of your spacecraft, you'll want to plot the in-plane and out-of-plane thrusting angles (with respect to the RNC frame of the spacecraft). That will give you an idea of how much the engines needs to gimbal by before ...

8

No. It is not more efficient to bypass LEO. You can't bypass getting out of earth's gravity well. 2.9 km/s is what it'd take to enter a Mars transfer orbit if the ship were outside of earth's gravity well moving 30 km/s in a circular 1 A.U. orbit. (1 A.U. or one astronomical unit is earth's average distance from the sun.) So, again, that 2.9 km/s is what ...

8

My Hohmann spreadsheet shows TMI from LEO is around 3.6 km/s. Here is a Non Hohmann transfer sheet from earth to Mars. Into the pink cells I type .9 A.U. for the transfer orbit's perihelion and 1.6 km/s into the transfer orbit's aphelion. Here's a pic of this transfer orbit: This trip is about 140 days from earth to Mars. TMI from LEO is about 5.2 km/s ...

8

When you've reached escape velocity relative to the planet, the star becomes the dominant gravitational influence: you're in orbit around the star. And as always when you're in orbit, your orbit's radius and speed are interchangable. So when you travel outward from the star, your speed will decrease, and when you move closer to the star, your speed will ...

8

Intuitively: going from one circular orbit to another requires two burns: one to raise the apogee and another to raise the perigee. To achieve escape, you need to raise the apogee until your orbit 'breaks' from an ellipse to an escape trajectory. So a longer apogee-raising burn, but no need for a perigee-raising burn.

8

This answer has the two-impulse Hohmann transfer $\Delta V$. It is: $$\sqrt{2x\over x+1}+\sqrt{1\over x}-\sqrt{2\over x\left(x+1\right)}-1$$ where $x$ is the ratio of the higher orbit radius to the lower orbit radius, assuming (without loss of generality) that the lower orbit radius is $1$ and $\mu$ is $1$. This answer notes that in the limit of very low ...

7

I don't think it has a particular name other than "worst-case Hohmann transfer".

7

I'll try to get you started anyway. From the frame of reference of the assisting body, the trajectory of the probe is hyperbolic, with the same $v_\infty$ going out as coming in, but in a different direction. The trajectory is simply bent. The bend angle is: $$\delta=2\sin^{-1}\left(1\over e\right)$$ where $e$ is the eccentricity of the hyperbola. You ...

7

Let's break down your problem into a very simplified launch phase from the ground to low Earth orbit at 250 km altitude. Then we'll use patch conics to get an estimate of the $\Delta v$ necessary to reach a low Mars orbit of 80 km altitude. One method that I like to use to get ball-park estimates of required launch $\Delta v$ is to first consider an ...

7

You can change orbits by slowly spiraling out, but then it isn't a Hohmann transfer anymore. In the extreme case of very low thrust, this will mean your orbit is almost circular at all times. You'll also burn more delta-v than with the Hohmann transfer, because on average you add orbital energy while your speed is lower. Wikipedia has this to say: It can ...

7

You didn't say anything about how you chose the transfer orbit, nor did you specify velocity relative to what. And of course, "appreciably" is not defined. So we will have to make some assumptions. First, I will assume that you chose a minimum energy transfer orbit between the two bodies. For bodies with this small of a ratio of radii, that minimum ...

7

Citing Wikipedia, some bi-elliptic transfers require a lower amount of total delta-v than a Hohmann transfer when the ratio of final to initial semi-major axis is 11.94 or greater, depending on the intermediate semi-major axis chosen. An intuitive approach to understanding it, is when you look at delta-V of insertion after Hohmann transfer. For near ...

6

It wasn't a Hohmann transfer. The rocket simply boosted it to a higher velocity from Earth in order to enable a Jupiter flyby that would send the spacecraft to Saturn. You can get the approximate orbit elements from the JPL HORIZONS Web-Interface, picking a day in the middle of the Earth to Jupiter transit (which by the way was 1.5 years): giving: $$SOE ... 6 Car accelerations vary a lot, but from this table, 3 m/s² seems like a safe bet. The Hohmann transfer burn towards the Moon from LEO is around 3100 m/s, so the "burn" is only going to take about 17 minutes. That is quick enough for a transfer orbit burn, so a low acceleration spiral is not required. As for the odometer, that is going to display half the ... 6 Let's start at the point which is common to the two maneuvers: we're at perigee of GTO - Hohmann transfer orbit from LEO (or even suborbital flight!) to GEO; an elongated orbit with perigee of ~200-300km and apogee of 36,000km. From then on, we can either circularize at apogee for GEO, or continue our burn at perigee for escape. change of apogee is very ... 5 The time in transfer will be a good approximation - replacing the long burns with impulsive. LEO altitude 160 km + 6,371km Earth average radius, so periapsis of the transfer orbit is 6531km from Earth center. Geostationary orbit radius: 42,164 km Semi-major axis is the average between the two. a=24347km T=2 \pi \sqrt{a^3 \over GM} The period of that ... 4 To Mars The Delta-V required to transfer from Low Earth Orbit to Low Mars Orbit is around 7 km/s (utilising the Oberth effect). So the payload being sent up to LEO must have this ∆V budget. Additionally your launcher needs about 8km/s to get up to LEO in the first place. Your question is specifically about payload mass. You've given a multitude of ... 4 The short term Mars missions do a non-standard path to get back. I believe many of them actually use a Venus flyby to get back to Earth, which you can see a bit on this Wikipedia article. In any case, one can return more quickly then waiting for an optimal transfer window, but at a much higher cost, even if there isn't a Venus flyby in the process. The ... 4 The best explanation I've heard is this. The problem is very similar, although on a larger level, to comparing throwing a ball to 500 km vs orbiting Earth at 200 km. It takes far more power to orbit Earth. The difference is less for Geostationary vs escape velocity, but the principal is the same. A more detailed explanation: It has to do with the Oberth ... 4 We'll only consider the simplified circular-circular transfer because it makes our lives easier. In this case the same considerations apply as with the bi-elliptic transfer: We only want to burn at apsis or periapsis - any other point will waste dV changing the argument of periapsis. This means we will always be raising or lowering the opposite Ap/Pe. We ... 3 Yes. Consider a simplifying case of a direct escape vs. a parking orbit departure from an airless, non-rotating body using instantaneous maneuvers. The most efficient way to get to the parking orbit is an initial horizontal \Delta V at the surface to get an orbit with a apoapsis at the parking orbit, followed by a circularization \Delta V at apoapsis to ... 3 When going from an inner to outer body in a Hohmann -- Earth to Mars, for example -- the spacecraft will be nearing the aphelion of its elliptical orbit as it enters the destination SOI, so will be going substantially slower than the outer body. From a fixed solar frame, the spacecraft will be ahead of and inside the planet, and the planet will overtake it ... 3 I whomped up a spreadsheet where you can set perihelion and aphelion of transfer orbit. Here's a screen capture from the spreadsheet where I set perihelion of transfer orbit at 1 A.U. and aphelion at 12 A.U. Of course SMA is (perihelion+aphelion)/2. Where the transfer orbit crosses Mars orbit, r is 1.524 A.U.. At Saturn, r is 9.537 A.U.. Knowing a, mu and ... 3 There are two ways of entering the system, depending on what side of the planet you fly by. The two are entirely equivalent, so you can chose what is going to be "prograde", and what is going to be "retrograde". Nothing special about the Triton case then. That is NOT to say that what direction you chose to enter is not important, it just mean that the ... 3 To complement @PearsonArtPhoto's information on upcoming missions, you can also look at the historical launch dates of Mars missions to get a sense of the launch windows actually used. While the launch window for a single mission isn't, logistically speaking, exactly the same thing as the span of a set of missions within a single 26-month opportunity cycle, ... 3 From Tsiolkovsky's rocket equation (see this Wikipedia article) you can find out the mass of the fuel used during the shuttle's launch. The equation is stated as it follows$${\Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}} Where ${\displaystyle \Delta v}$ is the maximum change of velocity of the vehicle (with no external forces acting), $m_{0}$ is ...

3

Turn and burn is more efficient. You get the most energy change in the orbit by burning in the velocity direction. The next question is what is the optimal time to start the burn ...

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