# Tag Info

71

Maybe some visual intuition for what actually happens in the Hohmann transfer helps? It's already very close to what you are describing. In the top arc, the spacecraft (yellow), is going a bit slower than Mars (red), so it's indeed "waiting" for the planet to catch up to it. It only touches the orbit of Mars in a point, but that's all we need if we time ...

38

It depends whether or not you want to orbit or land softly upon Mars, or just hit it. For the former, you have to match orbits with it, that probably means burning more fuel. For the latter, you can skip the Mars orbit injection and just crash. This is quite fuel efficient, especially as the reduced delta-V requirements mean that you need less fuel for the ...

20

Your delta-v analysis doesn't account for the landing delta-v. On Mars, only a fraction of a km/s has to be done propulsively, on Mercury the entire landing will be propulsive. You also don't account for transit time. BepiColombo launched in 2018 and won't be able to enter orbit until the end of 2025. MESSENGER similarly launched in 2004 and didn't enter ...

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 ...

17

could we launch a vehicle in space for Earth, stop right on Mars trajectory... Yes, you could have a trajectory that came to a stop (briefly) in the path of Mars like how a ball thrown upwards stops (instantaneously) before falling down, except in this case you'd have to be moving directly away from the Sun on a straight-line trajectory with eccentricity = ...

15

Circular orbits at different altitudes require different speeds, so if you start with a radial separation, the spacecraft and station will tend to drift further apart unless they accelerate themselves radially to close the distance. The effect is small at small distances, larger at long distances. To a first approximation, the separation is the difference ...

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. ...

12

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, $x$, ...

11

TL; DR: Trajectory optimization for continuous thrust is difficult and this field is very active in research. 2021 clarifications: Methodology For the least amount of fuel, the best is the thrust the least amount of time as possible and only when it's extremely efficient ($\eta \geq 0.98$). But that also implies that it will take an incredibly long amount ...

10

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 ...

9

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 ...

9

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 ...

9

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

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

It didn't have enough thrust. Small rocket engines are easier to build than large ones, and weigh a lot less. That probe (I assume you mean Chandrayaan 1) had a 440 N main thruster that it used to get to Lunar orbit. It weighed initially about 1350 kg all from wiki. That amounts to an acceleration of approximately 0.3 m/s/s. You need around 4000 m/s for ...

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 ...

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 be ...

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

Yes this is absolutely a thing. Delta-V can indeed be traded off in a continuous fashion for total flight time (i.e. uses more dV, but lowers transfer times, relative to a Hohmann transfer). Furthermore it's pretty directly analogous to a Hohmann transfer. I don't know a name for this but the heavily simplified version of this 'partial Hohmann' transfer is ...

7

There is no "stopping" in space - no matter how far you are from other celestial bodies, the force of gravity will always be tugging on you, pulling you in some direction. Within the Solar System, that tug is typically pulling you toward the Sun, unless you happen to be quite close to another planet or moon. If you were to try to get to the same path as Mars ...

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

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

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 ...

6

You can't land on the day side of Mercury nor on a peak of perpetual light because it's far too hot (800 deg F / 430 deg C), even if not as hot as on Venus. A crewed mission must land either on the night side or in a crater of long darkness. This crater can and probably will be on the north or south pole if there's perpetual darkness. Also, a crewed flight ...

5

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 ...

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 ...

5

There is a popular idea that has been put forward that one can return to Earth via Venus, and to achieve that one has to launch within a few weeks of landing on Mars. I suspect that the author is confusing that. You can take a look at http://clowder.net/hop/railroad/sched.html for when the times are, and see that the times pretty closely overlap. ...

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