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Studying again orbital maneuvers, I found the following plotting the cost of a Hohmman transfer between coplanar circular orbits (black line), in terms of $\Delta V$ against the orbit ratio $r_f/r_i$.

enter image description here

It surprises me to found a maximum $\Delta V$ at about $r_f/r_i=15.58$. As it can be seen in the figure, there is a range of $r_f/r_i$ values above the escape velocity cost, $\sqrt{2}$. For example, reaching a geostationary orbit from LEO (assuming $h=300$ km), $r_f/r_i=6.3139$ is more or less the same cost to put a spacecraft in a Moon's alike orbit $r_f/r_i=54.2977$.

Is there some intuitive explanation to this "apparently" counter-intuitive behaviour?.

Plot source: https://www.reddit.com/r/KerbalSpaceProgram/comments/1ajru7/i_was_curious_about_the_delta_v_requirements_for/

NOTE: only Earth gravity is being considered to take the plot

BONUS: the black line function can be defined mathematically as follows; The needed impulses are

$\Delta V_1=\sqrt{\frac{2 \mu}{r_i}-\frac{2 \mu}{r_i+r_f}}-\sqrt{\frac{\mu}{r_i}}$,

$\Delta V_2=\sqrt{\frac{\mu}{r_f}}-\sqrt{\frac{2 \mu}{r_f}-\frac{2 \mu}{r_i+r_f}}-$,

being the total impulse of the Hohmman transfer given by $\Delta V=\Delta V_1 + \Delta V_2$. If one divides the velocity increments by the initial velocity $V_i=\sqrt{\mu/r_i}$ it is obtained

$\frac{\Delta V_1}{V_i}=\sqrt{2-\frac{2r_i}{r_i+r_f}}-1$,

$\frac{\Delta V_2}{V_i}=\sqrt{\frac{r_i}{r_f}}-\sqrt{\frac{2r_i}{r_f}-\frac{2r_i}{r_i+r_f}}$,

defining $\lambda=r_f/r_i$ (note that this is the variable of the x-axis of the plot), and operating

$\frac{\Delta V_1}{V_i}=\sqrt{\frac{2\lambda}{1+\lambda}}-1$,

$\frac{\Delta V_2}{V_i}=\sqrt{\frac{1}{\lambda}}-\sqrt{\frac{2}{\lambda(1+\lambda)}}$,

and the total cost expressed as a function of $\lambda$ is

$\frac{\Delta V}{V_i}=\frac{\Delta V_1}{V_i}+\frac{\Delta V_2}{V_i}=\sqrt{\frac{2\lambda}{1+\lambda}}(1-\frac{1}{\lambda})+\sqrt{\frac{1}{\lambda}}-1$,

which is the analytical expression of the black line in the plot.

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  • $\begingroup$ What is the source of the plot? Can you add a link or citation? $\endgroup$ – uhoh May 14 '18 at 14:17
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    $\begingroup$ I have taken this one from reddit. This is a typical plot that appears in almost all Orbital Mechanics maneuvers section of any student textbook. $\endgroup$ – Julio May 14 '18 at 15:32
  • $\begingroup$ Thanks for the edit! That helps for readers like me who are less versed in the field but curious to learn more. It's also good to give credit to creators of images we use. $\endgroup$ – uhoh May 14 '18 at 21:54
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    $\begingroup$ Np. When I have time I will read in more detail the posted answers, and also edit the question to add the mathematical development explaining the black line of the plot. $\endgroup$ – Julio May 15 '18 at 10:06
  • $\begingroup$ "Geostationary" is a bit of a red herring, as depending on the parent objects mass, spin rate and the starting orbit getting there may be less than escape or impossible $\endgroup$ – JCRM May 16 '18 at 8:52
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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.

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    $\begingroup$ Maybe you can complete your question adressing that the second burn at the apogee for a highly eccentric orbit is extremely cheap since you are extremely far from the planet $\endgroup$ – Julio May 16 '18 at 9:36
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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 cheap for highly eccentric orbits. Escape burn is in essence making the orbit infinitely eccentric.

Why? Because the farther from Earth the weaker the gravity - in quadratic proportion! $F = G {{m_1 m_2} \over {r^2}}$ the $^2$ at $r$ means the force of gravity drops quadratically with distance. Much less gain of potential energy - so less kinetic energy / delta-V needed. Going from a 500km circular to a 500/1000km elliptic orbit takes vastly more energy than going from 500km/36,000km (GTO) to 500km/37,000km - because at 36,000km gravity is so much weaker that moving the rocket outwards is so much easier.

Add to that Oberth effect. $E_k = {1 \over 2} m v^2$ - that $^2$ is damn important here. If you're moving at 8km/s and add another 2km/s, you go from 64[units of energy] to 100[units] - a gain of 36. But if you go from 12km/s to 14km/s - same delta-V! - we go from 144[units] to 196 - a gain of 52 units of energy - kinetic at perigee, but potential at apogee. These things compound: changing apogee from altitude of GEO to "infinity"/escape while retaining the low perigee costs peanuts.

Both you already have a huge speed at perigee, so you benefit from Oberth, and you're fighting against rapidly diminishing gravity, the orbit growing by strides with minuscule investment. Delta-V needed is tiny.

  • costs of circularization for medium orbits is significant.

The orbital velocity of GEO is about 3 km/s. After all, you need to circle Earth in 24 hours, on a circle of 36,000km radius - you need to move pretty fast for that. Moon makes a circle that takes nearly a month, but the speed drops to about 1km/s. Move to the edge of the Hill sphere, sit at Earth-Sun L1 lagrangian point and your orbital speed around Earth drops to 0. But GEO is much lower - and so, still rather fast, even if much slower than the scary 8km/s of LEO.

And if you're at apogee of a strongly elliptical orbit, you're crawling at snail's pace - you're a rock tossed up and lingering before falling down. That means, to circularize, you need to spend nearly the full value of the circular orbit's orbital speed in your spacecraft's delta-V. You need to get from that crawl from the Hohmann transfer orbit from LEO to 3km/s through a long, hard burn. Which, had you performed it at perigee, for the same transfer orbit, would take you to a trajectory well beyond Mars.

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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 effect, which states that burns are more efficient if one is moving faster, for basic orbital maneuvers. The speed is fastest near a body, thus it is more efficient. The speed is quite a bit less at the far end of a GTO orbit, and thus it takes more fuel to stop. This doesn't even take in to account removing the inclination, which is required to be useful.

To show this a bit better, I played with Kerbal Space Program some. Here is the result:

Bottom line, only a bit more fuel is required to escape, but you lose the Oberth effect when you are far out and trying to circularize.

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  • $\begingroup$ "There isn't a really good explanation." I can understand saying that you can't think of one, or that you've never come across one. But can you really say none exists? $\endgroup$ – uhoh May 14 '18 at 15:18
  • $\begingroup$ @PearsonArtPhoto, I do not understand the relation with the Oberth effect since both a transfer from LEO to GEO or the Moon has the same departure orbit.... Moreover, GEO orbit is more closer to Moon orbit, implying an allegedly higher Oberth effect when doing the second impulse.... but maybe I misunderstood something $\endgroup$ – Julio May 14 '18 at 15:54
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    $\begingroup$ One way to see how important the Oberth effect is here is to do the same arithmetic at Jupiter. There you'll see a huge impact. $\endgroup$ – Mark Adler May 14 '18 at 21:19
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    $\begingroup$ Did a bit better of an example with KSP, hopefully that helps some. $\endgroup$ – PearsonArtPhoto May 21 '18 at 23:59
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    $\begingroup$ I love how KSP is actually a decent medium for answering things "pseudoaccurately" with visuals. $\endgroup$ – Magic Octopus Urn May 22 '18 at 1:16
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Though not stated, the question has an implicit assumption which is false, that the energy change scales monotonically with delta-v.

delta-v is related to momentum, not energy. There should never be any expectation that there should be a 1:1 correspondence between change of energy (a scalar) and change of momentum (a vector).

Consider an orbit-lowering maneuver which uses plenty of delta-v and yet reduces the energy of the orbit.

The question looks more subtle because there are multiple delta-v maneuvers involved, but by comparing the energy of escape vs bound orbit, it still mixes energy with momentum.

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  • $\begingroup$ Feel free to down vote, but please add a comment why. Please keep in mind though that $a > b$ does not need an "intuitive explanation" if there is no reason for $a$ to have any particularly simple relationship to $b$ to begin with. That would be faux intuition, or fauxtuition. $\endgroup$ – uhoh May 15 '18 at 7:04
  • $\begingroup$ Not my downvote, but physicist here: delta-v is a scalar. If you're going to argue "scalar versus vector", then delta-v would fall on the same scalar side as energy. $\endgroup$ – MSalters May 15 '18 at 13:26
  • $\begingroup$ @MSalters That's not right. Delta-v maneuvers are always vectors and always have a specified direction. One direction (prograde) you "raise" an orbit, another direction (retrograde) you "lower" it, and a third direction (perpendicular) you change the plane without affecting the shape or energy at all. $\endgroup$ – uhoh May 15 '18 at 14:43
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    $\begingroup$ But in this case (Hohmman transfers), the Delta-V is tangential to the orbit velocity (thus maximizing the resulting final velocity) and since $E \sim V^2$ I think that it is an intuitive (although wrong) thought to "assume" a monotonically relation between Delta-V with energy. I think the issue is that there are two impulses involved instead of one $\endgroup$ – Julio May 16 '18 at 9:16
  • $\begingroup$ @Julio Nope, that's also partly wrong. This question is about a bi-elliptic Hohmman transfer, where one of the burns is in the opposite direction of the other two. Even for a standard Hohmman transfer, if you are going to to a lower orbit (e.g. Venus from Earth) the burns are also retrograde and anti-parallel to the velocity. Momentum is a vector. Treat it like it's not at your own peril. $\endgroup$ – uhoh May 16 '18 at 11:15

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