How to intuitively explain that reaching geostationary orbits consume more Delta V than escape orbits?

Question

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 $r_f/r_i$.

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.