Descending into any orbit: Which requires less delta-v, a wide one, or a narrow-elliptic one?

Question

If a vessel is to enter an orbit around some body (approaching from the "outside" fast), and can choose between a very high circular orbit and a very elongated elliptical orbit of the same apoapsis, which will consume less fuel?

Intuitively, when reversing the situation, entering the wide orbit from low orbit around the body would require more fuel: first increase the apoapsis (entering the elliptical orbit) and then circularize it. So, in direction of outside the system -> low orbit, if the trend is reversed, the orbit with lower average altitude would consume more fuel. On the other hand, Oberth effect is anything but intuitive, and a maneuver performed at a high speed, close to the body's surface is supposed to be more efficient than one performed at a high orbit. So I can't decide which is the case.

_{Please answer urgently, I don't have much fuel and Minmus is drawing near.}

Answer 1

There are two different scenarios that I'll consider:

• Entering the circular orbit from high altitude.
• Entering the elliptical orbit from low altitude.

Entering the circular orbit from low altitude will always take more fuel than entering the elliptical orbit, since you have to enter an elliptical transfer orbit of the same size (using the same $\Delta v$), and then make an additional burn at apoapsis to circularize your orbit. (Using aerobraking to replace the initial burn will change this.)

Similarly, entering the elliptical orbit from high altitude will always take more fuel than entering the circular orbit, since the elliptical orbit will have a slower speed at apoapsis, requiring a greater $\Delta v$.

I'll make the assumption that the hyperbolic excess velocity ($v_\infty$) upon entry is the same for both cases, that all the orbital intersections are parallel (no plane changes), and that all burns are instantaneous.

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A circular orbit of radius $r_a$ requires a speed of $v_\text{circ}=\sqrt{\mu/r_a}$. Solving for the entry velocity using conservation of energy:

$$ \frac{v_\infty^2}{2} = \frac{v_\text{entry}^2}{2} - \frac{\mu}{r_a} \\ v_\text{entry} = \sqrt{v_\infty^2 + \frac{2\mu}{r_a}} $$

The $\Delta v$ required is simply the difference between the two:

$$ \Delta v_\text{circ} = \sqrt{v_\infty^2 + \frac{2\mu}{r_a}}-\sqrt{\frac\mu{r_a}} $$

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In the second scenario, we have a periapsis velocity of:

$$ v_p = \sqrt{\frac{2\mu}{r_a+r_p}\frac{r_a}{r_p}} $$

The entry velocity is almost the same, with $r_p$ in place of $r_a$, giving us:

$$ \Delta v_\text{ellip} = \sqrt{v_\infty^2 + \frac{2\mu}{r_p}}-\sqrt{\frac{2\mu}{r_a+r_p}\frac{r_a}{r_p}} $$

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Now, let's find under what conditions $\Delta v_\text{ellip} < \Delta v_\text{circ}$:

$$ \Delta v_\text{ellip} < \Delta v_\text{circ} \\ \sqrt{v_\infty^2 + \frac{2\mu}{r_p}}-\sqrt{\frac{2\mu}{r_a+r_p}\frac{r_a}{r_p}} < \sqrt{v_\infty^2 + \frac{2\mu}{r_a}}-\sqrt{\frac\mu{r_a}} $$

I'll make the substitutions $v_\infty^2\to \alpha\mu/r_a$ and $r_p\to\beta r_a$ to help us out a little.

$$ \sqrt{\frac{\alpha\mu}{r_a} + \frac{2\mu}{\beta r_a}}-\sqrt{\frac{2\mu}{r_a+\beta r_a}\frac{r_a}{\beta r_a}} < \sqrt{\frac{\alpha\mu}{r_a} + \frac{2\mu}{r_a}}-\sqrt{\frac\mu{r_a}} $$

Now let's take out a factor of $\sqrt{\mu/r_a}$ (which is positive, so it doesn't affect the inequality):

$$ \sqrt{\alpha + \frac{2}{\beta}}-\sqrt{\frac{2}{\beta\left(1+\beta\right)}} < \sqrt{\alpha + 2}-1 $$

Treating the left side as a function of $\beta$:

$$ f(\beta) = \sqrt{\alpha + \frac{2}{\beta}}-\sqrt{\frac{2}{\beta\left(1+\beta\right)}} $$

We have:

$$ f(1) = \sqrt{\alpha + 2} - \sqrt{\frac{2}{1\cdot 2}} = \sqrt{\alpha + 2} - 1 $$

Substituting into our inequality, we now must only prove:

$$ f(\beta) < f(1) $$

Note that since $0<r_p<r_a$, we know that $\beta\in(0,1)$. Looking at the derivative of $f$:

$$ f'(\beta) = -\frac{1}{\beta^2\sqrt{\alpha+2/\beta}}+\frac{\beta+(1+\beta)}{\beta^2(1+\beta)^2}\sqrt{\frac{\beta(1+\beta)}{2}} $$

We can check that $f'(\beta)>0$:

$$ \frac{1}{\beta^2\sqrt{\alpha+2/\beta}} < \frac{\beta+(1+\beta)}{\beta^2(1+\beta)^2}\sqrt{\frac{\beta(1+\beta)}{2}} $$

Since both sides are positive, we can square both sides without affecting the inequality:

$$ \frac{1}{\beta^4(\alpha+2/\beta)} < \frac{\left(\beta+(1+\beta)\right)^2\beta(1+\beta)}{2\beta^4(1+\beta)^4} \\ \frac{1}{(\alpha+2/\beta)} < \frac{(1+2\beta)^2\beta}{2(1+\beta)^3} \\ 2(1+\beta)^3 < (\alpha\beta+2)(1+2\beta)^2 \\ 2\beta^3+6\beta^2+6\beta+2 < 4\alpha\beta^3 + (4\alpha+8)\beta^2 + (\alpha+8)\beta + 2 \\ 0 < (4\alpha-2)\beta^3 + (4\alpha+2)\beta^2 + (\alpha+2)\beta \\ 0 < 4\alpha\beta^2(1+\beta) + 2\beta^2(1-\beta) + (\alpha+2)\beta $$

All three terms are positive when $\alpha>0$ and $\beta\in(0,1)$, so $f'(\beta)$ is always positive. This means that its maximum must occur on the right endpoint, $f(1)$, so that $f(\beta)<f(1)$ for all values under consideration.

Thus, the elliptical orbit is always more efficient.

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The greatest difference between the two orbits is when $\alpha=2$ (that is, $v_\infty=v_\text{esc}$ at apoapsis), where we have:

$$ \frac{\Delta v_\text{circ}}{\Delta v_\text{ellip}} = \frac 1{\sqrt{2}}\sqrt{1+\frac 1\beta} $$

Answer 2

Well lets think about this one logically. Say you're sitting in a spaceship out in the oort cloud, orbiting the sun at a very leisurely pace of about 3m/s, which is equivalent to jogging speed. It doesn't take much delta-v at all to stop dead, and fall directly towards the sun. With such a long fall, a slight adjustment will let us pick what planet we want to intercept.

We are now coming in from the outside, fast.

Clearly, it is MUCH easier to enter a narrow elliptical orbit. In order to enter a circular orbit you'd have to wait until you're at that orbital distance from the target planet, and then fire your rockets to overcome all the velocity gained falling from the oort cloud.

Aiming closer to the planet (MUCH closer), we can areobrake in the planet's atmosphere and bleed off a lot of the velocity that way. We aren't actually in a stable orbit, we'll either fly off into space again (if still travelling at escape velocity relative to the planet), or we'll subsequently crash into the planet without a correcting burn (but we can always do a correcting burn).

I know the Oberth effect isn't exactly intuitive, but think about it this way and I'm sure it will make sense; when we are coming in fast, the planet doesn't have much time to pull on us, so it can't increase our velocity much. So we zoom in really fast, and don't gain much velocity. When close to the planet, we fire our thrusters to slow down as much as we can. We now leave the planet's influence more slowly than we entered it, and the planet's gravity has more time to slow us down, so we lose more velocity leaving the planet, than we gained approaching the planet.

Now if while still quite far away from the planet, we'd fired our thusters, then because we have slowed our approach, the planet's gravity has more time to pull on us and increase our velocity. As we leave the planet's influence, the planet has exactly the same amount of time to decrease our velocity.

The Oberth affect is essentially about minimizing or maximizing the amount of time being 'pulled' strongly by the planet. This is because in accordance with basic Newtonian calculus, Velocity is Acceleration times Time, the more time we are being accelerated, the more velocity we gain. The less time we are being accelerated, the less velocity we gain. Rockets burn their fuel as close to Earth as they can, so they minimize the time in the strongest gravity nearest the Earth. Gravity is NOT like drag, it doesn't pull on you more the faster you're travelling - it pulls on you the same amount (at a given distance) however fast you're travelling. That is why Oberth effect works, and also why it may not be intuitive, because we often think of gravity as a "drag" on rockets, but it's not a drag like that caused by air or water, it's something readily modeled by simple calculus.