Thanks to answers to my question about gravity assists, I now pretty much understand how they work on the ecliptic, or near such. For that plane, the "null net speed change" path goes on a crash course right into the middle of the planet.

But what if we add the third dimension - about orbits of the sun that are 90o to the plane of the ecliptic or highly inclined? Which close fly-by would affect the orbital plane and direction, but not the speed? Polar? (wouldn't an approach from behind and passing the planet through the front serve as a fully-featured speed increase?)

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    $\begingroup$ I might just test this in KSP... $\endgroup$
    – PearsonArtPhoto
    Jun 13, 2015 at 1:22
  • $\begingroup$ The v in delta-v stands for velocity. Since velocity is composed of speed and direction, a zero delta-v gravity assist is not possible. $\endgroup$
    – user8406
    Jun 13, 2015 at 4:24

2 Answers 2


SF had asked how to preserve speed while changing direction. I'll try to answer that question. But I'll also talk about changing direction without changing speed. My goal is to give a feel for hyperbolas and swing bys.

Preserving direction wrt sun but changing speed

For my examples I'll use earth as the body doing the gravity assist.

From earth's point of view, the space ship's path is a hyperbola. Incoming Vinfinity and outgoing Vinfinity have the same speed but different directions. I attempt to illustrate the notion of Vinf on page 36 of my orbital mechanics coloring book.

enter image description here

Since incoming and outgoing Vinfinity are the same speed, the velocity vectors are two sides of an isosceles triangle. Delta V is the base of the triangle. Top angle δ is the hyperbola's turning angle:

enter image description here

From earth's point of view, direction is changed.

But it's possible to preserve direction from the sun's point of view.

Earth's path with to the sun is nearly horizontal, a flight path of 0º. It's speed is about 30 km/s. Let's imagine a space ship that has flight path 9º and speed 28 km/s when it crosses earth path:

enter image description here

In this case, incoming Vinf wrt earth is about 3 km/s.

To preserve direction, delta V vector must be parallel or anti-parallel to the space ship's vector:

enter image description here

For this to occur, the incoming Vinf and outgoing Vinf need to form symmetrical angles. The center angle is the hyperbola's turning angle, δ. So the angle between spaceship's vector and the incoming Vinf vector must be 90º-δ/2.

In this example the space ship would maintain direction with regard to the sun but it's speed would be boosted from 28 km/s to 29.6 km/s.

Let's change the scenario to a space ship in a circular 1 A.U. polar orbit. Speed is 30 km/s, flight path is zero but inclination is 90 degrees.

enter image description here

In this case incoming Vinf wrt earth would be 42 km/s. δ would be 90º. Delta V would be 60 km/s. This would flip the space ship to a 270º inclination orbit and the space ship's speed would remain 30 km/s but in the opposite direction.

This sort of gravity assist isn't possible unless we drastically reduced earth's diameter. Turning angle increases as hyperbola's perigee gets closer to earth. Closest perigee we can get without lithobraking is an altitude of 0 kilometers.

With 0 km altitude perigee and a vinf of 42 km/s, turning angle is only 4º.

Preserving speed wrt sun but changing direction

If the ship's new velocity vector has the same speed as the old, the vectors are sides of an isosceles triangle.

enter image description here

The delta V vector is the base of this isosceles triangle.

So to preserve speed, the angle Vinf forms with earth's velocity vector needs to be half of the hyperbola's turning angle.


Suppose we do a flyby of a planet. Then the spacecraft has constant energy in the reference frame of the planet.

In the reference frame of the planet, let the spacecraft's initial and final velocity be $\mathbf{v}_i$ and $\mathbf{v}_f$. In the reference frame of the sun, call the initial and final velocities $\mathbf{v}_i'$ and $\mathbf{v}_f'$.

Kinematically, we see that $$\mathbf{v}_i' = \mathbf{v}_i + \mathbf{v}_p$$ $$\mathbf{v}_f' = \mathbf{v}_f + \mathbf{v}_p$$ where $\mathbf{v}_p$ is the velocity of the planet in the Sun's frame.

We know that $v_i = v_f$ and you desire $v_i' = v_f'$, where non-bolded quantities denote magnitudes.

Using $$\mathbf{v}_i' \cdot \mathbf{v}_i' = \mathbf{v}_i \cdot \mathbf{v}_i + 2\mathbf{v}_i\cdot \mathbf{v}_p + \mathbf{v}_p \cdot \mathbf{v}_p$$ and likewise for $\mathbf{v}_f'$, we obtain $$v_i'^2 = v_f'^2 - 2(\mathbf{v}_f - \mathbf{v}_i)\cdot\mathbf{v}_p$$

Thus you are asking for orbits where the change in velocity, from the planet's reference frame, is perpendicular to the planet's velocity from the Sun's reference frame.

The spacecraft's orbit around the planet is a hyperbola in the planet's frame. The spacecraft's change in velocity is along the symmetry axis of the hyperbola. The spacecraft can take any orbit whose symmetry axis is perpendicular to the planet's motion.

An approach at 90 degrees to the ecliptic in the sun's reference frame is not quite right. In that sort of orbit, the spacecraft, as seen from the planet, has motion in the ecliptic of $-\mathbf{v}_p$, and thus its change in velocity will have some component pointing along the direction of motion of the planet.

There is a three-dimensional space of allowed orbits for the spacecraft. The requirement that the symmetry axis be perpendicular to the planet's motion removes only a single dimension from the space of all approaches.

Note, also, that your original assertion that all orbits along the ecliptic change the spacecraft's speed is incorrect.

  • $\begingroup$ "Note, also, that your original assertion that all orbits along the ecliptic change the spacecraft's speed is incorrect." What orbit would result in 0 net speed change? Also, I wondered if because it takes time to pass through the planet's gravity well, there would still be a net transfer of momentum parallel to its orbit, and that therefore a perpendicular orbit wouldn't be the solution. $\endgroup$
    – kim holder
    Jun 13, 2015 at 18:09
  • $\begingroup$ Any orbit for which the symmetry axis of the hyperbola in the planet frame is perpendicular to the planet's motion. Just draw a line in the ecliptic indicating the planet's motion, draw a line perpendicular to that, and draw a hyperbola in the ecliptic with that second line as the symmetry axis. Make sure the hyperbola's steepness is correct for its impact parameter. There you have an orbit in the ecliptic with zero speed change. For simplicity I am working in the limit where the planet's sphere of influence is very small compared to its separation from the sun. $\endgroup$ Jun 13, 2015 at 18:32
  • $\begingroup$ I cannot comment on your mention of a "perpendicular orbit" because it is unclear to me what you are using this term to mean. $\endgroup$ Jun 13, 2015 at 18:32
  • $\begingroup$ I am using 'perpendicular orbit' as you are, 90 degrees to the planet's motion. I see your point about the relative comparison making this factor insignificant. However, would it be the case that if the vessel approached at an angle just slightly to the side of the planet's forward motion, the influence of the planet's orbital angular momentum could be cancelled out? $\endgroup$
    – kim holder
    Jun 13, 2015 at 19:25
  • $\begingroup$ I cannot tell from your comment what frame of reference you are discussing. Velocity vectors are not perpendicular in any sort of absolute sense. This is something that depends on the frame you are analyzing. $\endgroup$ Jun 13, 2015 at 19:38

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