What near flyby wouldn't modify the speed of the vessel, just direction?

Question

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 90^o 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?)

Answer 1

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.

Answer 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.

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:

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:

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:

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.

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.

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.