Plotting a Heliocentric Gravity Turn

In the evolution of spaceflight one of the greatest qualitative changes will come with the invention and development of the thermonuclear rocket motor, a motor that derives its thrust from the fusion of light-element nuclei, such as Helium-3, Deuterium, and Tritium. At first fusion-powered rockets carrying loads of those elements in their tanks will replace chemically powered and fission-powered rockets in the same way as diesel-electric locomotives replaced steam-driven locomotives on the railroads across the middle of the Twentieth Century. But we certainly won’t stop there. Improvement will follow improvement until engineers develop a fusion rocket that can use Protium, the simplest and most common form of Hydrogen, as fuel and propellant. That development will end the Age of the Hohmann Transfer Ellipse and planetary billiards as effectively as the invention of the steamship ended the Age of Sail.

Built as much of forcefields as of matter, the new ships will not carry the huge propellant tanks borne by earlier ships. Smaller propellant tanks will suffice because the new ships will draw most of their propellant from the solar wind and, eventually, the interstellar medium. Inspired by the concept of Robert Bussard’s interstellar ramjet, into which they will eventually evolve, these ships will deploy broad magnetic fields to draw the flux of particles from the sun into their engines, where dynamic structures teased out of the roiling chaos of the quantum vacuum will catalyze the desired fusions. The exquisite beauty of these ships lies in the fact that they will reach the requisite ram speed by the simple expedient of coming clear of Earth’s magnetosphere. Having thus come into the solar wind, a ship will accelerate toward the sun and let the sun’s gravity deflect its trajectory toward its destination.

We call that maneuver a heliocentric gravity turn, though some will refer to it through the kenning Rubbing Odin’s Left Eye. In this essay I want to figure out how to calculate the two most fundamental parameters of that maneuver; first, the angle that the plane in which the ship’s trajectory lies makes with the ecliptic plane and, second, the angle through which the sun’s gravity must deflect the ship’s trajectory. We will calculate those angles from, respectively, one vertex angle and one side of a spherical triangle that we project onto the celestial sphere. In another essay I will figure out how to calculate the coordinates of the point on the sky at which an astrogator must aim their ship in order to meet those parameters.

Imagine that the time of launch has come for our ship and that we have turned our gaze (through appropriate filters) directly at the sun. In that circumstance our line of sight coincides with the line on which the plane of our ship’s trajectory crosses the Ecliptic plane. Our ship’s trajectory will follow that line closely, veering off it slightly to pass the sun. The sun’s gravity will so deflect the ship’s velocity that the ship will come out of the encounter flying along a straight line extending from the sun to the point on the sky that our destination will occupy when we arrive.

To establish the spherical triangle that we want to solve we imagine that we occupy the center of the sun and look outward. We see the Ecliptic and its associated coordinate grid laid out across the sky. Earth, lying on the Ecliptic (by definition of the Ecliptic), defines the vertex A of our triangle by its location at the time of launch. The location of our destination on the sky at the time of arrival defines the vertex B. For the vertex C we take the point on the Ecliptic that has the same Ecliptic longitude as does our destination at the time of arrival. That choice makes the vertex angle C=90E, which makes our solution of the triangle easier.

The side "
of that triangle, opposite the vertex A, has a magnitude equal to that of the
destination’s Ecliptic latitude, the angular distance between the points B and
C. The side $
consists of the segment of the Ecliptic between the points A and C, a segment
whose length equals the difference between Earth’s Ecliptic longitude at the
time of launch and the destination’s Ecliptic longitude at the time of arrival ($=2_{E}-2_{D}).
And the side (
represents the angle between the line from Earth to the sun at the time of
launch and the line from the sun to our destination at the time of arrival, the
complement of the angle through which we want the sun to deflect our ship’s
trajectory. We thus have the deflection as

(Eq’n 1)

Now we solve our triangle for the vertex angle A and the side (. Using the cosine rule, we get

(Eq’n 2)

because cosC=cos(90E)=0. To find the angle A we use the cosine rule in the form

(Eq’n 3)

which we rearrange into

(Eq’n 4)

In both cases we do not use negative signs on the angles and we interpret the arc-cosines as positive angles. We then use the location of the destination relative to Earth (either north or south of the Ecliptic, either east or west of Earth’s location) to determine how the plane of our ship’s trajectory tilts relative to the Ecliptic plane and whether our ship flies first over the sun’s northern or southern hemisphere.

If that seems a little too easy, take comfort in the assurance that the truly nasty part has yet to show up. A vast gulf of effort spans the difference between determining how the ship must turn as it passes the sun and determining how to aim the ship’s velocity and thrust to achieve that turn.

For now, though, let’s work an example. Assume that on the
date 2159 May 18 a ship leaves Earth bound for one of the planets of Alpha
Centauri. On that date, 1.593778 centuries past 2001 Jan 01, Alpha Centauri will
occupy the Ecliptic coordinates of latitude
N=-42.8838E
and longitude 2=239.891E.
On that same date, 58 days after it passes through the First Point of Aries (the
vernal equinox) and 138 days after the first day of the year, the sun will
occupy Ecliptic latitude zero (by definition) and Ecliptic longitude 58.15E,
which makes Earth’s Ecliptic longitude at launch
2_{E}=301.85E.
We thus have "=42.8838E
and $=61.959E,
so we calculate, through Equations 2, 1, and 4,

(Eq’n 5)

Alpha Centauri lies south of the Ecliptic and in this scenario has an Ecliptic longitude less than Earth’s, the difference having a magnitude less than 180 degrees, so the star’s location lies west of Earth’s location at launch. If we look toward Earth from the sun, with the Ecliptic north pole overhead, we have the plane of the ship’s trajectory slanting from southwest to northeast, from our lower right to our upper left, at an angle of 46.465 degrees. If we reverse that point of view, looking toward the sun from Earth with Ecliptic north overhead, then the plane in which our ship’s trajectory lies slants from our lower left to our upper right, tilting 43.535 degrees clockwise from the axis of Earth’s orbit. We must thus aim our ship at a point above the sun’s northern hemisphere (to our upper right), aiming for a point close enough to the sun that the sun’s gravity will bend the ship’s trajectory through an angle of 110.149 degrees. Determining precisely where the aiming point lies relative to the sun’s center gives us a problem that I will solve in another essay.

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