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 will replace chemically powered and fission-powered rockets as diesel-electric locomotives replaced steam-driven locomotives on the railroads across the middle of the Twentieth Century. 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. 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 and 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. 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. For now, though, we seek two angles that comprise, respectively, one angle and one side of a spherical triangle that we need to solve.
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. Looking through the center of the sun, we look toward one of the vertices of the triangle we must solve and we know that the angle in that vertex, angle A, is the angle by which we must tilt the plane of our ship's trajectory relative to the plane of the ecliptic. From that vertex we can construct the spherical triangle that we must solve.
Beginning at vertex A, we proceed along the ecliptic to vertex B, at the point on the ecliptic that marks the ecliptic longitude of the target (planet or star). That line gives us the side γ, whose angular length equals the target's longitude minus the sun's longitude at the time of launch. The angle in vertex B equals 90 degrees. We construct the side α by going from vertex B to the target, which marks vertex C, and note that the angular length of side α equals the target's ecliptic latitude. From vertex C back to vertex A we construct side β, whose angular length equals the angle through which the sun's gravity must deflect our ship's trajectory.
We know two sides of the spherical triangle, α and γ, and their included angle, B=90, so we can use the cosine rule to solve the triangle for side β. We have
(Eq'n 1)
because CosB=Cos(90)=0.
Now we can use the sine rule to find the angle A. We have
(Eq'n 2)
From that equation we get
(Eq'n 3)
If that seems a little too simple, take comfort in the assurance that the truly nasty part is yet to come. 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 Alpha Centauri. On that date Alpha Centauri will occupy the Ecliptic coordinates of latitude -42.8838° and longitude +60.8824°. On that same date the sun will occupy latitude zero degrees (by definition) and longitude 130.5876° (making the sun's longitude 229.4124°). So we have
α =-42.8838°
γ =+60.8824° - 229.4124° = -168.53°
from which we derive
β = 135.898°
and
A = -77.9147°.
Imagine looking down on the solar system from a point above the sun's north pole with longitude 90 at the twelve o'clock position (putting the First Point of Aries on your right). From that viewpoint you would see Alpha Centauri lying on a straight line passing just a smidgen east of one o'clock and Earth lying on straight line passing west of ten o'clock a little over halfway to eleven o'clock. Our ship's trajectory goes from Earth to slightly past the sun and thence to Alpha Centauri.
Now from a point above Earth's north pole look toward the sun as if preparing to launch the ship. Our calculation tells us that you must aim the ship at a point on the north part of a straight line that passes through the center of the sun and tilts 77.9147° clockwise from the axis of Earth's orbit; that is, the plane of the ship's trajectory makes an angle of 12.0853° with the ecliptic plane. That aiming point must lie close enough to the sun that the sun's gravity will bend the ship's trajectory through 135.898°. Determining precisely where the aiming point lies along that line gives us a problem that I will solve in another essay.
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