"But there are other means at our disposal. You are familiar with our artificial satellite Lunetta, circling the Earth these many years without any application of propulsive power.
Her orbit around is subjected to the same Keplerian Laws that determine the movements of Earth and Mars around the Sun. The centrifugal force imparted to her components when they were freighted up to her orbit and assembled sustains her weight and prevents her from falling to Earth. Her orbital velocity of 7.07 km/sec just suffices to keep her centrifugal force adequate to maintain that situation.
"In order to retain the required residual velocity of 3.03 km/sec beyond the field of terrestrial gravitation, we should only be obliged to accelerate the ship by 3.31 km/sec along Lunetta's orbit and in the direction of her motion. This additional slight velocity increment to the orbital velocity, 7.07 kilometers per second, of Lunetta, will give us what we need. You will note that such a maneuver calls for only a minute fraction of the power of a ship designed to rise from Earth to Lunetta.
"There is, however, a regrettable catch in the Lunetta business, for Mars' orbit lies in the same plane as the ecliptic, whereas Lunetta's does not — a practical and economic drawback to launching from Lunetta. Thus we must effect the departure of Operation Mars from an orbit similar to that of Lunetta, except that it must lie in the plane of the ecliptic.
We have actually selected a path of departure almost exactly equivalent to that of Lunetta and which, at the intersections of the planes, lies but a few miles from Lunetta's orbit." Spencer paused and ran a flamboyant bandanna handkerchief over his hairless skull.
Then he downed a glass of water and continued.
"We must remember that our problem is not limited to shooting our rocket ship to Mars on a one-way trip. You have doubtless realized from the foregoing that this might be done with our existing vessels if they were to depart from an orbital path around the Earth. But our expedition must not only reach Mars, it must remain there for quite a time and then return safely. This is much more difficult of achievement, so let us consider the demands made upon us by this problem.
"I have just shown you that 3.03 km/sec beyond the orbital velocity of the Earth is required to be imparted to a rocket ship after it has left the field of gravitation of the Earth, if it is to follow an elliptical free-flight path that will contact the orbit of Mars after one half a circle of the Sun, and that this velocity must be imparted in the same direction as that of Earth's orbital velocity. We also have learned that, if the start is made from an orbital path around the Earth in the plane of the ecliptic, in which path the ship has an initial velocity of 7.07 km/sec, we shall only have to accelerate it by 3.31 km/sec in order that it may leave the field of gravitation with the required residual velocity of 3.03 km/sec.
To attain this velocity increment of 3.31 km/sec is the first propulsive problem of our trip. When it has been completed, the ship will coast on without power on a Keplerian ellipse through the solar system. After 260 days, at the time it reaches its aphelion, it will intercept the Martian orbit. Here I should note that all these figures are valid for the mean distance of Mars from the Sun. That is, I have assumed Mars' orbit to be circular for purposes of simplification.
"If we so arrange matters that Mars arrives at the point of contact of this ellipse with its orbit after 260 days, the ship will enter Mars' field of gravitation. We must remember that Mars has an orbital velocity of 24.1 km/sec, while the velocity of the ship at this point will have been reduced to 21.5 km/sec by its battle against solar attraction during the voyage. Mars will, therefore, overhaul the ship from astern with a differential speed of 2.6 km/sec. If the ship at its aphelion were exactly in Mars' orbit, it would be caught by his gravity and crash upon Martian soil.
"But we are in a position to locate the ship's aphelion just inside of Mars' orbit. Thus Mars would attract it hyperbolically, according to the laws of motion of heavenly bodies.
The ship would approach most closely to Mars at the vertex of such hyperbola, and it would escape Martian gravity for all eternity on the second branch of the hyperbola. But, by the use of the rocket power plant, we may reduce the ship's velocity appropriately just before it reaches this vertex and cause it to enter a circular or slightly elliptical orbital path around Mars. Its radius around Mars will be about the same as the distance of the vertex of the hyperbola from the planet. The ship may orbit here indefinitely without propulsion.
In effect, it will have become an artificial satellite of Mars and will so remain until the situation is again altered by a propulsive maneuver.
"We have selected for our purposes an orbit at 1,000 kilometers from the surface of Mars. In order to go into this satellite orbit, a change in velocity of 2.01 km/sec has been computed as necessary. If the trip to Mars involved only this entrance into a satellite orbit, our ship would have two velocity changes only: 3.31 km/sec to leave the satellite orbit around Earth, and 2.01 km/sec for entering the satellite orbit around Mars, a total of 5.32 km/sec. Even this is far less than the velocity change necessary to reach Lunetta from Earth.
"The return trip requires exactly the same velocity changes. 2.01 km/sec are required to depart from the Martian satellite orbit so that the residual velocity beyond Mars's gravitational field is again 2.6 km/sec. If we impart this residual velocity to the ship in a direction opposed to that of the travel of Mars around the Sun, it will eventually follow the latter at a rate 2.6 km/sec less than his own. The ship's orbital speed, being now less than that of Mars, reduces centrifugal force so that solar attraction outweighs the former.
Thus the ship will move in a Keplerian ellipse which somewhat represents the continuation of the ellipse of arrival, and this ellipse contacts the orbit of the Earth at perihelion after 260 days. If the timing is such that the Earth reaches this point at the same time, the ship may be attracted into a hyperbolic path by the former. We can arrange matters so that this hyperbolic path will have its vertex at the same height as the circular path from which we originally took our departure. Now we must decelerate the ship just before it reaches this vertex and bring it down to the local orbital velocity, just as we did near Mars. We shall require a velocity change of exactly 3.31 km/sec for this purpose, identical with that of the departure. This is plain if the principle of the conservation of energy is considered. Once the ships are in such a satellite orbit, we can take off the crews with Sirius-class ships.
"To recapitulate: the trip to Mars and return will necessitate four propulsive maneuvers as follows:
1. Departure from satellite orbit around Earth. Velocity change: 3.31 km/sec.
2. Entrance into satellite orbit around Mars. Velocity change: 2.01 km/sec.
3. Departure from satellite orbit around Mars. Velocity change: 2.01 km/sec.
4. Entrance into satellite orbit around Earth. Velocity change: 3.31 km/sec.
"The sum of the velocity changes is therefore 10.64 km/sec.
"We need propellants to change the velocity of a rocket vessel, and their amount determines the size of the ships and their method of construction. Interplanetary filling stations, you'll agree, will be few and far between for some years, so we must carry along all propellants required, down to those for the final maneuver. Since the fuel which will be used in the last maneuvers represents ballast during the earlier ones, more propellant is required during the earlier maneuvers to achieve equal velocity increments.