at nearly the same point in its orbit and may encounter Earth and/or Mars multiple times during each cycle. VISIT-1 encounters Earth three times and Mars four times in 15 years. VISIT-2 encounters Earth five times and Mars two times in 15 years. Some possible Earth–Mars cyclers include the following: A detailed survey of Earth–Mars cycler trajectories was conducted by Ryan Russell and Cesar Ocampo from the University of Texas at Austin, Texas. They identified 24 Earth-Mars cyclers with periods of two to four synodic periods, and 92 cyclers with periods of five or six synodic periods. They also found hundreds of non-ballistic cyclers, ones which would require some powered maneuvers. Physics Earth orbits the Sun in one Earth year, Mars in 1.881. Neither orbit is perfectly circular; Earth has an orbital eccentricity of 0.0168, and Mars of 0.0934. The two orbits are not quite coplanar either, as the orbit of Mars is inclined by 1.85 degrees to that of Earth. The effect of the gravity of Mars on the cycler orbits is almost negligible, but that of the far more massive Earth needs to be considered. If we ignore these factors, and approximate Mars's orbital period as 1.875 Earth years, then 15 Earth years is 8 Martian years. In the diagram opposite, a spacecraft in an Aldrin cycler orbit that starts from Earth at point E1 will encounter Mars at M1. When it gets back to E1 just over two Earth years later, Earth will no longer be there, but it will encounter Earth again at E2, which is , of an Earth orbit, further round. The shape of the cycler orbit can be obtained from the conic equation: Where is 1 astronomical unit, is the semi-major axis, is the orbital eccentricity and (half of ). We can obtain by solving Lambert's problem with as the initial and final transfer angle. This gives: Solving the quadratic equation gives: with an orbital period of 2.02 years. The angle at which the spacecraft flies past Earth, , is given by: Substituting the values given and derived above gives a value for of . We can calculate the gravity assist from Earth: where is the heliocentric flyby velocity. This can be calculated from: where is the velocity of Earth, which is 29.8 km/s. Substituting gives us V = 34.9 km/s, and V = 8.73 km/s. The excess speed is given by: Which