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#21d Flight to Mars: the Return Trip
Index
19.Motion in a Circle
20. Newton's Gravity
21. Kepler's 3rd Law
21a.Applying 3rd Law
21b. Fly to Mars! (1)
21c. Fly to Mars! (2)
21d. Fly to Mars! (3)
22.Reference Frames
22a.Starlight Aberration
22b. Relativity
22c. Flight (1)
22d. Flight (2)
23. Inertial Forces
23a. The Centrifugal Force
23b. Loop-the-Loop
24a.The Rotating Earth
24b. Rotating Frames
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The Hohmann transfer ellipse may be shown to be the most efficient use of rocket thrust for reaching Mars. Other trajectories may get there faster, but will require more thrust to begin with and a larger thrust adjustment at the end, including perhaps a change in direction.
Any drawbacks? Just one: a very stringent requirement on the relative positions of Earth and Mars at the time of launch. As will be seen, the required conditions occur only once in about 26 months. For flying back from Mars to Earth the Hohmann ellipse can also be used, but again, the planets need to be positioned just right at the time of launch. If astronauts from Earth ever land on Mars, they must choose between waiting over one year for the right conditions to occur, or else taking a more direct but less economical ride home. In what follows this delay is calculated. That also involves the concept of the synodic period of Mars, its orbital period as seen from the orbiting Earth.
Hohmann Orbits--to and from Mars
The circles drawn on the left will be used to mark the motion of Earth (inner circle) and of Mars (outer circle) around their orbits. As was shown in section #21b, when a spacecraft is launched from Earth to Mars, the two should be at the points denoted by "1." After 0.70873 years, the craft arrives at Mars, which has meanwhile moved to the point marked "2." Where is Earth at that time? In one year, it goes around 360 degrees, so in 0.70873 years it covers |
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It will therefore reach position "2" on the inner circle, 75.140 past the position of Mars. Notice how Earth has overtaken Mars--at launch (position "1") it lagged behind, but now it is ahead. As Kepler's 3rd law shows, the closer a planet is to the Sun, the faster it completes its orbit, and Earth is closer than Mars. Suppose the spacecraft which has landed on Mars is a robot , which collects a sample and immediately takes off again for its return trip. Launching from point (2), after breaking free from the planet's gravity, it can again follow the Hohmann transfer ellipse, in a mirror image of the flight to Mars (drawing below).
Its trip starts with a reverse thrust of 2.545 km/s, reducing its orbital velocity from V3 to V2. Then, after 0.70873 years, it arrives again at the point marked (1) with velocity V1, which needs to be reduced to the Earth's orbital velocity V0 by a reverse thrust of 2.966 km/s. Unfortunately ... Earth will not be waiting there!
Let's see where Earth should be at the time of launch from Mars, in order for the return flight to meet it at the proper time. The return trip, being half the Hohmann ellipse, takes 0.70873 years, during which (as calculated above) Earth covers an arc of 255.140 in its orbit. To meet the return rocket when it reaches Earth's orbit at point (1), Earth must be--at the start of the return flight from Mars --255.140 behind point "1" in its orbit. That puts it at position (3), 75.140 behind the position of Mars, not at position (2) where Earth is 75.140 ahead of Mars. Since Earth and Mars constantly change their relative position, it stands to reason that if we sufficiently delay the return trip, Earth will move from position (2) relative to Mars to position (3), at which time the return trip can begin. Let us calculate that delay. To simplify the calculation of the delay, we calculate the relative rotation velocity between Earth and Mars around the Sun.
Each year, Earth increases its lead over Mars by (1 - 0.531293) = 0.468707 orbits
If Mars and Earth start out on the circle abreast of each other, after 2.13353 they will be again abreast. From the point of view of an observer on Earth, that is the time required by Mars for one full circle around the sky. It is known as the synodic period of Mars, and is about 25.6 months. The rest is easy. For the Earth to change its position relative to Mars from (2) to (3), Earth must advance (relative to Mars) by To advance by 3600 takes 2.13353 years, so to advance by the above angle takes
A more accurate calculation gives 459 days (ours contains approximations). When the return rocket arrives at Earth, it will be overtaking it, since its velocity V1 is exceeds the orbital velocity V0 of Earth by about 3 km/s. Before safely descending to the ground, the spacecraft also must get rid of the velocity v0 given to it by the pull of the Earth, about 11.3 km/s. However, if it reenters by skimming the atmosphere "just right," its extra kinetic energy will safely dissipate as heat, with no need for more rocket firings. AfterthoughtsOur diagrams only mark progress in circular orbits around the Sun, which keep approximately the same speed. The relative motion of Earth and Mars in the sky is much more variable, because the Earth-Mars distance changes all the time. Indeed, when Earth is closest to Mars and overtaking it, Mars will seem (for a while) to be moving backwards among the stars. The total period however remains 25.6 months, quite different from the actual orbital period of Mars which is 1.8822 years = 22.6 months. Considering all those complications, one can appreciate the subtlety of the work of Copernicus and Kepler, who extracted clean regular patterns of motion, from the much less regular ones of planets across the heavens. Questions from Users: Overtaking planets *** A "short stay on Mars" |
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Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 9-24-2004
Reformatted 24 March 2006
