Venus--Hesper--were we native
To that splendour or in Mars
We should see the globe we groan in
Fairest of their evening stars.
Could we dream of war and carnage
Craft and madness, lust and spite
Roaring London, raving Paris
In that peaceful point of light?
On a star so silver-fair
Yearn, and clasp the hands, and murmur:
"Would to God that we were there"?
Alfred, Lord Tennyson
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)
22c. Flight (1)
22d. Flight (2)
23. Inertial Forces
23a. The Centrifugal Force
24a.The Rotating Earth
24b. Rotating Frames
Using the orbital mechanics developed earlier, we can now plan (with approximations) a space mission to Mars. Assuming our rocket and spaceship are provided--how should they be aimed, and how long would the flight take?
As an approximation assume that the orbits of Earth and Mars are both circles in the same plane, centered on the Sun. The radius of the Earth's orbit, denoted by r1, is close to 150,000,000 kilometers, a distance known as the astronomical unit (AU). That is a convenient unit for measuring distances in the solar system and was used in section #10 on Kepler's Laws. Like there, here too all distances will be measured in AU and all times in years. With such units, for Earth r1 = 1 AU and the orbital period is T1 = 1 year. For Mars (subscript 2), r2 = 1.523691, T2 = 1.8822.
How not to go
First, Earth's gravity will bend the trajectory of any spaceship launched from here. To remove that factor assume the rocket has already been placed in a distant orbit around Earth, where Earth's gravity is weak and orbital motion is slow, allowing both to be neglected. In actual orbit planning they must always be taken into account as a correction.
Even then, that rocket is still orbiting the Sun together together with the Earth, to which it is loosely attached, moving at close to 30 km/sec, a velocity we will write as V0. This is much faster than what is needed to reach the Mars orbit (or can be easily provided by rockets!). If we fire when Mars is closest (drawing), V0 is transverse to the aiming direction, so that the spaceship will start in a direction quite different from the one pointing at Mars--and Mars is sure to move away long before it has covered the intervening distance. That is the second reason.
The third reason is that the entire system is dominated by the Sun's gravity. All objects travel in orbits or trajectories, which by Kepler's laws are parts of conic sections--in this cases, ellipses. In general, these are curved.
In addition, the direction in which the spaceship moves when it arrives should make it easy to match velocities with Mars. This leads to the so called Hohmann Transfer Ellipse (or transfer orbit), first proposed in 1925 by the German engineer Wolfgang Hohmann. That is an ellipse with perihelion P (point closest to the Sun) at the orbit of Earth and aphelion A (point most distant from the Sun) at the orbit of Mars (drawing). A similar transfer ellipse, between low Earth orbit (say, r = 1.1 RE = 1.1 Earth radii) and the synchronous orbit at 6.6 RE (see section #21a) is frequently used to inject communication satellites into their final orbits. We launch from P by giving the rocket an extra velocity in addition to V0, injecting it into the larger ellipse.
Mars should be in such a position relative to Earth at the time of launch, that it reaches point A at the same time as the spaceship does. To determine that position one needs to know the duration of the flight from P to A, and that is derived below, using Kepler's 3rd law.
where T is the orbital period and a the semi-major axis, half the length of the orbital ellipse (a=r in circular orbits). The constant is the same for all objects orbiting the Sun, including of course the Earth. Its exact value depends on the units in which T and a are measured. That value becomes very simple if T is measured in years and a in AU (as is done here). Inserting in the 3rd-law equation the values for Earth gives
In our units, then
Consequently in these units the constant also equals 1, and that value can be used for any planet. On multiplying both sides of the equation by a3 (second law of algebra)
This again holds for any orbit around the Sun, including the one of a spaceship in a transfer ellipse. The length PA of that ellipse is, in AU,
That is the "major axis" of the orbital ellipse, and half its length equals the semi-major axis a. Therefore
Taking the square root
That is the time for going from P to A and back to P. The one-way transit time to Mars is half that, i.e.0.70873 years or about 8.5 months.
Where should Mars be at the time of launch? From numbers cited at the beginning of this section, it takes Mars 1.8822 years for a full orbit of 3600. Therefore, assuming a circular orbit and uniform motion (a less accurate approximation for Mars than for Earth), in 0.70873 years it should cover
We therefore launch when Mars in its orbit is 135.5550 away from point A (drawing)
The next section calculates the velocity which a rocket must impart at point P in order to produce the transfer orbit, and the velocity change required at point A to match velocities with Mars. That calculation is longer and requires some algebra.
Next Stop: #21c. Flight to Mars: The Calculation
Timeline Glossary Back to the Master List
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 12-12-2004