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Navigator · Module 10

Launch windows: the road to Mars

You cannot aim at Mars, only at where Mars will be in eight months. Why Mars missions leave every 26 months, all at once.

12 min

NASA/JPL/Space Science Institute

Deze les is opgesteld door een lid van het Local Solar System Foundation team en kan verbeterd worden door een gekwalificeerde reviewer. Ben je gekwalificeerd in dit onderwerp? Stel een bewerking voor

In July 2020 three missions left Earth for Mars within twelve days: the United Arab Emirates’ Hope orbiter, China’s Tianwen-1 and NASA’s Perseverance rover. That was no coincidence. Mars can only be reached cheaply when the planets are lined up just right, and that happens about every 26 months.

An idealised Earth-to-Mars Hohmann transfer: circular orbits, planets enlarged about 10,000 times so you can see them.

You do not aim at a planet. You aim at where it will be.

Look

Earth and Mars both circle the Sun, Earth on the inside track and faster. A spacecraft leaving Earth does not fly in a straight line: it follows its own orbit around the Sun, a half-ellipse that starts at Earth’s orbit and ends at Mars’s.

  • The trip takes about 259 days on the cheapest path.
  • In that time Mars moves on too, so it must be about 44 degrees ahead of Earth at departure.
  • Earth overtakes Mars and the same lineup repeats only every 780 days, about 26 months.

Miss the window by a few weeks and the trip costs far more fuel, or becomes impossible for the rocket you have. Wait, and the next window is two years away.

Understand

Treating both orbits as circles (Earth at 1 AU, Mars at 1.524 AU) and using the Sun’s μ = 1.327 × 10¹¹ km³/s²:

  • The transfer ellipse has a = (1 + 1.524)/2 = 1.262 AU; half its period is 259 days.
  • Leaving Earth, the spacecraft needs 2.94 km/s more than Earth’s own 29.8 km/s around the Sun.
  • It arrives at Mars’s orbit 2.65 km/s slower than Mars and must be captured (next lesson).
  • Phase angle: in 259 days Mars covers 360° × 259/687 ≈ 136°, so it must start 180° − 136° ≈ 44° ahead.

Real orbits are slightly elliptical and tilted, so each window is different; mission designers search thousands of launch and arrival dates at once with “porkchop plots”. Paying a little more Δv buys a shorter trip: Perseverance took 203 days.

Master

Synodic period:

1/S = 1/P_Earth − 1/P_Mars  →  S = 1 / (1/365.25 − 1/686.98) ≈ 780 days

Departure from low Earth orbit is not simply 2.94 km/s: that is the hyperbolic excess speed v∞ Earth must be left with. Burning at 200 km altitude, the required speed is √(v∞² + 2μ_E/r), so the burn from a 7.79 km/s parking orbit is about 3.6 km/s, and the Oberth effect makes it cheaper than adding 2.94 km/s far from Earth. Launch vehicles quote this as C3 = v∞² ≈ 8.7 km²/s², the minimum for a Mars departure.

Probeer het eens

Earth's year is 365.25 days and Mars's 687 days. Use the synodic period formula 1/S = 1/P₁ − 1/P₂ to find how often the Earth-Mars geometry repeats. Then check a list of Mars missions: how many left Earth in 2020, and which month?

Een snelle quiz

3 snelle vragen. Kies een antwoord om te zien of je gelijk hebt.

  1. How often do launch windows to Mars open?

    1. A Every year
    2. B About every 26 months
    3. C Every 10 years
    4. D Any day
    Toon het antwoord

    B. About every 26 months

  2. In a Hohmann transfer to Mars, where must Mars be when the spacecraft leaves Earth?

    1. A Directly opposite, behind the Sun
    2. B About 44 degrees ahead of Earth
    3. C Exactly next to Earth
    4. D Behind Earth
    Toon het antwoord

    B. About 44 degrees ahead of Earth The trip takes about 259 days. In that time Mars moves about 136 degrees, so it must start about 44 degrees ahead to meet the spacecraft 180 degrees from the departure point.

  3. Why did the Perseverance rover need only about 7 months instead of 8.5?

    1. A It used a faster, slightly more expensive path than the ideal Hohmann transfer
    2. B Mars came closer that year because it stopped moving
    3. C It flew in a straight line
    4. D It used a gravity assist at the Moon
    Toon het antwoord

    A. It used a faster, slightly more expensive path than the ideal Hohmann transfer

De finishlijn

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