A parabola from Earth to Mars
- Round
- Theory
- Points
- 30
- Level
- Silver
- Taught in
- Section 3.6
Adapted from IOAA 2010 (Beijing), theory problem 16. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.
A spacecraft leaves Earth and is quickly brought to the speed, along the direction of Earth's motion round the Sun, that puts it on a parabola with the Sun at its focus. The parabola touches Earth's orbit. Take the orbits of Earth and Mars as circles in the same plane, of radius 1 au and 1.5 au, and count only the Sun's gravity for most of the flight.
(a) What angle
(b) Suppose Mars is very close to the crossing point when the spacecraft gets there. As seen from Mars, before the spacecraft is much affected by Mars's gravity, how fast is the spacecraft approaching, and from what direction relative to the Sun? [20 points]
- The parabola touches Earth's orbit, so at 1 au the spacecraft moves at right angles to the radius, at the escape speed
km/s. - At 1.5 au the speed is the escape speed there,
km/s. Angular momentum keeps the same, so km/s, and the part along the radius is km/s. - Mars's orbit is a circle, so its direction is at right angles to the radius, and
, which gives . - Mars moves at
km/s at right angles to the radius. Relative to Mars the spacecraft moves at km/s outwards along the radius and km/s along Mars's motion. - Its speed relative to Mars is
km/s. The velocity points almost straight away from the Sun, at from that direction, towards Mars's direction of motion.
Answer(a)
This problem also appears among the olympiad problems of Section 3.6, Escape and hyperbolic orbits, which teaches the ideas it uses. Equation and figure numbers refer to the course.