Escape and hyperbolic orbits

Builds on: Sections 3.2, 3.3 and 3.5
Reading time: 35 min
Exercises: 4
Olympiad problems: 1

In October 2017 a telescope in Hawaii found a small body moving through the inner Solar System too fast to be bound to the Sun. Named 'Oumuamua, it was the first object known to come from another star. Its orbit is a hyperbola with eccentricity 1.20 and perihelion 0.256 au, which means that far from the Sun it moves at about 26 km/s, and that the Sun turned its path through about 113°. In this section we work out such numbers for any open orbit, and see how spacecraft use the same turn round a planet to gain speed.

3.6.1The speed far away

Before you readMake a guess first

A rocket is launched straight up from Earth's surface, with no air, at 12.0 km/s, a little above the escape speed of 11.19 km/s. Roughly how fast is it moving when it is very far from Earth?

A body with positive energy ((3.6)) still has some kinetic energy when it is very far away. Its speed there is called the hyperbolic excess speed $v_{\infty}$ , and energy conservation links it to the speed at any distance.

ResultHyperbolic excess speed(3.14)

v^{2} = v_{\infty}^{2} + \frac{2 G M}{r} = v_{\infty}^{2} + v_{esc}^{2}

$v$	the speed at distance $r$ from the central body
$v_{\infty}$	the speed very far away, the same on the way in and on the way out
$v_{esc}$	the escape speed at distance $r$ , from (S.21)

For a hyperbola the vis-viva equation becomes

v^{2} = G M (2 / r + 1 / a)

, with

a

the distance from the centre of the hyperbola to its vertex, so

v_{\infty}^{2} = G M / a

. A parabola has

v_{\infty} = 0

Quick checkWhat is left after escaping

Take $G M = 3.986 \times 10^{14}$ m³ s⁻² and Earth's radius as 6371 km. A rocket leaves the ground at 12.0 km/s, with no air. How fast is it moving far from Earth?

km/s

Example 3.10Leaving Earth for Mars

The Hohmann transfer to Mars in Section 3.5 needs the spacecraft to leave Earth's neighbourhood with 2.95 km/s more than Earth's own speed round the Sun. The spacecraft starts on a circular orbit 300 km above the ground. What burn does it need? Take $G M = 3.986 \times 10^{14}$ m³ s⁻² and Earth's radius as 6371 km.

Solution

The 2.95 km/s is the speed the spacecraft must keep once it is far from Earth, so it is $v_{\infty}$ for its escape. At $r = 6671$ km, $2 G M / r = 119.50$ km² s⁻², so by (3.14)

v = \sqrt{8.70 + 119.50} = 11.32 km/s .

The circular orbit moves at 7.73 km/s, so the burn must add $11.32 - 7.73 = 3.59$ km/s along the spacecraft's direction of motion. It is timed so that, once the spacecraft is clear of Earth, it moves in the same direction as Earth round the Sun.

AnswerThe burn adds 3.59 km/s. Only 0.39 km/s of the final 11.32 km/s is above the escape speed of 10.93 km/s at that height, and yet it leaves the spacecraft 2.95 km/s once it is clear of Earth.

3.6.2The shape of a hyperbola

A hyperbola is fixed by its closest distance $r_{p}$ and its speed far away. Far from the central body the path becomes a straight line, called an asymptote, on the way in and another on the way out. The angle between them is the angle through which the central body turns the path.

ResultA hyperbolic pass(3.15)

\begin{aligned} e & = 1 + \frac{r_{p} v_{\infty}^{2}}{G M} \\ \sin \frac{δ}{2} & = \frac{1}{e} \\ b & = r_{p} \sqrt{1 + \frac{2 G M}{r_{p} v_{\infty}^{2}}} \end{aligned}

$e$	the eccentricity of the hyperbola, more than 1
$r_{p}$	the closest distance from the centre of the central body
$v_{\infty}$	the speed far away
$δ$	the angle between the directions of motion far away on the way in and on the way out
$b$	the aiming distance, the distance at which the body would pass the central body if it were not pulled

The first line comes from

r_{p} = a (e - 1)

with

a = G M / v_{\infty}^{2}

. The second comes from the direction of the asymptotes,

\cos θ = - 1 / e

, from (3.7). The third comes from angular momentum,

b v_{\infty} = r_{p} v_{p}

, and shows that gravity pulls in bodies aimed some way off, so a planet catches far more passing bodies than its size alone would.

For 'Oumuamua, $\sin (δ / 2) = 1 / 1.20$ gives $δ / 2 = 56.4 °$ , so the Sun turned its path through 113°. By (3.14) it passed perihelion at 87.3 km/s.

3.6.3Gravity assists

In the frame of a planet, a spacecraft that flies past arrives and leaves at the same speed $v_{\infty}$ , only turned through $δ$ . The planet, however, is moving round the Sun. Relative to the Sun the spacecraft's velocity is its velocity relative to the planet plus the planet's own velocity, and turning the first part towards the planet's direction of motion makes the total longer. This is a gravity assist. The spacecraft gains energy relative to the Sun, and the planet loses the same amount, far too little to notice. Voyager 2 used assists at Jupiter, Saturn and Uranus to reach Neptune in 12 years.

Figure 3.5A flyby of Jupiter

Closest approachSpeed far from Jupiter

Eccentricity
Speed relative to the Sun

The figure shows the path of a spacecraft past Jupiter J in Jupiter's own frame, a hyperbola with its closest point P and its two asymptotes dotted. Jupiter moves round the Sun upwards on the page at 13.06 km/s, shown by the arrow V, and the spacecraft arrives moving at right angles to that. The drawing is scaled so that P is always the same distance from J, so Jupiter is drawn smaller as the pass gets wider. The inset adds V to the spacecraft's velocity relative to Jupiter, dashed, on the way in and on the way out, and gives its velocity relative to the Sun, heavy on the way out. A closer pass turns the path further and gains more. A slower pass also turns further, but a pass that is too slow gains less, since the change of velocity is

2 v_{\infty} / e

. The figure can be set to , , which an exercise below uses, or , which hardly turns at all.

The change of the spacecraft's velocity is the difference between the outgoing and incoming velocities relative to the planet, which have the same size $v_{\infty}$ and the angle $δ$ between them. Its size is $2 v_{\infty} \sin (δ / 2)$ , so a pass can change the velocity relative to the Sun by at most $2 v_{\infty}$ .

Example 3.11How wide a target Jupiter isfinish it

A small body approaches Jupiter at $v_{\infty} = 10$ km/s. How far off its aim can it be and still graze Jupiter's surface? Jupiter has $G M = 1.267 \times 10^{17}$ m³ s⁻² and a radius of 71 492 km.

The start

A grazing pass has $r_{p} = R_{J}$ , and

\frac{2 G M}{R_{J} v_{\infty}^{2}} = \frac{2 \times 1.267 \times 10^{17}}{7.1492 \times 10^{7} \times 10^{8}} = 35.4 .

Your stepNow find the aiming distance in Jupiter radii.

Jupiter radii

Common errorsSection 3.6

Speeds do not subtract when a body escapes. The squares do, by (3.14), so a launch a little above the escape speed leaves a large speed far away.
In a gravity assist the spacecraft's speed relative to the planet is the same before and after. The gain appears only relative to the Sun, where the planet's own motion is added.
The aiming distance $b$ is larger than the closest distance $r_{p}$ , since gravity bends the path inwards. Using $r_{p}$ as the aim misses by a wide margin for a slow pass.
The angle $δ$ is the turn of the direction of motion, the angle between the two asymptotes as the body follows them. It is 0° for a pass with no pull and approaches 180° for a very close, slow pass.

ExercisesSection 3.64 questions

3.6.1
A comet falls in from very far away, where it was at rest relative to the Sun. What is the eccentricity of its orbit?
3.6.2
A body passes a star on a hyperbola with eccentricity 2. Through what angle is its path turned?
°
3.6.3
A spacecraft passes Jupiter at 1.1 Jupiter radii from its centre with $v_{\infty} = 15$ km/s. Through what angle is its path turned? Use the data of the last example.
°
3.6.4
A spacecraft approaches Venus with $v_{\infty} = 5$ km/s. By at most how much can a flyby of Venus change the spacecraft's velocity relative to the Sun?
km/s

Olympiad problemsPast IOAA problems, restated1 problem

2010A parabola from Earth to Mars30 points

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 $ψ$ does the spacecraft's path make with the orbit of Mars where it crosses it, ignoring the gravity of Mars? [10 points]

(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]

I solved it without the solution

Answer(a) $ψ = 35.26 °$ . (b) The spacecraft approaches at 20.21 km/s. Seen from Mars it comes from the side of the Sun, 10.7° from the direction of the Sun, on the side from which Mars is moving away.