Energy and the vis-viva equation

Builds on: Sections 3.1 and S4
Reading time: 40 min
Exercises: 5
Olympiad problems: 2

The International Space Station can often be seen after dusk as a bright point that crosses the sky in a few minutes. It circles Earth about 420 km up at 7.66 km/s, once every 92.8 minutes, on an orbit that is very nearly a circle, so its speed hardly changes. A comet falling in from the edge of the Solar System, or a spacecraft on its way to Mars, has a different speed at every point of its orbit. In this section we find that speed from the energy, with a single formula called the vis-viva equation, which most olympiad orbit problems use.

3.2.1The energy of an orbit

Before you readMake a guess first

Two spacecraft pass the same point above Earth at the same speed. One moves horizontally and the other is climbing at 45°. Both are on closed orbits. Which one has the longer period?

A body on an orbit trades kinetic energy for potential energy as it moves, speeding up as it falls towards the central body and slowing down as it climbs away, while its total energy ((S.20)) stays the same. For a circular orbit (S.21) gave that total as $- G M m / 2 r$ . For an ellipse the radius is replaced by the semi-major axis, so the energy depends only on the size of the orbit and not on its shape.

ResultThe energy of an orbit(3.4)

E = \frac{1}{2} m v^{2} - \frac{G M m}{r} = - \frac{G M m}{2 a}

$E$	the total energy of a body of mass $m$ on its orbit
$v, r$	its speed and its distance from the central body at any one moment
$M$	the mass of the central body, much larger than $m$
$a$	the semi-major axis of the orbit

A bigger orbit has a larger, less negative, energy. Raising an orbit therefore takes energy, even though the body moves more slowly on the bigger orbit.

Show the derivation

At the pericentre and the apocentre the velocity is at right angles to the line from the central body, so angular momentum gives $v_{a} = v_{p} r_{p} / r_{a}$ ((3.3)). Energy is the same at both points, so

\frac{1}{2} (v_{p}^{2} - v_{a}^{2}) = G M (\frac{1}{r_{p}} - \frac{1}{r_{a}}) .

Putting in $v_{a}$ , the left side is $\frac{1}{2} v_{p}^{2} (r_{a}^{2} - r_{p}^{2}) / r_{a}^{2}$ and the right side is $G M (r_{a} - r_{p}) / (r_{p} r_{a})$ . Dividing both by $r_{a} - r_{p}$ and using $r_{p} + r_{a} = 2 a$ gives

v_{p}^{2} = \frac{G M}{a} \frac{r_{a}}{r_{p}} .

The energy for each kilogram is then

\begin{aligned} \frac{E}{m} & = \frac{G M r_{a}}{2 a r_{p}} - \frac{G M}{r_{p}} \\ = \frac{G M}{r_{p}} \frac{r_{a} - 2 a}{2 a} = - \frac{G M}{2 a}, \end{aligned}

since $r_{a} - 2 a = - r_{p}$ .

3.2.2The vis-viva equation

Setting the two forms of the energy in (3.4) equal and dividing by $m / 2$ gives the speed at any distance from the central body. The name means "living force", an old word for kinetic energy.

ResultThe vis-viva equation(3.5)

v^{2} = G M (\frac{2}{r} - \frac{1}{a})

$v$	the speed of the orbiting body when it is at distance $r$
$r$	its distance from the centre of the central body
$a$	the semi-major axis of the orbit
$M$	the mass of the central body, replaced by the total mass $M + m$ when the two masses are comparable

For a circle,

r = a

and the equation gives

v^{2} = G M / r

, as in (S.19). For orbits round the Sun, with

r

and

a

in astronomical units, it becomes

v = 29.78 km/s \times \sqrt{2 / r - 1 / a}

, since 29.78 km/s is Earth's average orbital speed. At the pericentre it gives

v_{p}^{2} = (G M / a) (1 + e) / (1 - e)

For Earth, with $e = 0.0167$ , the speed at perihelion is $29.78 \times \sqrt{1.0167 / 0.9833} = 30.28$ km/s and the speed at aphelion is $29.78 \times \sqrt{0.9833 / 1.0167} = 29.29$ km/s. Their ratio is the 1.034 found in Section 3.1 from angular momentum.

Quick checkSpeed on an orbit twice the size of Earth's

An asteroid has an orbit round the Sun with a semi-major axis of 2 au. How fast does it move when it is 1 au from the Sun?

km/s

Example 3.3How fast does Halley's Comet move?

Halley's Comet has a semi-major axis of 17.83 au and an eccentricity of 0.967. How fast does it move at perihelion and at aphelion?

Solution

By (3.1), the perihelion distance is $17.83 \times 0.033 = 0.588$ au and the aphelion distance is $17.83 \times 1.967 = 35.07$ au. At perihelion, by (3.5),

\begin{aligned} v_{p} & = 29.78 \times \sqrt{\frac{2}{0.588} - \frac{1}{17.83}} \\ = 29.78 \times \sqrt{3.399 - 0.0561} = 54.5 km/s . \end{aligned}

At aphelion the two terms are $2 / 35.07 = 0.05703$ and $0.0561$ , so

v_{a} = 29.78 \times \sqrt{0.00094} = 0.91 km/s .

AnswerHalley's Comet moves at 54.5 km/s at perihelion and at only 0.91 km/s at aphelion, which is a ratio of 60, as (3.3) gave in Section 3.1. At aphelion the two terms of the vis-viva equation nearly cancel, so they must be worked out to several figures.

Example 3.4The speed at the bottom of a transfer orbitfinish it

A satellite is on a transfer orbit with its perigee 300 km and its apogee 35 786 km above Earth's surface, the orbit of an exercise in Section 3.1. Earth's radius is 6371 km and $G M$ for Earth is $3.986 \times 10^{14}$ m³ s⁻². How fast does the satellite move at perigee?

The start

The perigee is $r = 6671$ km from Earth's centre, and the semi-major axis is $a = 24 414$ km, so $2 / r = 2.998 \times 10^{- 7}$ m⁻¹ and $1 / a = 4.096 \times 10^{- 8}$ m⁻¹.

Your stepNow use the vis-viva equation to find the speed.

km/s

3.2.3Ellipse, parabola or hyperbola

The sign of the energy decides what kind of path a body follows. A body with negative energy is bound, and moves on an ellipse that brings it back again and again. A body with exactly zero energy moves at the escape speed of (S.21) at every point, and follows a parabola, an open curve that carries it ever further away while its speed falls towards zero. A body with positive energy follows a hyperbola, and still has some speed left when it is very far away. To decide which, we compare the speed $v$ of the body with the escape speed $\sqrt{2 G M / r}$ at its distance.

ResultThe kind of orbit from the speed(3.6)

\begin{aligned} v^{2} < \frac{2 G M}{r} & ellipse \\ v^{2} = \frac{2 G M}{r} & parabola \\ v^{2} > \frac{2 G M}{r} & hyperbola \end{aligned}

$v, r$	the speed of the body and its distance from the central body at any one moment
$M$	the mass of the central body

The test does not depend on the direction of motion. The direction decides how the orbit is turned and how close the body comes to the central body, which Section 3.3 takes up.

Common errorsSection 3.2

In the vis-viva equation $r$ is measured from the centre of the central body. For a satellite the height above the ground must have Earth's radius of 6371 km added to it.
The energy depends only on the semi-major axis, so two orbits with the same $a$ round the same body have the same energy and the same period, however different their shapes.
Raising an orbit takes energy, and yet the body ends up moving more slowly. Energy goes into the potential energy, which, between two circular orbits, grows by twice as much as the kinetic energy falls.
The shortcut $v = 29.78 km/s \times \sqrt{2 / r - 1 / a}$ holds only round the Sun, with $r$ and $a$ in astronomical units. Round Earth or any other body, $G M$ must be put in with consistent units, such as metres and seconds.

ExercisesSection 3.25 questions

3.2.1
A satellite on a circular orbit fires its engine forwards for a moment, so that it speeds up. Afterwards, how do its energy and its period compare with before?
3.2.2
Earth's orbit has a semi-major axis of 1 au and an eccentricity of 0.0167. How fast does Earth move at aphelion?
km/s
3.2.3
A spacecraft passes 1 au from the Sun at 35 km/s. What is the semi-major axis of its orbit?
au
3.2.4
A satellite of mass 1000 kg is on the transfer orbit of the example above, with a semi-major axis of 24 414 km. What is its total energy? Take $G M$ for Earth as $3.986 \times 10^{14}$ m³ s⁻².
× 10⁹ J
3.2.5
A body is found 2 au from the Sun moving at 30 km/s. What kind of path is it on? The escape speed at 2 au is $42.1 / \sqrt{2} = 29.8$ km/s.

Olympiad problemsPast IOAA problems, restated2 problems

2011Is Voyager leaving the Solar System?10 points

Adapted from IOAA 2011 (Katowice), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

On 9 March 2011 the Voyager probe was 116.406 au from the Sun and moving at 17.062 km/s. Is it on an elliptical, a parabolic or a hyperbolic orbit? What is the apparent magnitude of the Sun as seen from Voyager? Take the Sun's apparent magnitude from Earth as −26.74.

I solved it without the solution

AnswerVoyager is on a hyperbolic orbit and will leave the Solar System. The Sun shines on it at magnitude −16.41, still about 30 times as bright as the full Moon seems from Earth.

2010A walk round an asteroid10 points

Adapted from IOAA 2010 (Beijing), theory problem 6. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

A spacecraft lands on a spherical asteroid that does not rotate. The asteroid is 2.2 km across and its average density is 2.2 g/cm³. Can an astronaut walk once round its equator in 2.2 hours? Answer yes or no, and explain why with formulae and numbers.

I solved it without the solution

AnswerThe answer is no. The walk needs 0.873 m/s, which is faster than the 0.863 m/s of a circular orbit at the surface, so the astronaut would leave the ground. Even an orbit just above the surface takes 2.23 hours to go round.