The shape of an orbit

Builds on: Sections 3.1 and 3.2
Reading time: 35 min
Exercises: 5
Olympiad problems: 2

A stone thrown across a field follows what school physics calls a parabola. That is only nearly true. Gravity pulls towards the centre of Earth, 6371 km below, so the pulls at the two ends of the field are not quite parallel. The stone is really moving on a long, thin ellipse with Earth's centre at one focus, and the ground gets in its way long before it could go round. Thrown at 7.91 km/s horizontally, with no air to slow it, the stone would clear the ground all the way round and be in orbit. In this section we write down the equation of the curve that any orbiting body follows, and use it to find the whole orbit from a single position and velocity.

3.3.1The equation of an orbit

Before you readMake a guess first

A comet on a parabola passes perihelion 1 au from the Sun. How far from the Sun is it when it has gone a quarter of the way round the Sun from perihelion, so that the line to it has turned through 90°?

We describe the position of an orbiting body by its distance $r$ from the central body and the angle $θ$ between the line to the body and the line to the pericentre, measured at the central body in the direction of motion. This angle is called the true anomaly. It is 0° at the pericentre and 180° at the apocentre. The distance at $θ = 90 °$ , where the line to the body is at right angles to the major axis, is called the semi-latus rectum $ℓ$ . Every orbit, closed or open, then has the same equation.

ResultThe equation of an orbit(3.7)

r = \frac{ℓ}{1 + e \cos θ}

$r$	the distance of the body from the central body
$θ$	the true anomaly, the angle at the central body from the pericentre to the body
$ℓ$	the semi-latus rectum, the distance at $θ = 90 °$
$e$	the eccentricity, less than 1 for an ellipse, equal to 1 for a parabola and more than 1 for a hyperbola

At the pericentre

r_{p} = ℓ / (1 + e)

, so

ℓ = r_{p} (1 + e)

. For an ellipse this is

ℓ = a (1 - e^{2})

, and for a parabola

ℓ = 2 r_{p}

. On a hyperbola the denominator reaches zero when

\cos θ = - 1 / e

, and the body heads off to infinity along that direction.

The semi-latus rectum is set by the angular momentum alone, while the semi-major axis is set by the energy alone ((3.4)). Together they fix the eccentricity.

ResultThe semi-latus rectum and angular momentum(3.8)

ℓ = \frac{h^{2}}{G M}, h = \frac{L}{m} = r v_{⟂}

$h$	the angular momentum of each kilogram of the orbiting body
$v_{⟂}$	the part of its velocity at right angles to the line from the central body
$M$	the mass of the central body

For an ellipse,

ℓ = a (1 - e^{2})

gives

e = \sqrt{1 - ℓ / a}

. A circle has the largest angular momentum that an orbit of a given energy can have, and a body dropped straight down has none.

Show the derivation

Newton's law gives the acceleration $d v / d t = - G M \hat{r} / r^{2}$ , where $\hat{r}$ is the unit vector from the central body to the orbiting one. The vector $h = r \times v$ is constant, and its size is $r^{2} d θ / d t$ . Then

\begin{aligned} \frac{d}{d t} (v \times h) & = - \frac{G M}{r^{2}} \hat{r} \times h \\ = G M \frac{d \hat{r}}{d t}, \end{aligned}

since $\hat{r}$ turns at the rate $d θ / d t = h / r^{2}$ . Both sides are the rate of change of something, so $v \times h = G M (\hat{r} + e)$ , where $e$ is a constant vector. Taking the dot product with $r$ , and using $r \cdot (v \times h) = (r \times v) \cdot h = h^{2}$ , gives

h^{2} = G M (r + r e \cos θ),

where $θ$ is the angle between $r$ and $e$ . Rearranged, this is $r = (h^{2} / G M) / (1 + e \cos θ)$ , which is (3.7) with $ℓ = h^{2} / G M$ . The constant vector $e$ points to the pericentre, and its size is the eccentricity.

Quick checkMercury a quarter of the way round

Mercury's orbit has a semi-major axis of 0.387 au and an eccentricity of 0.206. How far is Mercury from the Sun when its true anomaly is 90°?

3.3.2Finding the orbit from a launch

A position and a velocity at one moment fix the whole orbit. The speed and the distance give the semi-major axis through the vis-viva equation ((3.5)). The part of the velocity at right angles to the radius gives the angular momentum, and with it the semi-latus rectum by (3.8). That part is $v \cos γ$ , where $γ$ is the flight-path angle, the angle between the velocity and the local horizontal, the line at right angles to the radius. The eccentricity then follows, and (3.7) gives the true anomaly of the starting point.

Figure 3.2A launch from Earth's surface

Launch speedAngle above the horizontal

Path
Semi-major axis and eccentricity
Highest point
Lands

The figure shows Earth to scale with its centre O and a launch point L on the surface. A body leaves L at the chosen speed and angle above the horizontal, with no air to slow it, and the heavy line is its path, which is part of a conic with O at one focus. For an ellipse the empty focus F and the highest point A are marked, and the part of the ellipse inside Earth is dashed. At 7.91 km/s horizontally the path is a circle just above the ground, and at 11.19 km/s in any direction it is a parabola. The figure can be set to , which is the long thin ellipse of a thrown stone, , from the equator, which a problem below works out, , or .

The velocity can also be split along the radius and at right angles to it. Both parts follow from the orbit and the true anomaly, and the flight-path angle is $\tan γ = v_{r} / v_{⟂}$ .

ResultThe parts of the velocity(3.9)

\begin{aligned} v_{r} & = \frac{G M}{h} e \sin θ \\ v_{⟂} & = \frac{G M}{h} (1 + e \cos θ) \end{aligned}

$v_{r}$	the speed straight away from the central body
$v_{⟂}$	the speed at right angles to the line from the central body
$h, θ$	the angular momentum of each kilogram and the true anomaly

The body moves away from the central body while

θ

runs from 0° to 180°, and towards it on the way back. The flight-path angle is zero only at the pericentre and the apocentre.

Example 3.5An orbit from a position and a velocity

A satellite is 7000 km from Earth's centre, moving at 8.0 km/s at 5° above the horizontal. Find the semi-major axis and the eccentricity of its orbit, and the heights of its perigee and apogee above the ground. Take $G M = 3.986 \times 10^{14}$ m³ s⁻² and Earth's radius as 6371 km.

Solution

By (3.5),

\begin{aligned} \frac{G M}{a} & = \frac{2 G M}{r} - v^{2} \\ = 1.1389 \times 10^{8} - 0.64 \times 10^{8} \\ = 4.989 \times 10^{7} m^{2} s^{- 2}, \end{aligned}

so $a = 7990$ km.

The angular momentum of each kilogram is

\begin{aligned} h & = r v \cos γ = 7.0 \times 10^{6} \times 8000 \times \cos 5 ° \\ = 5.579 \times 10^{10} m^{2} s^{- 1}, \end{aligned}

so by (3.8), $ℓ = h^{2} / G M = 7808$ km. Then $ℓ / a = 0.9772$ and

e = \sqrt{1 - 0.9772} = 0.151 .

By (3.1), the perigee is $a (1 - e) = 6783$ km from Earth's centre and the apogee $a (1 + e) = 9198$ km.

AnswerThe orbit has a semi-major axis of 7990 km and an eccentricity of 0.151. Its perigee is 412 km and its apogee 2827 km above the ground. By (3.7),

\cos θ = (ℓ / r - 1) / e = 0.764

, so the satellite is 40° past perigee and still climbing.

Common errorsSection 3.3

The true anomaly is measured at the central body, from the pericentre. An angle measured at the centre of the ellipse is a different angle, and putting it into (3.7) gives a wrong distance.
The flight-path angle is measured from the local horizontal, at right angles to the radius. Only $v \cos γ$ enters the angular momentum, while the whole speed $v$ enters the vis-viva equation.
A body moving horizontally is at its pericentre if it moves faster than a circular orbit there, and at its apocentre if it moves more slowly.
The equation of the orbit knows nothing about the ground. For a body launched from the surface, the conic passes through Earth, and the point where it meets $r = R$ again is where the body lands.

ExercisesSection 3.35 questions

3.3.1
At which true anomaly is a body's distance from the central body equal to the semi-latus rectum?
3.3.2
A comet on a parabola passes perihelion 0.5 au from the Sun. How far is it from the Sun when its true anomaly is 90°?
au
3.3.3
A body passes a star on a hyperbola with eccentricity 1.5. Far away, its direction from the star makes what angle with the direction of the pericentre?
°
3.3.4
Earth's orbit has $e = 0.0167$ , and $G M / h$ for it is $29.78 / \sqrt{1 - e^{2}}$ km/s. How fast is Earth moving away from the Sun when its true anomaly is 90°, in early April?
km/s
3.3.5
A satellite moves horizontally at 8.5 km/s when it is 7000 km from Earth's centre, where a circular orbit needs 7.546 km/s. What is the eccentricity of its orbit?

Olympiad problemsPast IOAA problems, restated2 problems

2015The eccentricity of a satellite from its motion10 points

Adapted from IOAA 2015 (Magelang), theory problem 5. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

An observer wants an approximate value of the eccentricity of an artificial satellite's orbit. When the satellite is at apogee, it is seen to move through $Δ θ_{1} = 2^{'} 44^{″}$ in a short time. When the line from Earth to the satellite is at right angles to the major axis, so that the true anomaly is 90°, it moves through $Δ θ_{2} = 21^{'} 17^{″}$ in the same length of time. Take the observer to be at the centre of Earth. Find the eccentricity of the orbit.

I solved it without the solution

AnswerThe eccentricity is $e = 0.642$ .

2021From the equator to the North Pole15 points

Adapted from IOAA 2021 (Bogotá, held online), theory problem 11. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

(a) What is the smallest speed at which a projectile must be launched from Earth's surface at the equator so that it reaches the North Pole? [12 points]

(b) What is the eccentricity of the projectile's path? [3 points]

Ignore Earth's rotation and its air, and take Earth to be a sphere.

I solved it without the solution

AnswerThe smallest speed is 7.20 km/s, launched at 22.5° above the horizontal towards the north, and the path is an ellipse with eccentricity $\sqrt{2} - 1 = 0.4142$ . The figure above shows this launch.