Where a body is on its orbit

Builds on: Sections 3.1, 3.2 and 3.3
Reading time: 40 min
Exercises: 5
Olympiad problems: 3

Halley's Comet last passed perihelion in February 1986 and will be back in July 2061. It reached aphelion, beyond the orbit of Neptune, in December 2023, halfway in time between the two. Yet in each 75-year trip round the Sun it spends only about 2.2 years inside the orbit of Jupiter, where it moves fast and can be seen, and the rest of the time crawling through the outer Solar System. Kepler's second law says why in words. To turn it into numbers we need a way to find where a body is on its orbit at any time, and in this section we build it.

3.4.1The mean and eccentric anomalies

Before you readMake a guess first

Earth passes perihelion in early January. A quarter of a year later, has Earth gone more or less than a quarter of the way round its orbit, measured by the angle at the Sun?

Time on an orbit runs steadily, while the angle at the Sun does not. We therefore count time with an angle that does grow steadily, the mean anomaly $M = 2 π (t - τ) / T$ , where $τ$ is the moment of pericentre and $T$ the period. It is the angle that a body moving at a steady rate round a circle with the same period would have turned through since the pericentre.

To link it to the true anomaly we draw a circle of radius $a$ round the centre of the ellipse, called the auxiliary circle. From the body we draw a line at right angles to the major axis until it meets the circle. The angle at the centre of the ellipse between the pericentre and that point is the eccentric anomaly $E$ . The ellipse is the auxiliary circle squashed in the direction of the minor axis, by the factor $b / a$ , so the body's distance from the focus is simple in terms of $E$ .

Figure 3.3The three anomalies

Eccentricity

e

Time since perihelion

Watch

Mean anomaly $M$
Eccentric anomaly $E$
True anomaly $θ$
Distance SX

The figure shows an orbit with the Sun S at one focus, its centre C, and the auxiliary circle of radius

a

round C, drawn dotted. The planet X moves at its true speed, and Q is the point of the circle straight above or below it, so the angle E at C is the eccentric anomaly and the angle θ at S is the true anomaly. The hollow point Y goes round the circle at a steady rate with the same period, and its angle from CP is the mean anomaly M. All three are 0° at P and 180° at A. On the way from P to A, θ runs ahead of E, which runs ahead of M, and on the way back the order is reversed. The figure can be set to , which the exercises use, or to , when it has turned through 142° round the Sun but Y only 5°.

3.4.2Kepler's equation

The second law says that the area swept since the pericentre grows steadily with time. On the auxiliary circle the matching area is easy to find, and squashing it by $b / a$ gives the area on the ellipse. Setting that area in proportion to the time gives Kepler's equation.

ResultKepler's equation(3.10)

\begin{aligned} M & = E - e \sin E \\ r & = a (1 - e \cos E) \\ \tan \frac{θ}{2} & = \sqrt{\frac{1 + e}{1 - e}} \tan \frac{E}{2} \end{aligned}

$M$	the mean anomaly $2 π (t - τ) / T$ , in radians
$E$	the eccentric anomaly, in radians
$e, a$	the eccentricity and the semi-major axis
$r, θ$	the distance from the focus and the true anomaly

From a time, we find

M

, solve the first line for

E

, and then find

r

and

θ

. From a position, we go the other way, with no equation to solve. The first line needs radians, since it compares

E

, an arc length on a circle of radius 1, with the pure number

e \sin E

Show the derivation

On the auxiliary circle, the area swept from the focus S as the point Q moves from the pericentre P is the sector CPQ, of area $\frac{1}{2} a^{2} E$ , less the triangle CSQ, whose base is CS = $a e$ and whose height is $a \sin E$ . That area is $\frac{1}{2} a^{2} (E - e \sin E)$ . Squashing the circle by $b / a$ into the ellipse squashes every area by the same factor, so the area swept on the ellipse is $\frac{1}{2} a b (E - e \sin E)$ . By the second law it is the fraction $(t - τ) / T$ of the whole area $π a b$ , which gives

\frac{1}{2} a b (E - e \sin E) = π a b \frac{t - τ}{T},

and so $E - e \sin E = 2 π (t - τ) / T = M$ . The position of the body measured from the centre is $(a \cos E, b \sin E)$ , and the focus is at $(a e, 0)$ , so $r^{2} = a^{2} (\cos E - e)^{2} + b^{2} \sin^{2} E$ , which simplifies to $a^{2} (1 - e \cos E)^{2}$ .

Kepler's equation cannot be solved for $E$ with a formula. For an eccentricity well below 1, we can start with $E = M$ and repeat $E \leftarrow M + e \sin E$ , which comes closer each time. With $M = 60 ° = 1.0472$ rad and $e = 0.5$ , the steps give 1.4802, 1.5451 and 1.5470 rad, and the answer is 1.5471 rad, or 88.6°. For an eccentricity close to 1 this is slow, and Newton's method, $E \leftarrow E - (E - e \sin E - M) / (1 - e \cos E)$ started from $E = π$ , is much faster.

Example 3.6How long Halley's Comet spends inside Jupiter's orbit

Halley's Comet has $a = 17.83$ au and $e = 0.967$ , so its period is 75.3 years. For how long in each orbit is it closer to the Sun than Jupiter, at 5.2 au?

Solution

We go from the position to the time, with no equation to solve. By (3.10), the comet is 5.2 au from the Sun when

\cos E = \frac{1 - 5.2 / 17.83}{0.967} = 0.7325,

so $E = 42.9 ° = 0.749$ rad. Then $M = E - e \sin E = 0.749 - 0.967 \times 0.681 = 0.090$ rad. The comet takes this long to go from perihelion to 5.2 au, and as long again to come in, so it is inside 5.2 au for the fraction $2 \times 0.090 / 2 π = 0.0288$ of each period.

AnswerHalley's Comet is inside Jupiter's orbit for

0.0288 \times 75.3 = 2.2

years of each 75.3-year orbit.

Example 3.7When Earth is a quarter of the way roundfinish it

Earth's orbit has $e = 0.0167$ and its period from perihelion to perihelion is 365.26 days. How many days after perihelion is Earth's true anomaly 90°?

The start

By (3.10), $\tan (E / 2) = \sqrt{0.9833 / 1.0167} \times \tan 45 ° = 0.9834$ , so $E = 89.04 ° = 1.5541$ rad.

Your stepNow find the mean anomaly, and from it the time.

days

3.4.3Falling straight in

A body let go from rest falls straight towards the central body. Its path is an ellipse squashed flat, with eccentricity 1 and the central body at one end, so its semi-major axis is half the starting distance. The fall takes half the period of that ellipse, from the far end to the near end.

ResultThe time to fall from rest(3.11)

t = \frac{π}{2} \sqrt{\frac{r_{0}^{3}}{2 G M}}

$t$	the time to fall to the centre of the central body
$r_{0}$	the starting distance, where the body is at rest
$M$	the mass of the central body

The time is half the period of an orbit with

a = r_{0} / 2

, which is

1 / (4 \sqrt{2}) = 0.177

times the period of a circular orbit of radius

r_{0}

. Partway through the fall,

r = a (1 - \cos E)

and

M = E - \sin E

hold, which is Kepler's equation with

e = 1

Quick checkIf Earth stopped

If Earth suddenly stopped moving round the Sun, how many days would it take to fall into the Sun?

days

Common errorsSection 3.4

Kepler's equation needs $M$ and $E$ in radians. Using degrees for one and radians for the other is a common slip, since $e \sin E$ is a number and not an angle in degrees.
The eccentric anomaly is measured at the centre of the ellipse and the true anomaly at the focus. They agree only at the pericentre and the apocentre.
The time from pericentre to apocentre is always half the period, whatever the eccentricity. The two halves of the orbit on either side of the major axis are mirror images.
A body falling from rest moves on an ellipse with semi-major axis half the starting distance. Using the full starting distance as $a$ gives a time $2 \sqrt{2}$ times too long.

ExercisesSection 3.45 questions

3.4.1
A body is a quarter of a period past pericentre. What is its mean anomaly?
3.4.2
An orbit has $e = 0.5$ . A sixth of a period after pericentre, what is the eccentric anomaly? Work in radians and give the answer in degrees.
°
3.4.3
Mars has a period of 686.98 days and an eccentricity of 0.0934. How long does it take to go from perihelion to aphelion?
days
3.4.4
At which eccentric anomaly is a body exactly a distance $a$ from the focus?
3.4.5
If the Moon stopped in its orbit, 384 400 km from Earth, how long would it take to fall to Earth's centre? Take $G M$ for Earth as $3.986 \times 10^{14}$ m³ s⁻² and ignore the Moon's own mass.
days

Olympiad problemsPast IOAA problems, restated3 problems

2011A comet's fall from the Oort Cloud10 points

Adapted from IOAA 2011 (Katowice), theory problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Most comets that pass through the inner Solar System only once come straight in from the Oort Cloud. Estimate how long a comet takes to make this journey, if it was at aphelion in the Oort Cloud, 35 000 au from the Sun.

I solved it without the solution

AnswerThe journey takes about 1.16 million years.

2024An asteroid that falls into a star10 points

Adapted from IOAA 2024 (Rio de Janeiro), 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.

An asteroid of mass $m$ is seen at a distance $d$ from a star of mass $M$ , moving at speed $v = \sqrt{G M / d}$ . The distance $d$ is much larger than the radius of the star. Give each answer in terms of $M$ , $d$ and constants.

(a) If the asteroid is moving straight towards the star, how long will it take to hit it? [8 points]

(b) If it is instead moving straight away from the star, how long will it take to hit it? [2 points]

I solved it without the solution

Answer(a) $t = (π / 2 - 1) \sqrt{d^{3} / G M}$ . (b) $t = (3 π / 2 + 1) \sqrt{d^{3} / G M}$ .

2009A projectile launched very fast45 points

Adapted from IOAA 2009 (Tehran), 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 projectile leaves Earth's surface at sea level with speed $v_{0} = \sqrt{G M / R}$ , at an angle $θ = 30 °$ above the local horizontal, where $M$ and $R$ are Earth's mass and radius. Ignore the air and Earth's rotation.

(a) Show that the path is an ellipse with semi-major axis $a = R$ .

(b) What is the greatest height of the projectile above the surface, in Earth radii?

(d) What is the eccentricity of the ellipse?

(e) How long is the projectile in flight?

I solved it without the solution

Answer(a) $a = R$ . (b) $0.5 R$ . (c) $2 π R / 3 = 2.094 R$ . (d) $e = 0.5$ . (e) $(π + 1) \sqrt{R^{3} / G M}$ , which is 55.6 minutes, or 0.659 of the period of an orbit with $a = R$ . The launch figure in Section 3.3 shows this path.