Maths toolkit

Builds on: School algebra, trigonometry and graphs
Reading time: 1 h 40 min
Exercises: 11
Olympiad problems: 4

In 1838 Friedrich Bessel published the first reliable measurement of the distance to a star. He chose 61 Cygni, a pair of faint stars in the Swan that can just be seen with the naked eye from a dark site, because they drift across the sky faster than almost any other star visible to the eye, by more than 5″ a year. A star that moves quickly across the sky is likely to be close to us, and its small yearly shift caused by Earth's motion around the Sun should then be large enough to measure. Its , half of that yearly shift, turned out to be about 0.3″, the angle covered by a coin 2 cm across seen from 14 km away. Turning angles as small as this into distances, and distances and angles into speeds, takes a small kit of mathematics that is used in every chapter of this course. In this section we gather that kit in one place, from radians and logarithms to vectors, derivatives and integrals, the geometry of a sphere, estimation and uncertainties.

Entry checkCan you skip this section?7 questions

If you can already answer these 7 questions, you know what this section teaches and can go straight on. Anything you get wrong shows which subsection to read.

Write 0.5° in radians.

rad

If $\log x = 2.5$ , what is $x$ ?

What is the angle between the vectors $(1, 1, 0)$ and $(0, 1, 1)$ ?

A quantity $y$ is proportional to $x^{3 / 2}$ . By about what percentage does $y$ grow when $x$ grows by 2%?

per cent

What is the radius of the parallel of latitude 60° on a sphere of radius 6400 km?

Without a calculator, which of these is closest to $a b / c^{2}$ , where $a = 6.7 \times 10^{- 11}$ , $b = 2.0 \times 10^{30}$ and $c = 1.5 \times 10^{11}$ ?

A length is measured as 10.0 cm with an uncertainty of 0.2 cm. What is the percentage uncertainty of its cube?

per cent

S3.1Angles and small angles

Before you readMake a guess first

For angles up to what size does $\sin θ$ agree with the angle $θ$ itself, measured in radians, to within 1%?

We measure angles in degrees, and each degree is divided into 60 arcminutes and each arcminute into 60 arcseconds. One arcminute is written 1′ and one arcsecond 1″, so a degree holds 3600″. The full Moon is about 0.52° across, or 31′, and the parallax of 61 Cygni is about 0.3″.

For calculations a different unit is far more convenient. We draw a circle of radius $r$ centred on the corner of the angle. The angle cuts off an arc of the circle, and the size of the angle in radians is the length $s$ of that arc divided by the radius. One radian is therefore the angle that cuts off an arc as long as the radius. A whole turn cuts off the whole circumference, $2 π r$ , so 360° is $2 π$ radians.

ResultArc length(S.2)

s = r θ

$s$	length of the arc
$r$	radius of the circle
$θ$	angle at the centre of the circle, in radians

One radian is 57.30°, or 3438′, or 206 265″. To turn degrees into radians we multiply by

π / 180

The average distance between Earth and the Sun is one astronomical unit (au), $1.496 \times 10^{8}$ km. By (S.2), a length of 1 au seen from 206 265 au away covers an angle of 1/206 265 rad, which is 1″. That distance is one parsec (pc), and a kiloparsec (kpc) is 1000 pc. A star whose parallax is $p$ arcseconds is therefore $1 / p$ parsecs away.

Many angles in astronomy are small, and small angles are easy to work with because three lengths in Figure S.1 become almost equal. In a circle of radius 1 the arc AB is the angle $θ$ itself, the line BH is $\sin θ$ and the tangent AT is $\tan θ$ . As the angle shrinks, B moves down towards A, and the arc, the line BH and the tangent close up into what is very nearly the same short straight line. At the same time OH, which is $\cos θ$ , gets closer and closer to 1.

Figure S.1Why small angles are simpleinteractive

Angle

θ

Angle in radians
BH, arc AB and AT
Using $θ$ for $\sin θ$
Using $θ$ for $\tan θ$

The figure shows a circle of radius

r

centred on O, and a second radius OB at the angle

θ

above OA. The heavy arc AB has length

r θ

. The line BH, dropped at right angles to OA, has length

r \sin θ

, and the tangent AT at A, which meets the line from O through B at T, has length

r \tan θ

. The slider sets

θ

, and the readout gives the three lengths for a circle of radius 1. At they differ clearly, at the arc is 1% longer than BH, and by the three can no longer be told apart.

ResultSmall-angle approximations(S.3)

\begin{aligned} \sin θ & \approx θ \\ \tan θ & \approx θ \\ \cos θ & \approx 1 - \frac{1}{2} θ^{2} \end{aligned}

$θ$	a small angle, in radians

For angles below about 10° each approximation is correct to better than 1%. All three hold only when

θ

is in radians.

Together these are called the small-angle approximation, and they are what make tiny angles easy to use. An object of size $D$ seen side on from a distance $d$ covers an angle $θ$ with $\tan (θ / 2) = D / (2 d)$ , and for a small angle this becomes simply $θ = D / d$ . Section 1.1 uses this relation, (1.1), to turn angular sizes into sizes and distances.

Quick checkUse the small-angle approximation

Find $\sin 2 °$ without using the sine key of a calculator.

Example S.3The space station overhead

The International Space Station is about 109 m across, and it orbits about 420 km above the ground. How large does it look when it passes overhead?

Solution

The station is small compared with its distance, so the angle it covers is

θ = \frac{109 m}{420 \times 10^{3} m} = 2.595 \times 10^{- 4} rad .

Multiplying by 206 265 turns this into arcseconds,

θ = 2.595 \times 10^{- 4} \times 206 265 = {53.5}^{″} .

AnswerThe station looks about 54″ across, a little less than an arcminute. The eye cannot make out details much finer than about 1′, so to the naked eye the station is a bright point of light, while a small telescope shows its shape.

S3.2Logarithms and exponentials

Before you readMake a guess first

The orbits of the planets range in size from 0.39 au for Mercury to 30 au for Neptune, and their periods from 88 days to 165 years. On a graph of period against orbit size, with axes that start at zero and are just long enough for Neptune, where do Mercury, Venus, Earth and Mars appear?

Astronomy deals with numbers that range over many powers of ten, and logarithms turn each power of ten into a step of one. The logarithm of a number $y$ is the power $x$ to which 10 must be raised to give $y$ ,

y = 10^{x} ⟺ x = \log y,

so that $\log 100 = 2$ , $\log 1000 = 3$ and $\log 0.01 = - 2$ . Multiplying a number by 10 adds 1 to its logarithm, and a number between 1 and 10 has a logarithm between 0 and 1. Logarithms to the base $e = 2.718$ are called natural logarithms and written $\ln$ . They appear wherever a quantity grows or shrinks in proportion to its own size, as we see below.

ResultRules of logarithms(S.4)

\begin{aligned} \log (a b) & = \log a + \log b \\ \log (a / b) & = \log a - \log b \\ \log a^{n} & = n \log a \\ \ln y & = 2.303 \log y \end{aligned}

$a, b, y$	positive numbers
$n$	any number

The first three rules hold for natural logarithms too. There is no rule for the logarithm of a sum.

The rules turn multiplication into addition and powers into multiplication. Chapter T4 uses them for the magnitude scale of brightness, on which each factor of 100 in brightness is a step of 5 magnitudes.

Quick checkUse the result above

Given that $\log 2 = 0.301$ , find $\log 8000$ without a calculator.

Many laws in astronomy have the form $y = k x^{n}$ , in which $y$ is proportional to a power of $x$ . This is called a power law. Taking logarithms of both sides turns it into $\log y = \log k + n \log x$ , so a graph of $\log y$ against $\log x$ is a straight line whose slope is the power $n$ . Logarithmic axes draw this graph for us, as Figure S.2 shows for the planets. They are labelled with the numbers themselves but spaced by their logarithms, so that 0.1, 1, 10 and 100 are equally far apart.

ResultA power law on logarithmic axes(S.5)

y = k x^{n} ⟹ \log y = \log k + n \log x

$x, y$	the two quantities, both positive
$k$	a constant
$n$	the power, which is the slope of the line on logarithmic axes

The slope must be worked out from the logarithms, as

(\log y_{2} - \log y_{1}) / (\log x_{2} - \log x_{1})

for two points on the line.

Figure S.2Kepler's third law on two kinds of axesinteractive

Axes

The figure shows the orbital periods of the eight planets against the semi-major axes of their orbits, taken from the JPL table of planetary orbits that the sky charts of this course also use. The semi-major axis is half the longest diameter of an orbit, and Chapter T3 explains it. On the four inner planets are crowded into the corner and the points follow a curve. On each factor of ten takes the same distance along an axis, all eight planets can be seen, and the points lie on a straight line. The thin line is

P = a^{3 / 2}

, with

P

in years and

a

in astronomical units.

Example S.4The power in Kepler's third law

Earth takes 1.0000 year to go round an orbit with a semi-major axis of 1.0000 au, and Jupiter takes 11.86 years for one of 5.203 au. If the period is proportional to a power of the semi-major axis, what is the power?

Solution

If $P$ is proportional to a power of $a$ , all the planets lie on one straight line on logarithmic axes, and by (S.5) the slope between any two of them is the power,

n = \frac{\log 11.86 - \log 1}{\log 5.203 - \log 1} = \frac{1.0741}{0.7163} = 1.500 .

AnswerThe period grows as the 3/2 power of the semi-major axis, so

P^{2}

is proportional to

a^{3}

. This is Kepler's third law. With

P

in years and

a

in astronomical units,

P^{2} = a^{3}

holds to better than 0.2% for every planet in the table.

Some quantities change by the same factor in every equal step of time or distance, rather than by the same amount. Such a quantity changes in proportion to its own size, and it follows an exponential law.

ResultExponential decay(S.6)

N = N_{0} e^{- t / τ}

$N$	the quantity at time or distance $t$
$N_{0}$	its value at $t = 0$
$τ$	the time or distance in which $N$ falls by a factor $e$

N

halves in a time

τ \ln 2 = 0.693 τ

. Taking natural logarithms gives the time at which

N

reaches a chosen value,

t = τ \ln (N_{0} / N)

The constant $τ$ is called the e-folding time, and the time in which $N$ halves is its half-life. When $N$ falls off with height, as the density of an atmosphere does, $τ$ is a length called the scale height.

Quick checkUse the result above

The number of Sun-like stars per cubic parsec in the disc of the Milky Way falls off with height above the middle of the disc, with a scale height of 300 pc. At what height has it fallen to half its value in the middle?

S3.3Vectors

Before you readMake a guess first

The star 61 Cygni A is coming towards us at 64.8 km/s and moving across our line of sight at 87.3 km/s. How fast is it moving through space, relative to the Sun?

A speed tells us how fast something moves, and a velocity tells us in which direction as well. A quantity with both a size and a direction, such as a velocity, a force or the position of one star seen from another, is called a vector. We write vectors in bold, as $v$ , and the size of $v$ as $v$ . A vector can be described by its components, the parts of it that lie along each of a set of axes at right angles to each other. Two vectors are added by adding their components.

ResultSize and direction of a vector(S.7)

\begin{aligned} v & = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ \tan θ & = \frac{v_{y}}{v_{x}} \end{aligned}

$v_{x}, v_{y}$	components of the vector along two axes at right angles
$v$	size of the vector
$θ$	angle of the vector from the $x$ axis

In three dimensions a third term

v_{z}^{2}

goes under the square root. The signs of

v_{x}

and

v_{y}

show which quadrant the angle is in.

On the sky we give a direction as a position angle, measured from north through east, so that a position angle of 90° points due east. Measured this way, the angle of a vector with an eastward component $v_{E}$ and a northward component $v_{N}$ has $\tan θ = v_{E} / v_{N}$ .

The velocity of a star relative to the Sun is a vector, and we can see two of its components. The part along the line of sight is the radial velocity $v_{r}$ , which is measured from the Doppler shift of the star's spectrum (Section S4) and is negative for a star that is coming towards us. The part across the line of sight is the tangential velocity $v_{t}$ . It shows as a slow drift of the star across the sky, its proper motion $μ$ , usually given in arcseconds per year. A star at distance $d$ moving at $v_{t}$ across the line of sight covers a length $v_{t}$ in each unit of time, so by (S.2) its direction changes by $μ = v_{t} / d$ radians in that time. In the units that astronomers use, this becomes the following result.

ResultTangential velocity from proper motion(S.8)

v_{t} = 4.74 μ d

$v_{t}$	tangential velocity, in km/s
$μ$	proper motion, in arcseconds per year
$d$	distance of the star, in parsecs

The number 4.74 comes from the units. Seen from 1 pc, a length of 1 au covers 1″, so a star 1 pc away with a proper motion of 1″ a year moves 1 au a year across the line of sight, which is 4.74 km/s.

Figure S.3The velocity of 61 Cygni

The figure shows the Sun S and the star 61 Cygni A at X, a distance

d

apart. The heavy arrow is the velocity

v

of the star relative to the Sun. It is the sum of two components at right angles, the radial velocity

v_{r}

along the line of sight towards the Sun and the tangential velocity

v_{t}

across it, which are drawn in their true proportions, and the two faint lines that meet at the tip of

v

complete the rectangle of the components. The velocity makes the angle

θ

with the direction towards the Sun. The dashed line continues the straight path of the star to C, the point where it is closest to the Sun, and SC meets the path at a right angle. The distance

d

is not drawn to the same scale as the velocities.

Example S.561 Cygni through space

The HYG star catalogue gives 61 Cygni A a distance of 3.4865 pc, a radial velocity of −64.8 km/s, and a proper motion of 4.1551″ a year eastwards and 3.2589″ a year northwards. How fast does the star move relative to the Sun, and at what angle to our line of sight?

Solution

The two parts of the proper motion are at right angles on the sky, so by (S.7) the whole proper motion is

μ = \sqrt{{4.1551}^{2} + {3.2589}^{2}} = {5.281}^{″} a year,

and by (S.8) the tangential velocity is

v_{t} = 4.74 \times 5.281 \times 3.4865 = 87.3 {km s}^{- 1} .

The radial and tangential velocities are also at right angles, so the speed of the star is

v = \sqrt{{64.8}^{2} + {87.3}^{2}} = 108.7 {km s}^{- 1} .

The angle $θ$ between the velocity and the direction towards the Sun, in Figure S.3, has

\tan θ = \frac{v_{t}}{| v_{r} |} = \frac{87.3}{64.8}, θ = 53.4 ° .

Answer61 Cygni A moves at 108.7 km/s relative to the Sun, along a line that makes an angle of 53.4° with the direction towards us. The minus sign of the radial velocity shows that it is coming closer.

Example S.6The closest approach of 61 Cygnifinish it

If 61 Cygni A keeps moving in a straight line at the same velocity, how close to the Sun will it come?

The start

In Figure S.3 the path of the star runs from X through C, and the line from the Sun meets it at C at a right angle. The triangle SXC therefore has a right angle at C, its longest side SX is the present distance, 3.4865 pc, and its angle at X is $θ = 53.4 °$ .

Your stepNow find the distance SC, in parsecs.

Two vectors can also be combined to give a single number, their dot product. It is the most direct way to find the angle between two directions, and Section S3.5 uses it to find the distance between two places on Earth.

ResultDot product(S.9)

\begin{aligned} a \cdot b & = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z} \\ = a b \cos θ \end{aligned}

$a_{x}, a_{y}, a_{z}$	components of $a$ , and those of $b$ likewise
$a, b$	sizes of the two vectors
$θ$	angle between them

Vectors at right angles have a dot product of zero. For two unit vectors, vectors of size 1 that mark directions, the dot product is the cosine of the angle between them.

Quick checkUse the dot product

What is the angle between an edge of a cube and the diagonal through the cube from one corner to the opposite one? The edge runs along $(1, 0, 0)$ and the diagonal along $(1, 1, 1)$ .

Two vectors can also be combined to give a third vector, their cross product $a \times b$ . It is at right angles to both of them, pointing the way a right-handed screw moves when it is turned from $a$ towards $b$ , and its size is the area of the parallelogram with $a$ and $b$ as two of its sides. Section S4 uses it for angular momentum, and on a sphere the cross product of the directions of two points gives the pole of the great circle through them.

ResultCross product(S.10)

\begin{aligned} | a \times b | & = a b \sin θ \\ (a \times b)_{x} & = a_{y} b_{z} - a_{z} b_{y} \\ (a \times b)_{y} & = a_{z} b_{x} - a_{x} b_{z} \\ (a \times b)_{z} & = a_{x} b_{y} - a_{y} b_{x} \end{aligned}

$a, b$	sizes of the two vectors
$θ$	angle between them
$a_{x}, \dots, b_{z}$	components of the two vectors

The cross product of two parallel vectors is zero, and swapping the two vectors reverses its direction.

Quick checkUse the cross product

The vectors $(2, 0, 0)$ and $(1, 3, 0)$ are two sides of a parallelogram. What is its area?

S3.4Derivatives and integrals

Before you readMake a guess first

If the orbit of a planet were 1% larger, about how much longer would its year be?

The derivative of a quantity $y$ with respect to $x$ , written $d y / d x$ or $y^{'}$ , is the rate at which $y$ changes as $x$ changes. On a graph of $y$ against $x$ it is the slope of the curve at each point, that is, the slope of the tangent, the straight line that touches the curve there. A smooth curve has its peaks and troughs where its derivative is zero, because there it is momentarily flat, so the largest and smallest values of a quantity lie at those points or at the ends of its range. The same idea works for a vector that changes with time. The derivative $d r / d t$ of the position $r$ of a planet is its velocity, which points along the tangent to the planet's path.

An integral does the opposite job. It adds up a quantity in many small pieces. The integral of $f (x)$ from $x = a$ to $x = b$ , written $\int_{a}^{b} f (x) d x$ , is the area under the graph of $f$ between $a$ and $b$ , found by cutting the area into thin strips of width $d x$ and height $f (x)$ and adding them up. If $F$ is a function whose derivative is $f$ , the integral is $F (b) - F (a)$ .

ResultDerivatives and integrals of common functions(S.11)

\begin{array}{lll} f (x) & f^{'} (x) & \int f (x) d x \\ x^{n} & n x^{n - 1} & x^{n + 1} / (n + 1) \\ 1 / x & - 1 / x^{2} & \ln x \\ e^{x} & e^{x} & e^{x} \\ \ln x & 1 / x & x \ln x - x \\ \sin x & \cos x & - \cos x \\ \cos x & - \sin x & \sin x \end{array}

$f, f^{'}$	a function and its derivative
$n$	any number except $- 1$ in the integral of $x^{n}$
$x$	the variable, which must be in radians in $\sin x$ and $\cos x$

Any constant may be added to an integral, and it is fixed by one known value, such as the position at

t = 0

. A function of a function is differentiated with the chain rule,

\frac{d}{d x} f (g (x)) = f^{'} (g) g^{'} (x)

, so that, for example, the derivative of

e^{- t / τ}

with respect to

t

- e^{- t / τ} / τ

One of the most useful applications of the derivative is to small changes. If $x$ changes by a small amount $Δ x$ , the curve is nearly straight over that short stretch, so $y$ changes by very nearly the slope times $Δ x$ . For a power law this gives a rule that needs no calculus at all once it is known.

ResultSmall changes(S.12)

\begin{aligned} Δ y & \approx \frac{d y}{d x} Δ x \\ \frac{Δ y}{y} & \approx n \frac{Δ x}{x} for y = k x^{n} \\ (1 + ε)^{n} & \approx 1 + n ε \end{aligned}

$Δ x$	a small change in $x$
$Δ y$	the change in $y$ that it causes
$k, n$	a constant and the power in a power law
$ε$	a number much smaller than 1

The rules hold when

Δ x

is small compared with

x

. The last line, the binomial approximation, is the second line written another way.

Quick checkUse the result above

The radius of a star is measured 2% too large. About how much too large is the volume worked out from it?

per cent

Example S.7A star whose density falls steadily

In a simple model of a star of radius $R$ , the density falls steadily from $ρ_{c}$ at the centre to zero at the surface, as $ρ = ρ_{c} (1 - r / R)$ . Find the mass of the star, and compare its mean density with its central density.

Solution

A thin shell of radius $r$ and thickness $d r$ has a volume of $4 π r^{2} d r$ , so its mass is $4 π r^{2} ρ d r$ . Adding up the shells from the centre to the surface with (S.11),

\begin{aligned} M & = \int_{0}^{R} 4 π r^{2} ρ_{c} (1 - \frac{r}{R}) d r \\ = 4 π ρ_{c} {[\frac{r^{3}}{3} - \frac{r^{4}}{4 R}]}_{0}^{R} \\ = 4 π ρ_{c} (\frac{R^{3}}{3} - \frac{R^{3}}{4}) \\ = \frac{π}{3} ρ_{c} R^{3} . \end{aligned}

The mean density is the mass divided by the volume, $\frac{4}{3} π R^{3}$ ,

\bar{ρ} = \frac{π ρ_{c} R^{3} / 3}{4 π R^{3} / 3} = \frac{ρ_{c}}{4} .

AnswerThe mass is

π ρ_{c} R^{3} / 3

, and the central density is four times the mean density. The Sun's mean density is 1410 kg/m³, so in this model the density at its centre would be about 5640 kg/m³. The real Sun is far more concentrated towards its centre than this, as Chapter T7 explains.

When a function has no simple integral, or is known only at measured points, its integral can be estimated numerically. We cut the area under the graph into strips of equal width $h$ and treat each strip as a trapezium, whose area is $h$ times the mean of its two edges.

ResultTrapezium rule(S.13)

\int_{a}^{b} f (x) d x \approx h [\frac{1}{2} y_{0} + y_{1} + \dots + y_{n - 1} + \frac{1}{2} y_{n}]

$y_{0}, \dots, y_{n}$	values of $f$ at equally spaced points from $a$ to $b$
$h$	the spacing of the points, $(b - a) / n$

The estimate improves as the strips get narrower. With measured points the same rule gives, for example, the distance travelled from a table of speeds.

Quick checkUse the trapezium rule

A spacecraft's speed is 2.0, 3.0 and 5.0 km/s at times 0, 100 and 200 s. About how far does it travel in these 200 s?

S3.5Circles on a sphere

Before you readMake a guess first

Ankara and Beijing both lie at latitude 39.9° N. An aircraft flies between them by the shortest route. Which way does the route run?

A sphere is the surface made of all the points at the same distance $R$ from a centre O. A flat plane through O cuts the sphere in a great circle, the largest circle that can be drawn on it, with radius $R$ . A plane that misses O cuts it in a small circle. On Earth the equator is a great circle, and each meridian, a half circle from pole to pole, is half of a great circle. The parallels of latitude other than the equator are small circles. The shortest route between two points on a sphere runs along the great circle through them.

The latitude $φ$ of a place is its angle north or south of the equator, measured at the centre of Earth, which we treat here as a sphere. Its longitude $λ$ is its angle east or west of the meridian through Greenwich. A parallel of latitude is a circle around Earth's axis, and Figure S.4 shows that its radius is shorter than Earth's.

ResultRadius of a parallel of latitude(S.14)

r = R \cos φ

$r$	radius of the parallel
$R$	radius of the sphere
$φ$	latitude of the parallel

Along a parallel, two places whose longitudes differ by

Δ λ

radians are

R \cos φ Δ λ

apart. Along a meridian, two places whose latitudes differ by

Δ φ

radians are

R Δ φ

apart.

Figure S.4The radius of a parallel of latitude

The figure shows a section through Earth along a meridian, with centre O and the poles N and S, drawn for the latitude of Ankara, 39.93° N. The equator, seen edge on, is the dotted line through O. The point P on the surface is at latitude

φ

, the angle at O between the equator and the radius OP. The parallel of latitude through P, also seen edge on, is the heavy line, and its radius QP, measured from the point Q on the axis, is

r = R \cos φ

, here 0.767 of Earth's radius.

Quick checkUse the radius of a parallel

Earth turns once relative to the stars in 86 164 s, and its radius is 6378 km. How fast does Earth's rotation carry the ground at Mumbai, at latitude 19.08° N?

m/s

To find the distance between two places that are not on the same parallel or meridian, we can use vectors. Seen from the centre of Earth, a place at latitude $φ$ and longitude $λ$ lies in the direction of the unit vector

(\cos φ \cos λ, \cos φ \sin λ, \sin φ),

where the $z$ axis points to the North Pole and the $x$ axis to the point on the equator at longitude 0°. The first two components come from the radius of the parallel, $\cos φ$ , split along the $x$ and $y$ axes. By (S.9) the angle $θ$ between two places at the centre of Earth has

\begin{aligned} \cos θ & = \sin φ_{1} \sin φ_{2} \\ + \cos φ_{1} \cos φ_{2} \cos (λ_{2} - λ_{1}), \end{aligned}

and the distance between them along the great circle is $R θ$ . Section 1.5 finds the same formula from a triangle drawn on the sphere and uses it for the distance between two stars.

Example S.8From Ankara to Beijing

Ankara is at 39.93° N, 32.86° E and Beijing at 39.90° N, 116.40° E. Taking Earth as a sphere of radius 6378 km, how long is the shortest route between them, and how much longer is the route along the parallel of latitude 39.9° N?

Solution

The difference in longitude is $116.40 ° - 32.86 ° = 83.54 °$ . The angle between the two places at the centre of Earth has

\begin{aligned} \cos θ & = \sin 39.93 ° \sin 39.90 ° \\ + \cos 39.93 ° \cos 39.90 ° \cos 83.54 ° \\ = 0.4779, \end{aligned}

so $θ = 61.45 °$ , or 1.0725 rad, and the shortest route is $6378 \times 1.0725 = 6840$ km long. Along the parallel the radius is $6378 \cos 39.9 ° = 4893$ km, and the route is

4893 \times 83.54 \times \frac{π}{180} = 7134 km .

AnswerThe shortest route is about 6840 km long. The route along the parallel is about 7130 km, some 290 km longer, because a parallel of latitude is a small circle and only a great circle gives the shortest path.

A region on a sphere, seen from the centre, covers a solid angle, which does for areas what an angle in radians does for lengths. The solid angle of the region is its area divided by $R^{2}$ , measured in steradians, and the whole sphere covers $4 π$ steradians. The simplest region is a cap, the part of a sphere within some angle $θ$ of a point on it, such as the part of the sky within 30° of the , the point directly overhead.

ResultArea and solid angle of a cap(S.15)

\begin{aligned} A & = 2 π R^{2} (1 - \cos θ) \\ Ω & = 2 π (1 - \cos θ) \end{aligned}

$A$	area of the cap
$R$	radius of the sphere
$θ$	angular radius of the cap, measured at the centre of the sphere
$Ω$	solid angle of the cap, in steradians

For a small cap,

Ω \approx π θ^{2}

. One steradian is

(180 / π)^{2} = 3283

square degrees, and the whole sky is 41 253 square degrees.

Show the derivation

We cut the cap into thin rings around its centre. The ring at angle $t$ from the centre of the cap has radius $R \sin t$ , by (S.14) with $φ = 90 ° - t$ , and width $R d t$ , so its area is $2 π R \sin t \times R d t$ . Adding the rings from $t = 0$ to $t = θ$ with (S.11),

A = 2 π R^{2} \int_{0}^{θ} \sin t d t = 2 π R^{2} (1 - \cos θ) .

Dividing by $R^{2}$ gives the solid angle. For a small cap, $1 - \cos θ \approx θ^{2} / 2$ by (S.3), so $Ω \approx π θ^{2}$ , the area of a flat disc of radius $θ$ .

Quick checkUse the area of a cap

The full Moon is 0.52° across. What fraction of the whole sky does it cover? Give the answer in millionths.

millionths

S3.6Estimating

Before you readMake a guess first

About how many seconds are there in a year?

Many olympiad problems ask for an estimate, and every exact answer should be checked against one. An estimate keeps only the first figure or two of each number and deals with the powers of ten separately. To multiply $3.0 \times 10^{8}$ by $4.0 \times 10^{7}$ , for example, we multiply the leading figures, $3 \times 4 = 12$ , and add the powers, $10^{8 + 7} = 10^{15}$ , to get $12 \times 10^{15}$ , or $1.2 \times 10^{16}$ . The power of ten that a number is nearest to on a logarithmic scale is called its order of magnitude, and an estimate that gets the order of magnitude right is often all a problem needs.

When all we know is that a quantity lies somewhere between two values, such as between 10 and 1000, the best single guess is their geometric mean, the square root of their product, here 100. On a logarithmic scale it lies halfway between the two.

A few numbers come up so often that they are best learnt by heart.

Quantity	Value
A year	$3.156 \times 10^{7}$ s, close to $π \times 10^{7}$ s
The astronomical unit	$1.496 \times 10^{11}$ m
Time for light to cross 1 au	499 s, about 500 s
The parsec	$3.086 \times 10^{16}$ m, or 206 265 au, or 3.26 light years
One radian	57.30°, or about $2 \times 10^{5}$ arcseconds
$\log 2$ and $\log 3$	0.301 and 0.477
$\sqrt{10}$ and $π^{2}$	3.16 and 9.87
$2^{10}$	1024, close to $10^{3}$

Example S.9How many Earths would fill the Sun

The Sun's radius is $6.96 \times 10^{8}$ m and Earth's is $6.38 \times 10^{6}$ m. Estimate without a calculator how many times larger the Sun's volume is than Earth's.

Solution

The ratio of the radii is $6.96 \times 10^{8} / 6.38 \times 10^{6}$ . Since 6.96/6.38 is a little more than 1, at about 1.09, the ratio is about $1.09 \times 10^{2} = 109$ . Volume goes as the cube of the radius, and $109^{3} = {1.09}^{3} \times 10^{6}$ . As ${1.09}^{2}$ is about 1.19, and $1.19 \times 1.09$ is about 1.30, the ratio of the volumes is about $1.3 \times 10^{6}$ .

AnswerThe Sun's volume is about 1.3 million times Earth's. A calculator gives

1.298 \times 10^{6}

, the same to two significant figures.

Example S.10The density of the Sunfinish it

Estimate the mean density of the Sun from its mass of $2.0 \times 10^{30}$ kg and its radius of $7.0 \times 10^{8}$ m.

The start

The volume is $\frac{4}{3} π R^{3}$ . The cube of the radius is $7^{3} \times 10^{24} = 343 \times 10^{24}$ m³, or about $3.4 \times 10^{26}$ m³, and $\frac{4}{3} π$ is about 4.2, so the volume is about $1.4 \times 10^{27}$ m³.

Your stepNow estimate the density, in units of 10³ kg/m³.

× 10³ kg/m³

Estimates also help with equations that cannot be rearranged to give the unknown, such as Kepler's equation, $E - e \sin E = M$ , which Chapter T3 uses. We start from a rough value, put it into one side of the equation and use the result as the next value, and we repeat this until the values stop changing. Written as $E = M + e \sin E$ with $M = 1$ and $e = 0.1$ , and started from $E = 1$ , the equation gives 1.0841, then 1.0884, then 1.0886 twice, so $E = 1.0886$ to four decimal places. This is called solving by iteration. It works when the change from one step to the next keeps shrinking, as it does here, and if the changes grow instead, the equation has to be rearranged another way.

Quick checkSolve by iteration

With a calculator in radian mode, solve $x = 0.5 \cos x$ by iteration, starting from $x = 0$ , to three decimal places.

S3.7Uncertainties

Before you readMake a guess first

A star's parallax is measured as 0.100″, with an uncertainty of 0.010″, so its distance is $1 / 0.100 = 10$ pc. Roughly how uncertain is the distance?

Every measured value has an uncertainty, an estimate of how far it may be from the true value. We write the result as $x \pm Δ x$ , for example a parallax of 0.100″ ± 0.010″, and we call $Δ x / x$ the relative uncertainty, here 10%. The uncertainty is given to one or two significant figures, and the value is rounded to the same decimal place.

When a result is worked out from measured values, their uncertainties carry through to it. For a result that depends on a single measured quantity, the rule for small changes, (S.12), gives the uncertainty directly. When the result depends on several quantities whose errors are independent of each other, the errors partly cancel, because the error in one is as likely to push the result down as the error in the other is to push it up. Their uncertainties are then combined in quadrature, by taking the square root of the sum of their squares.

ResultCombining uncertainties(S.16)

\begin{aligned} Δ z & = \sqrt{Δ x^{2} + Δ y^{2}} \\ \frac{Δ z}{z} & = \sqrt{{(\frac{Δ x}{x})}^{2} + {(\frac{Δ y}{y})}^{2}} \\ \frac{Δ z}{z} & = | n | \frac{Δ x}{x} \end{aligned}

$x, y$	measured values, with uncertainties $Δ x$ and $Δ y$
$z$	the result, with uncertainty $Δ z$
$n$	a power

The first line is for a sum or difference,

z = x \pm y

, the second for a product or quotient,

z = x y

z = x / y

, and the third for a power,

z = x^{n}

. The first two lines need the errors of

x

and

y

to be independent, and the last two need uncertainties that are small compared with the values.

Adding the uncertainties instead, as $Δ z = Δ x + Δ y$ for a sum and likewise for the relative uncertainties of a product, gives the largest error the result could have, called the maximum error. It is quicker and more cautious, and the IOAA syllabus allows either method. For random errors the uncertainty quoted is usually the standard deviation, the typical scatter of repeated measurements about their mean. Half of a set of measurements lie above their median and half below it, so the median, unlike the mean, is hardly moved by a few wild values. Chapter D covers these and the rest of the statistics that data analysis needs.

Example S.11The density of a planet

The mass of a planet is known to 1% and its radius to 2%. How well is its mean density known?

Solution

The density, $M / \frac{4}{3} π R^{3}$ , is proportional to $M R^{- 3}$ . By the third line of (S.16), $R^{- 3}$ is uncertain by $3 \times 2 % = 6 %$ , and by the second line the uncertainties of $M$ and $R^{- 3}$ combine to

\frac{Δ ρ}{ρ} = \sqrt{1^{2} + 6^{2}} % = 6.1 % .

AnswerThe density is known to about 6%. The uncertainty of the radius dominates because the radius enters as its cube, and measuring the mass more precisely would hardly help.

Quick checkUse the result above

A student times 10 complete cycles of a variable star's brightness and finds 52.3 ± 0.2 days. What is the uncertainty of the period?

days

These rules assume that the uncertainties are small. A parallax uncertain by a quarter of its value gives distances that spread much farther above the true distance than below it, and then the rules fail, as the last of the past problems below shows.

Common errorsSection S3

A calculator in degree mode gives $\sin 0.01 = 1.745 \times 10^{- 4}$ instead of 0.0100, because it takes the 0.01 to be in degrees. The approximation $\sin θ \approx θ$ needs $θ$ in radians.
The logarithm of a sum cannot be split up. $\log (a + b)$ is not $\log a + \log b$ , and only products, quotients and powers turn into sums, differences and multiples.
The slope of a line on logarithmic axes has to be worked out from the logarithms of the values. Reading it from the printed numbers, as $(11.86 - 1) / (5.203 - 1) = 2.58$ for Jupiter and Earth, gives a meaningless answer.
The arctangent key gives angles only between −90° and +90°. For a vector whose $x$ component is negative the angle is 180° more than the key shows, and the signs of the two components tell which quadrant is right.
A distance along a parallel of latitude is $R \cos φ Δ λ$ . Using $R Δ λ$ , as if the parallel were a great circle, gives too large a distance, and the error grows with the latitude.
A result should not be quoted to more figures than its uncertainty allows. A distance written as 10.374 ± 0.5 pc claims a precision that the uncertainty denies, and it should be 10.4 ± 0.5 pc.
The rules for products, quotients and powers hold only for small uncertainties. When a measurement is uncertain by a large fraction of its value, a quantity worked out from it, such as a distance from a parallax, has an uneven spread, and its average is pulled away from the true value.

ExercisesSection S311 questions

S3.1
Near opposition, when Jupiter is opposite the Sun in the sky and nearest to us, it is about 47″ across. What is this angle in radians?
rad
S3.2
A pea 5 mm across just covers the full Moon, which is 0.52° across. How far from the eye is the pea?
cm
S3.3
Given that $\log 2 = 0.301$ and $\log 3 = 0.477$ , find $\log 0.006$ without a calculator.
S3.4
A quantity falls off as $e^{- t / τ}$ , with $τ = 10.0$ days. How long does it take to fall to 1% of its starting value?
days
S3.5
A star is coming towards us at 30 km/s and moving across the line of sight at 40 km/s. What angle does its velocity make with the direction towards us?
°
S3.6
A ball thrown at speed $v$ and angle $θ$ above level ground lands a distance $(v^{2} / g) \sin 2 θ$ away. At what angle does it go farthest?
S3.7
The Moon's orbit grows by 3.8 cm a year. The Moon is 384 400 km away and goes round Earth in 27.32 days. By how much does the month grow longer each year, in milliseconds?
ms
S3.8
In another model of a planet, the density falls as $ρ = ρ_{c} (1 - r^{2} / R^{2})$ . What is the mean density, as a fraction of $ρ_{c}$ ?
S3.9
From Ankara, at latitude 39.93° N, the stars within 39.93° of the never rise. What percentage of the whole sky do they fill?
per cent
S3.10
Light takes 499 s to cross 1 au. About how long does sunlight take to reach Jupiter, 5.20 au from the Sun?
minutes
S3.11
The total mass of a pair of stars in orbit around each other is proportional to $a^{3} / P^{2}$ , where $a$ is the size of the orbit and $P$ its period (Chapter T8). If $a$ is known to 3% and $P$ to 1%, how well is the mass known?
per cent

Olympiad problemsPast IOAA problems, restated4 problems

2012The Sun seen from Barnard's Star10 points

Adapted from IOAA 2012 (Rio de Janeiro), theory problem 11. We have restated it in our own words. The original gives no data, so we have added the values that the official solution uses. The solution was last checked on 16 October 2026.

Jupiter pulls the Sun round a small orbit, so that seen from Barnard's Star the Sun would swing to and fro across the sky. What is the angular amplitude of this motion, the largest angle by which the Sun moves from its mean position, and what is its period? Take Jupiter's orbit as a circle of radius 5.204 au, the Sun as 1047 times as massive as Jupiter, and the distance of Barnard's Star as 1.834 pc.

I solved it without the solution

AnswerThe Sun swings up to 0.0027″ from its mean position, less than three thousandths of an arcsecond, once every 11.87 years. From Barnard's Star its circle is seen at a slant, as an ellipse whose long half-axis is this angle. Measuring motions as small as this is one way to find giant planets around nearby stars (Chapter T9).

2018Where Vega and Altair are heading3 points

Adapted from IOAA 2018 (Beijing), theory problem 10, part (c). We have restated it in our own words and kept the original data. The solution was last checked on 16 October 2026.

In the year 2000, Vega was moving across the sky by 200.94 mas a year eastwards and 286.23 mas a year northwards, and Altair by 536.23 mas a year eastwards and 385.29 mas a year northwards, where 1 mas is a thousandth of an arcsecond. Find the position angle of the proper motion of each star, that is, the direction in which it moves on the sky, measured from north through east.

I solved it without the solution

AnswerVega is heading towards position angle 35.07° and Altair towards 54.30°, both towards the north-east of the sky. Section 1.5 restates parts (a) and (b) of the same problem, which find the angle and the distance between the two stars.

2010Sun-like stars above the disc10 points

Adapted from IOAA 2010 (Beijing), theory problem 10. We have restated it in our own words and kept the original data. The solution was last checked on 16 October 2026.

The number of stars like the Sun, which astronomers call main-sequence stars of type G (Chapter T7), per unit volume in the disc of the Milky Way falls off exponentially with height above the middle plane of the disc, with a scale height of 300 pc. What fraction of its value in the middle plane does it have at heights of 0.5 kpc and 1.5 kpc?

I solved it without the solution

AnswerAt 0.5 kpc the number per unit volume is 0.189 of its value in the middle plane, just under a fifth, and at 1.5 kpc it is 0.0067, about a hundred and fiftieth.

2018Four ways to average a distance10 points

Adapted from IOAA 2018 (Beijing), theory problem 2. We have restated it in our own words and kept the original data. The solution was last checked on 16 October 2026.

An observer measures the parallaxes of many stars in a star cluster that is 5 kpc away. Random errors spread the measured parallaxes symmetrically about the true value, with a standard deviation of 0.05 mas, and there are no systematic errors. All the stars in the cluster have the same luminosity. Four students use the table of parallaxes to estimate the distance of the cluster in four different ways.

Student A turns each parallax into a distance and takes the average of the distances, $R_{A}$ .
Student B takes the average of the parallaxes and turns it into a distance, $R_{B}$ .
Student C turns each parallax into a distance and takes the median of the distances, $R_{C}$ .
Student D takes the median of the parallaxes and turns it into a distance, $R_{D}$ .

Say whether each of the following statements is true or false, and give the correct relation for each one that is false. Each part is worth 2 points, and $R$ is the true distance.

(a) If star $i$ has the smallest measured parallax and star $j$ the largest, then most likely $R_{i} - R > R - R_{j}$ .

(b) $R_{A} = R$ , so that A's estimate is likely to be close to the true distance.

(d) $R_{C}$ is less than $R$ , so that C's estimate is likely to be too small.

(e) $R_{D} = R$ .

I solved it without the solution

AnswerThe answers are (a) true, (b) false, with $R_{A} > R$ , (c) true, (d) false, with $R_{C} = R$ , and (e) true. Of the four methods, only averaging the distances gives an answer that is systematically too large.