Physics toolkit

Builds on: Section S3 and school physics
Reading time: 1 h 40 min
Exercises: 10
Olympiad problems: 4

Like every body in orbit, the Moon is falling towards Earth all the time. In each second it drops about 1.4 mm towards Earth from the straight line it would follow if nothing pulled on it, while an apple let go near the ground falls 4.9 m in its first second, about 3600 times as far. The Moon is 60 times as far from the centre of Earth as the apple, and 3600 is 60 squared. From this comparison Isaac Newton concluded that one force pulls the apple down and holds the Moon in its orbit, and that it weakens as the square of the distance. Olympiad astronomy uses a small set of ideas of this kind from school physics. In this section we collect them, from gravity and circular motion to energy and angular momentum, gases and pressure, lenses and diffraction, the Doppler effect, and photons and atoms.

Entry checkCan you skip this section?6 questions

If you can already answer these 6 questions, you know what this section teaches and can go straight on. Anything you get wrong shows which subsection to read.

How fast does a satellite move in a circular orbit 420 km above the ground? Take $G M = 3.986 \times 10^{14}$ m³ s⁻² for Earth and its radius as 6378 km.

km/s

A spinning star shrinks to a tenth of its radius, keeping its mass, its angular momentum and the way its mass is spread inside it. How does its period of rotation change?

How many particles are there in each cubic metre of a gas at a pressure of $1.0 \times 10^{5}$ Pa and a temperature of 300 K? Take $k = 1.381 \times 10^{- 23}$ J K⁻¹.

per m³

What is the smallest angle, in arcseconds, between two stars that a perfect telescope 1.0 m across can separate in light of wavelength 550 nm?

″

A spectral line whose rest wavelength is 500.00 nm is seen at 500.10 nm. What is the radial velocity of its source?

km/s

What is the energy, in electronvolts, of a photon of wavelength 600 nm?

S4.1Gravity and circular motion

Before you readMake a guess first

If Earth kept its mass but shrank to half its present radius, how would the weight of a person standing on it change?

Every pair of bodies attracts each other with a force that grows with their masses and weakens as the square of the distance between them. A sphere, such as a planet or a star, attracts a body outside it as if all its mass were at its centre. Dividing the force on a falling body by the body's own mass gives its acceleration, so near the surface of a planet every body falls with the same acceleration $g$ , whatever its mass.

ResultNewton's law of gravitation(S.17)

\begin{aligned} F & = \frac{G M m}{r^{2}} \\ g & = \frac{G M}{R^{2}} \end{aligned}

$F$	the force with which each body pulls the other
$G$	the gravitational constant, $6.674 \times 10^{- 11}$ N m² kg⁻²
$M, m$	the masses of the two bodies
$r$	the distance between their centres
$g$	the acceleration of a falling body at the surface of a sphere of mass $M$ and radius $R$

For Earth, with

M = 5.972 \times 10^{24}

kg and

R = 6378

km,

g = 9.80

m s⁻².

A body that moves round a circle at a steady speed is still accelerating, because the direction of its velocity keeps changing. The acceleration points towards the centre of the circle and is called the centripetal acceleration. A body moving at speed $v$ round a circle of radius $r$ goes round once in a period $T = 2 π r / v$ , and its angular speed, the angle it turns through each second, is $ω = 2 π / T$ .

ResultCircular motion(S.18)

a = \frac{v^{2}}{r} = ω^{2} r = \frac{4 π^{2} r}{T^{2}}

$a$	the acceleration towards the centre
$v$	the speed
$r$	the radius of the circle
$ω$	the angular speed, in radians per second
$T$	the period, the time for one turn

Example S.12Newton's test with the Moon

The Moon goes round Earth once in 27.32 days on a nearly circular orbit of radius 384 400 km. Compare its acceleration towards Earth with $g$ at the surface, weakened by the square of the Moon's distance in Earth radii.

Solution

The period is $27.32 \times 86 400 = 2.360 \times 10^{6}$ s, so by (S.18) the Moon's acceleration is

a = \frac{4 π^{2} \times 3.844 \times 10^{8}}{(2.360 \times 10^{6})^{2}} = 2.72 \times 10^{- 3} {m s}^{- 2} .

The Moon is $384 400 / 6378 = 60.27$ Earth radii away, and ${60.27}^{2} = 3632$ , so the inverse-square law predicts

\frac{g}{3632} = \frac{9.80}{3632} = 2.70 \times 10^{- 3} {m s}^{- 2} .

AnswerThe two accelerations agree to within 1%, so the force that makes an apple fall is the same force that keeps the Moon in its orbit.

For a body in a circular orbit, gravity supplies exactly the acceleration towards the centre that circular motion needs. Setting $G M / r^{2}$ equal to $v^{2} / r$ gives the speed of the orbit, and $T = 2 π r / v$ then gives its period.

ResultCircular orbits(S.19)

\begin{aligned} v & = \sqrt{\frac{G M}{r}} \\ T^{2} & = \frac{4 π^{2}}{G M} r^{3} \end{aligned}

$v$	the speed of the orbit
$r$	the radius of the orbit
$M$	the mass of the central body
$T$	the period of the orbit

The second line is Kepler's third law for a circular orbit, which Section S3 found from the periods of the planets. When the two bodies have comparable masses, both go round their common centre of mass, and

M

becomes the total mass

M_{1} + M_{2}

, with

r

the distance between them. Chapter T3 extends both lines to elliptical orbits.

Figure S.6Newton's cannoninteractive

Launch speed

Compared with a circular orbit
Path of the ball

The figure shows Earth drawn to scale, with a launch point L 1000 km above the ground, and the path of a ball fired horizontally from L, ignoring the air. The dotted circle is the path of a circular orbit at the height of L. The slider sets the launch speed. At the ball falls back to the ground, at it circles Earth, at it swings out on an ellipse and comes back to L, and from about it escapes.

Quick checkUse the period of a circular orbit

The International Space Station orbits 420 km above the ground. How long does it take to go round Earth once? Take $G M = 3.986 \times 10^{14}$ m³ s⁻² for Earth and its radius as 6378 km.

minutes

Example S.13Weighing Jupiter with Io

Io, the innermost of the four large moons of Jupiter, goes round the planet once in 1.769 days on an almost circular orbit of radius 421 700 km. What is the mass of Jupiter?

Solution

Io's own mass is tiny compared with Jupiter's, so by (S.19) the mass of Jupiter is $M = 4 π^{2} r^{3} / G T^{2}$ . The period is $1.769 \times 86 400 = 1.528 \times 10^{5}$ s, and

\begin{aligned} M & = \frac{4 π^{2} \times (4.217 \times 10^{8})^{3}}{6.674 \times 10^{- 11} \times (1.528 \times 10^{5})^{2}} \\ = 1.90 \times 10^{27} kg . \end{aligned}

AnswerJupiter has a mass of

1.90 \times 10^{27}

kg, about 318 times the mass of Earth. Timing their moons is how astronomers first found the masses of the planets that have moons.

S4.2Energy and angular momentum

Before you readMake a guess first

How much faster must a ball be thrown to escape from Earth altogether than to put it into a circular orbit just above the ground?

A moving body has kinetic energy $\frac{1}{2} m v^{2}$ . Lifting a body against gravity takes work, which is stored as gravitational potential energy. Near the ground the potential energy grows by $m g h$ when a body rises by a height $h$ . Over large distances the pull weakens, and integrating the force as in Section S3.4 shows that the work needed to take a body from a distance $r$ to very far away is $G M m / r$ . We take the potential energy to be zero very far away, so at a distance $r$ it is $- G M m / r$ . It is negative because energy has to be added to set the body free. When gravity is the only force at work, the sum of the kinetic and potential energy stays the same.

ResultEnergy in a gravitational field(S.20)

\begin{aligned} K & = \frac{1}{2} m v^{2} \\ U & = - \frac{G M m}{r} \\ E & = K + U \end{aligned}

$K$	the kinetic energy of a body of mass $m$ moving at speed $v$
$U$	its gravitational potential energy at a distance $r$ from the centre of a body of mass $M$ , taken as zero very far away
$E$	its total energy, which stays constant when gravity is the only force

A body whose total energy is negative is bound to

M

and cannot get away. A body whose total energy is zero or positive can travel to any distance.

A body just escapes when it has enough kinetic energy to reach a great distance with none to spare, which means $E = 0$ . In a circular orbit, $v^{2} = G M / r$ gives a kinetic energy of $G M m / 2 r$ , half the size of the potential energy, so the total energy of the orbit is negative.

ResultEscape speed and the energy of a circular orbit(S.21)

\begin{aligned} v_{esc} & = \sqrt{\frac{2 G M}{r}} = \sqrt{2} v_{circ} \\ E_{circ} & = - \frac{G M m}{2 r} \end{aligned}

$v_{esc}$	the escape speed from a distance $r$
$v_{circ}$	the speed of a circular orbit at that distance
$E_{circ}$	the total energy of a body of mass $m$ in that orbit

From Earth's surface the escape speed is 11.2 km/s.

Quick checkUse the escape speed

A spacecraft is at Earth's distance from the Sun, $1.496 \times 10^{11}$ m. How fast must it move relative to the Sun to leave the Solar System? Take the Sun's mass as $1.988 \times 10^{30}$ kg.

km/s

A body moving round a centre also has angular momentum. For a body of mass $m$ at distance $r$ , moving at speed $v$ at right angles to the line from the centre, it is $m v r$ . In general it is the cross product of the position and the momentum (Section S3.3), and its size is $m r v_{⟂}$ , where $v_{⟂}$ is the part of the velocity at right angles to the line from the centre. Angular momentum stays the same as long as the force on the body points along that line, as gravity does. A planet on an elliptical orbit therefore moves fastest when it is nearest the Sun.

ResultAngular momentum(S.22)

\begin{aligned} L & = r \times m v \\ L & = m r v_{⟂} \end{aligned}

$L$	the angular momentum, of size $L$
$r$	the position of the body measured from the centre
$m, v$	the mass and the velocity of the body
$v_{⟂}$	the part of the velocity at right angles to $r$

L

stays constant when the force points along

r

. At the nearest and farthest points of an orbit the velocity is at right angles to

r

, so there

r_{1} v_{1} = r_{2} v_{2}

A spinning body has angular momentum too. For a sphere of mass $M$ and radius $R$ spinning with angular speed $ω$ , it is $k M R^{2} ω$ , where the number $k$ depends on how the mass is spread inside the sphere and is $2 / 5$ for a sphere of uniform density. If a spinning star shrinks while keeping its mass, its angular momentum and the way its mass is spread, then $ω R^{2}$ stays the same, and the star spins faster as it shrinks.

Example S.14Earth near and far from the Sun

Earth's orbit has its nearest point to the Sun at 0.9833 au, in early January, and its farthest at 1.0167 au, in early July. How much faster does Earth move in January than in July?

Solution

At these two points the velocity is at right angles to the line to the Sun, so by (S.22) $r_{1} v_{1} = r_{2} v_{2}$ , and

\frac{v_{Jan}}{v_{Jul}} = \frac{1.0167}{0.9833} = 1.034 .

AnswerEarth moves 3.4% faster in early January than in early July.

Example S.15The Sun as a neutron starfinish it

The Sun turns once in about 25 days at its equator, and its radius is $6.957 \times 10^{8}$ m. If it shrank to a radius of 10 km, keeping its mass, its angular momentum and the way its mass is spread, how long would it take to turn once?

The start

The angular momentum $k M R^{2} ω$ stays the same, and $k$ and $M$ do not change, so $ω R^{2}$ stays the same. The period $T = 2 π / ω$ therefore changes in proportion to $R^{2}$ .

Your stepNow find the new period, in milliseconds.

S4.3Gases and pressure

Before you readMake a guess first

Mount Everest is 8.8 km high. About how large is the air pressure at its top, compared with the pressure at sea level?

Pressure is force per unit area, measured in pascals, and 1 Pa is 1 N m⁻². In a gas or liquid at rest, the pressure at each depth holds up the weight of everything above it, so it grows with depth. Across a thin layer of thickness $d z$ , the pressure below must be larger than the pressure above by the weight of the layer per unit area, $ρ g d z$ . This balance is called hydrostatic equilibrium. It holds in the atmosphere of a planet and, with a $g$ that changes with depth, inside every star (Chapter T7).

ResultHydrostatic equilibrium(S.23)

\frac{d p}{d z} = - ρ g

$p$	the pressure
$z$	the height
$ρ$	the density
$g$	the acceleration of gravity

In a layer of constant density the pressure grows by

ρ g h

over a depth

h

Example S.16The weight of the air

The air pressure at sea level is 101 325 Pa. Taking Earth as a sphere of radius 6378 km, with $g = 9.81$ m s⁻², find the mass of the atmosphere.

Solution

The pressure at the ground is the weight of the air above each square metre, so the weight of the whole atmosphere is the pressure times the area of Earth's surface. Its mass is

\begin{aligned} M & = \frac{4 π R^{2} p}{g} = \frac{4 π \times (6.378 \times 10^{6})^{2} \times 101 325}{9.81} \\ = 5.28 \times 10^{18} kg . \end{aligned}

AnswerThe atmosphere has a mass of about

5.3 \times 10^{18}

kg, less than a millionth of the mass of Earth. The true value is a little smaller, because much of the land stands well above sea level, where the pressure is lower.

A gas is made of particles, atoms or molecules, which fly about and collide with each other and with anything in their way. Their impacts make the pressure. In most gases in astronomy, from the air to the inside of the Sun, the particles are far enough apart for the gas to behave as an ideal gas.

ResultIdeal gas law(S.24)

p = n k T = \frac{ρ k T}{μ m_{H}}

$n$	the number of particles per unit volume
$k$	the Boltzmann constant, $1.381 \times 10^{- 23}$ J K⁻¹
$T$	the temperature, in kelvin
$ρ$	the density
$μ$	the mean molecular mass, the average mass of a particle in units of $m_{H}$
$m_{H}$	the mass of a hydrogen atom, $1.673 \times 10^{- 27}$ kg

For air

μ

is about 29, and for the hot, fully ionised gas inside the Sun it is about 0.6.

Quick checkUse the ideal gas law

The gas at the visible surface of the Sun has a pressure of about $1.0 \times 10^{4}$ Pa and a temperature of 5800 K. How many particles are there in each cubic metre?

per m³

Temperature measures the energy of the random motion of the particles. In a gas at temperature $T$ , the mean kinetic energy of a particle is $\frac{3}{2} k T$ , whatever the particle, so at the same temperature light particles move faster than heavy ones.

ResultThermal speed(S.25)

\begin{aligned} \frac{1}{2} m \overset{―}{v^{2}} & = \frac{3}{2} k T \\ v_{rms} & = \sqrt{\frac{3 k T}{m}} \end{aligned}

$m$	the mass of one particle
$\overset{―}{v^{2}}$	the mean of the square of the particles' speeds
$v_{rms}$	the root-mean-square speed, the square root of $\overset{―}{v^{2}}$

A planet can keep a gas for billions of years only if

v_{rms}

is well below its escape speed, which Chapter T3 makes precise.

Quick checkUse the thermal speed

A nitrogen molecule has a mass of $28 m_{H}$ . What is the root-mean-square speed of nitrogen molecules in air at 288 K?

m/s

Putting the ideal gas law into the condition for hydrostatic equilibrium gives the pressure in an atmosphere at a steady temperature. The pressure falls off exponentially with height, as in Section S3.2.

ResultAn isothermal atmosphere(S.26)

\begin{aligned} p & = p_{0} e^{- z / H} \\ H & = \frac{k T}{μ m_{H} g} \end{aligned}

$p_{0}$	the pressure at height $z = 0$
$H$	the scale height

The density falls off in the same way. For Earth's air at 288 K,

H = 8.4

km.

Show the derivation

The ideal gas law gives $ρ = μ m_{H} p / k T$ . Putting this into (S.23) gives

\frac{d p}{d z} = - \frac{μ m_{H} g}{k T} p = - \frac{p}{H} .

A quantity whose rate of change is proportional to itself changes exponentially, and the solution that starts at $p_{0}$ is $p = p_{0} e^{- z / H}$ .

S4.4Lenses and diffraction

Before you readMake a guess first

A camera lens of focal length 500 mm photographs the full Moon, which is 0.52° across. About how large is the Moon's image on the camera's sensor?

A lens bends light because light travels more slowly in glass than in air. A converging lens brings the parallel rays from a distant point, such as a star, together at a single point in its focal plane, a distance $f$ behind the lens called its focal length. Rays from a second star, a small angle $θ$ away from the first, meet in the same plane at a distance $f θ$ from the first image, so the angles on the sky become distances on the detector. For an object at a distance $u$ in front of the lens, the image forms at a distance $v$ behind it.

A telescope forms an image with a large lens or mirror called the objective, of focal length $F$ , and a small lens called the eyepiece, of focal length $f$ , is used to look at the image. An object seen through the eyepiece covers $F / f$ times the angle it covers to the naked eye, and $F / f$ is called the magnification. The amount of light that a telescope collects grows with the area of its objective, and the ratio of its focal length to its diameter is its focal ratio.

ResultLenses and telescopes(S.27)

\begin{aligned} \frac{1}{u} + \frac{1}{v} & = \frac{1}{f} \\ s & = f θ \\ magnification & = \frac{F}{f} \end{aligned}

$u, v$	the distances of the object and of its image from the lens
$f$	the focal length of the lens, or of the eyepiece of a telescope
$s$	the distance in the focal plane between the images of two distant points an angle $θ$ apart, with $θ$ in radians
$F$	the focal length of a telescope's objective

Chapter T5 treats telescopes and detectors in full.

Quick checkUse the magnification of a telescope

A telescope has an objective of focal length 1200 mm. What magnification does it give with an eyepiece of focal length 25 mm?

Even a perfect telescope does not show a star as a point. Light is a wave, and a wave that passes through an opening spreads out, by an angle of about its wavelength divided by the width of the opening. This spreading is called diffraction. A circular opening of diameter $D$ turns the image of a star into a small bright disc surrounded by faint rings, called the Airy pattern, and its first dark ring lies at an angle of $1.22 λ / D$ from the centre. By the Rayleigh criterion, two equal stars can just be told apart when the centre of the image of one falls on the first dark ring of the other.

ResultDiffraction limit(S.28)

θ_{min} = 1.22 \frac{λ}{D}

$θ_{min}$	the smallest angle between two stars that can be told apart, in radians
$λ$	the wavelength of the light
$D$	the diameter of the telescope's objective

On the ground, the moving air blurs the image of a star to about 1″ even at a good site. In visible light that is more than the diffraction limit of any telescope wider than about 14 cm.

Figure S.7When two stars can be told apartinteractive

Separation of the stars

Separation
Brightness between the peaks

The figure shows the brightness along a line through the images of two equal stars A and B, seen through a telescope of diameter

D

in light of wavelength

λ

. Each star alone, drawn thin, gives the bright centre and faint rings of an Airy pattern, and the heavy curve is their sum. The slider sets the separation of the stars in units of

λ / D

. At the two peaks stand clearly apart, at , the Rayleigh limit, the dip between them falls to about three quarters of the peak brightness, and at the two images have merged into one.

Example S.17How sharp the Hubble Space Telescope is

The mirror of the Hubble Space Telescope is 2.4 m across. What is the smallest angle between two stars that it can separate in light of wavelength 550 nm?

Solution

By (S.28),

θ_{min} = 1.22 \times \frac{550 \times 10^{- 9}}{2.4} = 2.80 \times 10^{- 7} rad,

and multiplying by 206 265 turns this into 0.058″.

AnswerHubble can separate stars 0.058″ apart, about 17 times closer together than the 1″ that the air allows at a good site on the ground.

S4.5The Doppler effect

Before you readMake a guess first

A star moves across the sky at 50 km/s, at right angles to our line of sight. Are the lines in its spectrum shifted?

The light of a star that moves away from us is stretched to longer wavelengths, and the light of one that comes towards us is squeezed to shorter ones. This is the Doppler effect. In Figure S.8 each crest of a wave leaves the source from a point a little farther along the source's path than the crest before it, so the crests crowd together ahead of a moving source and spread apart behind it. The change in wavelength, as a fraction of the rest wavelength that the source sends out, is the speed of the source along the line of sight divided by the speed of the waves. This fraction is called the redshift $z$ , and it is negative for a source that comes towards us.

Figure S.8Crests from a moving sourceinteractive

Speed of S

S moves at
Spacing of the crests at A
Spacing of the crests at B

The figure shows the crests of the waves sent out by a source S that moves to the right. Each circle is a crest, centred on the point where S was when it sent that crest out. The slider sets the speed of S as a fraction of the speed of the waves. With S the crests are evenly spaced all round. When S moves at , the crests ahead of it, towards the observer A, are squeezed to 0.7 of that spacing, and those behind it, towards the observer B, are spread to 1.3 times that spacing, as the heavy segments show. At the squeeze is stronger still. For light, A sees the lines in the spectrum of the source shifted towards the blue, and B sees them shifted towards the red.

ResultDoppler shift(S.29)

\begin{aligned} z = \frac{λ - λ_{0}}{λ_{0}} & = \frac{v_{r}}{c} \\ 1 + z & = \sqrt{\frac{1 + v_{r} / c}{1 - v_{r} / c}} \end{aligned}

$λ_{0}$	the rest wavelength, which the source sends out
$λ$	the wavelength that we receive
$z$	the redshift
$v_{r}$	the radial velocity of the source, positive when it moves away from us
$c$	the speed of light, $2.998 \times 10^{8}$ m s⁻¹

The first line holds for speeds much less than

c

, which covers the stars of our Galaxy. The second holds at any speed of a source moving straight towards us or away from us. In frequency the shift has the opposite sign, so a source that comes towards us is seen at a higher frequency.

Example S.18The spectrum of 61 Cygni

61 Cygni A comes towards us at 64.8 km/s (Example S.5). Where in its spectrum is the hydrogen line Hα, whose rest wavelength in air is 656.28 nm?

Solution

By (S.29), with $v_{r} = - 64.8$ km/s,

Δ λ = λ_{0} \frac{v_{r}}{c} = 656.28 nm \times \frac{- 64.8}{299 800} = - 0.142 nm .

AnswerThe line is shifted 0.142 nm towards the blue, to 656.14 nm. A spectrograph that measures wavelengths to 0.001 nm finds radial velocities to about 0.5 km/s.

Quick checkUse the exact Doppler formula

Gas in a jet from near a black hole moves straight away from us, and its redshift is $z = 0.5$ . How fast is it moving, as a fraction of the speed of light?

S4.6Photons and atoms

Before you readMake a guess first

Which carries more energy, a photon of red light or a photon of blue light?

Visible light, with wavelengths from about 400 nm to 700 nm, is one small band of the electromagnetic spectrum. The spectrum runs from radio waves, with wavelengths of a millimetre to many metres, through infrared, visible light, ultraviolet and X-rays to gamma rays, whose wavelengths are far smaller than an atom. Light is given out and absorbed in packets of energy called photons. A photon of frequency $ν$ , and so of wavelength $λ = c / ν$ , carries the energy $h ν$ , where $h$ is Planck's constant. The energies of photons are often given in electronvolts, and 1 eV, the energy that an electron gains in crossing a voltage of 1 V, is $1.602 \times 10^{- 19}$ J.

ResultEnergy of a photon(S.30)

\begin{aligned} E & = h ν = \frac{h c}{λ} \\ E & \approx \frac{1240 eV nm}{λ} \end{aligned}

$E$	the energy of the photon
$h$	Planck's constant, $6.626 \times 10^{- 34}$ J s
$ν, λ$	the frequency and the wavelength of the light

A photon of green light, of wavelength 500 nm, carries 2.48 eV.

Quick checkUse the energy of a photon

A green laser pointer gives out 1.0 mW of light of wavelength 532 nm. How many photons does it send out each second?

photons per second

An atom can hold its electrons only at certain energies, called its energy levels. An electron that drops from a higher level to a lower one gives out a photon whose energy is the difference between the two levels. An electron can be lifted from the lower level to the higher one only by absorbing a photon of exactly that energy. For hydrogen, the simplest atom, the levels follow a simple rule.

ResultEnergy levels of hydrogen(S.31)

\begin{aligned} E_{n} & = - \frac{13.6 eV}{n^{2}} \\ \frac{h c}{λ} & = 13.6 eV (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \end{aligned}

$E_{n}$	the energy of level $n$ , where $n = 1, 2, 3$ and so on, measured from an electron that has just been set free
$n_{1}, n_{2}$	the lower and the upper level of a jump
$λ$	the wavelength of the photon given out or absorbed in the jump

An electron in the lowest level,

n = 1

, needs 13.6 eV to leave the atom. This is the ionisation energy of hydrogen.

Figure S.9The energy levels of hydrogeninteractive

Upper level

Lower level

Energy of the photon
Wavelength in a vacuum
Line

The figure shows the energy levels of the hydrogen atom drawn to scale, from

n = 1

at −13.6 eV up to

n = \infty

at 0 eV, where the electron is free. The heavy arrow is a jump from an upper level down to a lower one, and the buttons below choose the two levels. The jump from level 2 to level 1 gives the line in the ultraviolet, and the jumps from levels 3 and 4 down to level 2 give the Balmer lines and in visible light. The wavelengths are those in a vacuum. In air they are about 0.03% shorter, which is why Hα is usually quoted as 656.3 nm.

Example S.19The wavelength of Hα

Find the wavelength of the photon that a hydrogen atom gives out when its electron drops from level 3 to level 2.

Solution

By (S.31), the photon carries

Δ E = 13.6 eV \times (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = 1.8889 eV,

and by (S.30) its wavelength is $λ = 1240 / 1.8889 = 656.5$ nm.

AnswerThe photon has a wavelength of 656.5 nm in a vacuum, in the red. This is the Hα line, the strongest line of hydrogen in visible light, and it gives the glowing gas of many nebulae their red colour in photographs.

The light of a hot, dense body, such as the surface of a star, has a continuous spectrum, with all wavelengths present. A hot, thin gas gives out light only at the wavelengths of its own jumps, which show as bright emission lines, as in the glowing clouds of hydrogen around young stars. A cooler, thin gas in front of a hot source takes those same wavelengths out of the light, leaving dark absorption lines. The dark lines in the spectrum of the Sun come from the cooler gas of its outer layers, and Chapter T4 shows how such lines tell us what stars are made of and how hot they are.

Mass is itself a form of energy, with $E = m c^{2}$ . A helium nucleus has about 0.7% less mass than the four protons that are fused to make it, and the energy that this missing mass carries away is what makes the Sun shine (Chapter T7).

Common errorsSection S4

The distance in Newton's law and in the formulas for orbits runs between the centres of the two bodies. For a satellite 420 km above the ground, $r$ is 6798 km.
The formula $m g h$ holds only for heights much smaller than the radius of the planet. Over larger heights $g$ weakens, and the work needed comes from $- G M m / r$ instead.
A planet's speed is inversely proportional to its distance from the Sun only at the nearest and farthest points of its orbit, where its velocity is at right angles to the line to the Sun.
Temperatures in the ideal gas law and in $k T$ must be in kelvin. A temperature of 15 °C is 288 K.
The diffraction limit $1.22 λ / D$ is an angle in radians when $λ$ and $D$ are in the same units. Multiplying it by 206 265 turns it into arcseconds.
The Doppler shift depends only on the radial velocity. Motion across the line of sight leaves the lines of a spectrum almost where they were.
A redshift of more than about 0.1 needs the exact second line of (S.29). The first line gives speeds that are too large, and above $z = 1$ it would give speeds greater than that of light.

ExercisesSection S410 questions

S4.1
Jupiter has a mass of $1.898 \times 10^{27}$ kg and a radius of $6.991 \times 10^{7}$ m. What is the acceleration of a falling body at the height of its cloud tops?
m/s²
S4.2
A geostationary satellite goes round Earth once in a sidereal day, 86 164 s, so it stays above one point on the equator. What is the radius of its orbit? Take $G M = 3.986 \times 10^{14}$ m³ s⁻² for Earth.
km
S4.3
Why do astronauts float inside the International Space Station, 420 km above the ground?
S4.4
The Moon has a mass of $7.342 \times 10^{22}$ kg and a radius of 1737 km. What is its escape speed?
km/s
S4.5
A comet passes its nearest point to the Sun, 0.5 au away, at 50 km/s. How fast does it move at its farthest point, 30 au from the Sun?
km/s
S4.6
What is the root-mean-square speed of hydrogen atoms in a gas at 6000 K?
km/s
S4.7
The James Webb Space Telescope has a mirror 6.5 m across. What is its diffraction limit, in arcseconds, for infrared light of wavelength 2.0 µm?
″
S4.8
Earth moves round the Sun at 29.8 km/s. A star lies in the plane of Earth's orbit. By how much does its measured radial velocity change between the day when Earth moves straight towards it and the day, half a year later, when Earth moves straight away from it?
km/s
S4.9
The lines given out when the electron of a hydrogen atom drops to level 3 form the Paschen series. What is the wavelength in a vacuum of Paschen α, the line from level 4 to level 3?
nm
S4.10
What is the longest wavelength of light that can ionise a hydrogen atom whose electron is in the lowest level?
nm

Olympiad problemsPast IOAA problems, restated4 problems

2009Escaping from the centre of a cloud10 points

Adapted from IOAA 2009 (Tehran), theory problem 6. We have restated it in our own words and kept the original data. The solution was last checked on 16 October 2026.

A cloud of gas from which a star will form has a uniform density, a mass $M$ and a radius $R$ . An object is launched from the centre of the cloud and moves through it without colliding with any of its particles. An object that fell freely from the surface of the cloud would reach the centre at a speed of $\sqrt{G M / R}$ . Find the speed at which the object must be launched from the centre to escape from the cloud altogether.

I solved it without the solution

AnswerThe object must be launched at $v = \sqrt{3 G M / R}$ , which is $\sqrt{3 / 2}$ , about 1.22, times the escape speed from the surface of the cloud.

2013Separating a planet from its star10 points

Adapted from IOAA 2013 (Volos), 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.

A giant planet orbits a star at an average distance of 5 au, and the system is 250 pc from us. What is the smallest diameter of a telescope that could separate the planet from the star? Assume that the telescope works in visible light of wavelength about 500 nm, outside Earth's atmosphere, and that its optics are perfect.

I solved it without the solution

AnswerThe telescope must be at least about 6.3 m across. In practice the star outshines the planet so strongly that much more is needed to see the planet, as Chapter T9 explains.

2007A cloud near the centre of the Galaxy2 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 1.12. We have restated it in our own words and kept the original data. The solution was last checked on 16 October 2026.

Atomic hydrogen at rest gives out radio waves of frequency $ν_{0} = 1420.406$ MHz, the 21 cm line. In a gas cloud near the centre of the Galaxy the line is seen at $ν = 1421.65$ MHz. How fast is the cloud moving along our line of sight, and is it coming towards us or going away?

I solved it without the solution

AnswerThe cloud is coming towards us at about 262.5 km/s.

2009Photons from the faintest stars10 points

Adapted from IOAA 2009 (Tehran), theory problem 2. We have restated it in our own words and kept the original data, adding the Sun's apparent magnitude and the solar constant from the IOAA 2025 data sheet. The solution was last checked on 16 October 2026.

A main-sequence star like the Sun has an apparent magnitude of 6, about the faintest that the naked eye can see. Estimate how many photons from it enter the eye each second. Assume that the pupil of the eye is 6 mm across and that all the star's light has a wavelength of 550 nm. The Sun has an apparent magnitude of −26.74, and its light brings $1361$ W m⁻² to Earth.

I solved it without the solution

AnswerAbout 8500 photons a second, or roughly $10^{4}$ , enter the eye from a star at the limit of naked-eye visibility.