Pearson Edexcel Physics

Unit 5: Thermodynamics, Radiation, Oscillations and Cosmology

5.6 Astrophysics and Cosmology

• 
  

  154) Understand that a gravitational field (force field) is defined as [math]g = \frac{F}{m}[/math] a region where a mass experiences a force

• ⇒ Gravitational field:
• A region of space where a mass experiences a force due to the presence of another mass.
• – If you place a mass (like a ball) in that region, it will feel a gravitational pull.
• – The force always acts towards the mass creating the field (like the Earth).
• A field is a way of describing how a force acts at a distance—without physical contact.
• – For gravity, the field surrounds any object with mass.
• – The strength and direction of the field tell you what force a mass would feel if it were placed there.
• 
• Figure 1 Gravitational field
• Characteristics of Gravitational Fields

• 155) Understand that gravitational field strength is defined as [math]g = \frac{F}{m}[/math]  and be able to use this equation

• ⇒ The attraction of gravity:
• Every particle with mass attracts every other particle with mass. So, if the Big Bang spread tiny particles across space, they would all attract each other
• The acceleration generated by these tiny forces might be exceedingly small, but the Universe has plenty of time. Slowly but surely, the particles move towards each other and clump together. This effect is greater for particles nearer to each other, as they attract more strongly than those separated by larger distances.
• These clumps of matter will continue to attract other nearby particles or lumps that have formed, and continue to accrete (come together under the influence of gravity) into larger and larger bodies of material.
• This collection of matter might become a planet, or if enough material gets packed together densely enough, nuclear fusion may start and it becomes a star.
• Like electric fields created by charged particles, a massive particle will also generate a radial gravitational field around itself.
• A particle that has mass will experience a force when it is in a gravitational field. Unlike electric fields, gravity is always attractive.
• The force that a body will experience is the strength of the gravitational field (g) multiplied by the amount of mass (m), as given by the equation:
• [math]F = mg[/math]
• 
• Figure 2 The attraction of gravity
• From this force equation, we can also see how quickly a massive particle would accelerate. From Newton’s second law we know that [math]F = ma[/math], so we can equate the two equations:
• [math]F = ma = mg[/math]
• So,
• [math]\begin{gather}
  a = \frac{mg}{m} = g \\
  g = \frac{F}{m}
  \end{gather}[/math]
• Therefore, what we have previously referred to as the acceleration due to gravity is the same as the gravitational field strength, g. On Earth, these are [math]9.81 ms^{-2} or 9.81 N kg^{-1}[/math]
• Example:
• What force will a two-tonne elephant feel when it is in the Moon’s surface gravitational field, which has a strength of 1.62 [math]N \, kg^{-1}[/math]?
• [math]\begin{gather}
  F = mg \\
  F = (2000)(1.62) \\
  F = 3240 \ \text{N}
  \end{gather}[/math]
• 
  

  156) Be able to use the equation [math]F = \frac{G m_1 m_2}{r^2}[/math]  (Newton’s law of universal gravitation)

• ⇒ Gravitational forces:
• Newton was the first scientist to publish the equation that gives us the gravitational force between two masses, m₁ and m₂, which are separated by a distance, r, between their centres of gravity:
• [math]\begin{gather}
  F = \frac{G m_1 m_2}{r^2} \\
  \text{where } G \text{ is the gravitational constant,} G = 6.67 \times 10^{-11} \ \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2}
  \end{gather}[/math]
• Example:
• What force will exist between two neutrons in the Crab Nebula which are two metres apart?
• [math]\begin{gather}
  F = \frac{G m_1 m_2}{r^2} \\
  F = \frac{(6.67 \times 10^{-11})(1.67 \times 10^{-27})(1.67 \times 10^{-27})}{2^2} \\
  F = 4.65 \times 10^{-65} \ \text{N}
  \end{gather}[/math]
• How quickly will each accelerate towards the other (ignoring the effects of all other particles)?
• [math]\begin{gather}
  a = \frac{F}{m} \\
  a = \frac{4.65 \times 10^{-65}}{1.67 \times 10^{-27}} \\
  a = 2.78 \times 10^{-38} \ \text{m/s}^2
  \end{gather}[/math]
• ⇒ Weighing the earth:
• Newton thought about how the Moon knows that the Earth is there, when he suggested that gravity keeps the Moon in orbit around the Earth.
• The answer to Newton’s problem is still not fully understood by scientists, but his formula can be used to great effect.
• It was of vital importance to NASA scientists when they made calculations in order to send the six Apollo Moon missions about fifty years ago.
• Even before space travel, it was possible to use data about the Moon’s orbit to work out the mass of the Earth.
• The time period of the Moon’s orbit around the Earth is 27.3 days, or [math]T= 2.36 × 10^6 s[/math]. The average orbital radius for the Moon is 384 000 km, or [math]r = 2.36 × 10^6 m[/math] From these data, we can calculate the mass of the Earth:
• [math]\begin{gather}
  \text{Gravitational attraction between Moon and Earth:} \, F = \frac{G m_E m_M}{r^2} \\
  \text{Centripetal force on the Moon in orbit:} \, F = \frac{m_M v^2}{r}
  \end{gather}[/math]
• Gravity is the cause of the centripetal force, so these are equal:
• [math]\begin{gather}
  \frac{G m_E m_M}{r^2} = \frac{m_M v^2}{r} \\
  \frac{G m_E}{r^2} = \frac{v^2}{r} \\
  \frac{G m_E}{r} = v^2 \\
  m_E = \frac{r v^2}{G}
  \end{gather}[/math]
• 
  

  157) Be able to derive and use the equation [math]g = \frac{G m}{r^2}[/math]  for the gravitational field due to a point mass

• ⇒ Gravitational Field strength:
• The radial field produced by a point mass naturally has its field lines closer together nearer the mass, as a result of its geometry (figure 3).
• 
• Figure 3 Radial field line of force
• This means that the strength of the field decreases with increasing distance from the mass causing it. The decrease is significant. In outer space, where it is the furthest possible from a galaxy or other particles, there are regions where there is almost no gravity.
• This can be explained mathematically by the formula which tells us the strength of a gravitational field at a certain distance, r, from a mass, M.
• We have already seen the force on a mass, m, caused by a gravitational field is
• [math]F = mg[/math]
• Also, the gravitational force on a mass, m, because of another mass, M, is given by Newton’s expression:
• [math]F = \frac{G M m}{r^2}[/math]
• o These two expressions are calculating the same force, so must themselves be equal:
• [math]F = ma[/math]
• So,
• [math]\begin{gather}
  mg = \frac{G M m}{r^2} \\
  g = \frac{G M}{r^2}
  \end{gather}[/math]
• The field strength is independent of the object being acted upon.
• 
  

  158) Be able to use the equation [math]V_{\text{grav}} = -\frac{G m}{r}[/math] for a radial gravitational field

• ⇒ Gravitational Potential:
• Potential energy is the stored energy that an object has due to its position.
• The potential at a point in any type of field is the amount of energy which is needed to get to that position in the field for any object which is affected by the field.
• Therefore, for a gravitational field it is expressed as an amount of energy per unit mass (Jkg-¹), as mass is what is affected by the field.
• The definition of gravitational potential is the amount of work done per unit mass to move an object from an infinite distance to that point in the field.
• As gravitational fields are always attractive, objects will always gain energy on moving into a point in the field, and so gravitational potential is always a negative quantity.
• Gravitational potential can be calculated at a distance r from a mass M from the equation:
• [math]V_{\text{grav}} = -\frac{G m}{r}[/math]
• The gravitational field strength tells us how quickly the potential is changing over distance, and the mathematical connection between the two is:
• [math]g = -\frac{dV_{\text{grav}}}{dx}[/math]
• For small distances, in which g does not change significantly, this can be calculated using the actual distance change:
• [math]\begin{gather}
  g = -\frac{\Delta V_{\text{grav}}}{\Delta x} \\
  g \Delta x = -\Delta V_{\text{grav}}
  \end{gather}
  [/math]
• As the potential was defined per unit mass, and is already a negative value, the magnitude of the actual gravitational potential energy change for an object of mass m will then be given by:
• [math]E = mg∆x = m × ∆V_{grav}[/math]
• Example:
• Calculate the gravitational potential caused by the Earth, at the distance of a geostationary satellite, which orbits at a height, h, of 35 800 km above the surface of the Earth. The radius of the Earth, R_E. is 6400 km.
• [math]\begin{gather}
  r = R_E + h \\
  r = 42200 \\
  r = 4.22 \times 10^7 \ \text{m} \\
  V_{\text{grav}} = -\frac{GM}{r} \\
  V_{\text{grav}} = -\frac{(6.67 \times 10^{-11})(6.02 \times 10^{24})}{4.22 \times 10^7} \\
  V_{\text{grav}} = -9.52 \ \text{MJ/kg}
  \end{gather}[/math]
• 
  

  159) Be able to compare electric fields with gravitational fields

• Electrical and Gravitational fields:
• They are of exactly the same form as those we used to calculate the force between charged particles, electric potential and electric field strength.
• The only differences are the symbols we use to represent the quantities causing the fields, and the constants of proportionality.
• The similarities come from the fact that both types of field are radial from a point. The constants of proportionality depend upon the way the forces interact with the Universe, and with the unit system that we use in the calculations.

•  Although the force and field strength of the gravitational and electrical fields share the same form, they do differ in some significant ways.
• Gravitational forces are always attractive but electric forces are not. As electrical charges can be either positive or negative, the electric field can be in either direction to or from a charge.
• A charged particle can be shielded from an electric field, but a massive particle cannot be shielded from a gravitational field. In addition, the electromagnetic force is significantly stronger than the gravitational force.
• 
  

  160) Be able to apply Newton’s laws of motion and universal gravitation to orbital motion

• Newton’s Laws of Motion:
• ⇒  First Law (Inertia):
• “An object stays in motion at constant velocity unless acted on by a net force.”
• – In orbit, the object (like a satellite) is constantly changing direction, so it’s accelerating, even if its speed is constant.
• – That means there must be a net force acting on it → this force is gravity.
• ⇒  Second Law:
• “The net force on an object equals the mass times its acceleration.”
• – For an object in circular orbit:
• [math]\begin{gather}
  F = ma \\
  F = \frac{mv^2}{r}
  \end{gather}[/math]
• where:
• – m is the mass of the orbiting object
• – v is its orbital speed
• – r is the radius of the orbit
• This force is provided by gravity, so we equate it with the gravitational force (see below).
• ⇒  Third Law:
• “If object A exerts a force on object B, object B exerts an equal and opposite force on object A.”
• – The planet pulls the satellite → the satellite pulls the planet with an equal and opposite force.
• – This principle also explains why both bodies technically orbit around a common center of mass.
• Newton’s Law of Universal Gravitation
• Formula:
• [math]F = \frac{GMm}{r^2}[/math]
• Where:
• – F = gravitational force
• – G = universal gravitational constant ([math]6.674 × 10^{-11} Nm^2/kg^2[/math])
• – M, m = masses of the two objects
• – r = distance between their centers
• Applying Both to Orbital Motion
• Equating Gravitational Force with Centripetal Force:
• For stable circular orbit:
• [math]F = \frac{G M m}{r^2} = \frac{m v^2}{r}[/math]
• – Mass m cancels out:
• [math]\begin{gather}
  \frac{G M m}{r} = v^2 \\
  v = \sqrt{\frac{G M}{r}}
  \end{gather}[/math]
• So orbital speed depends only on:
• – The mass of the central object M (e.g., Earth)
• – The distance r from the center
• Orbital Period (T):
• Use the relation:
• [math]v = \frac{2πr}{T}[/math]
• Substitute into earlier equation:
• [math]\left(\frac{2 \pi r}{T}\right)^2 = \frac{G M}{r}[/math]
• Solving gives:
• [math]T^2 = \frac{4 \pi^2 r^3}{G M}[/math]
  This is Kepler’s Third Law, derived from Newton’s laws.
• 
  

  161) Understand what is meant by a black body radiator and be able to interpret radiation curves for such a radiator

• 162) Be able to use the Stefan-Boltzmann law equation [math]L = σAT^4[/math]  for black body radiators

• ⇒ Black Body Radiation:
• A black body radiator is a theoretical body which completely absorbs all radiation that lands on it. We can imagine a cavity with a tiny hole in one side, and the hole itself is the black body.
• It is considered to absorb all radiation that lands on the hole, as it goes into the box (or cavity) and bounces around with no chance of escaping out of the hole before it is absorbed inside the box.
• No radiation is reflected, hence the name ‘black body’.
• 
• Figure 4 A box with a tiny hole in the surface is considered to be a perfect black body.
• All objects emit electromagnetic radiation. Infrared cameras can detect radiation from our bodies, but with our eyes we can only see visible EM radiation from hot objects; for example, the red glow from a fire.
• We know that good absorbers of radiation are also good emitters, and a black body radiator is considered to be a perfect emitter.
• It will emit EM radiation at all wavelengths, and how much of each wavelength will depend on the black body’s temperature.

• ⇒ The Stefan-Boltzmann Law:
• With the naked eye, we are only able to distinguish six different levels of how bright stars appear to us.
• This is insufficient for scientific use, as many stars of differing brightness would appear identical to our eyes. Astronomers therefore use a more precise measure to classify the actual brightness of stars: their output power, which is known as luminosity.
• We define luminosity as the rate at which energy of all types is radiated by an object in all directions. This depends upon both the object’s size and, more importantly, its temperature.
• The electromagnetic radiation is emitted across a very large range of wavelengths.
• 
• Figure 5 Black body radiation curves for different temperatures
• A perfect black body radiator will emit energy across the entire electromagnetic spectrum, following a distribution like those in figure 5. This distribution is given by the Stefan-Boltzmann law. This tells us that the output power from a black body is proportional to its surface area and the fourth power of its temperature in kelvin.
• [math]L = σAT^4[/math]
• Where the Stefan-Boltzmann constant, [math]σ = 5.67 x 10^{-8} Wm^{-2} K^{-4}[/math]. For a sphere, this would become:
• [math]L = 4πr^2 σAT^4[/math]
• Working on the assumption that a star acts like a black body emitter, which is a very good approximation, this equation describes the luminosity of a star.

• 
  

  163) Be able to use Wien’s law equation [math]\lambda_{\max} T = 2.898 \times 10^{-3} \ \text{m K}[/math]  for black body radiators

• ⇒ Wien’s Law:
• To calculate luminosity, we needed to know the temperature of the star.
• There are various methods for determining the temperatures of stars, but we will focus on one that uses the wavelengths of light given off by a star.
• When we examine the range of wavelengths emitted by a star, known as its spectrum, we find that some wavelengths are given off with more intensity than others.
• 
• Figure 6 The spectrum of light emitted by the Sun.
• We saw from the Stefan-Boltzmann law that as the temperature of a black body increases, it emits more energy. Fig D shows how the amount of energy emitted at different wavelengths changes with temperature. At higher temperatures the curve has a more pronounced peak, and the wavelength of the peak output gets shorter as the temperature rises. The relationship between the peak output wavelength and temperature is described by Wien’s law:
• [math]\lambda_{\max} T = 2.898 \times 10^{-3} \ \text{m K}[/math]
• The number [math]\lambda_{\max} T = 2.898 \times 10^{-3} \ \text{m K}[/math] is known as Wien’s constant.
• Example:
• Looking at the spectrum of light from Betelgeuse, in the constellation of Orion, its peak wavelength is at [math]9.35 × 10^{-7} m[/math] What is the surface temperature of Betelgeuse?
• [math]\begin{gather}
  \lambda_{\max} T = 2.898 \times 10^{-3} \\
  T = \frac{2.898 \times 10^{-3}}{\lambda_{\max}} \\
  T = \frac{2.898 \times 10^{-3}}{9.35 \times 10^{-7}} \\
  T = 3100 \ \text{K}
  \end{gather}[/math]
• 
  

  164) Be able to use the equation, intensity [math]I = \frac{L}{4 \pi l^2}[/math] where L is luminosity and d is distance from the source

• 165) Understand how astronomical distances can be determined using trigonometric parallax

• 166) Understand how astronomical distances can be determined using measurements of intensity received from standard candles (objects of known luminosity)

• ⇒ Trigonometric Parallax:
• To measure the distance to relatively close stars, astronomers use a method which is commonly used in surveying, known as trigonometric parallax.
• As the Earth moves around the Sun, a relatively close star will appear to move across the background of more distant stars.
• This optical illusion is used to determine the distance of the star. The star itself does not move significantly during the course of the observations.
• 
• Figure 7 Trigonometric parallax measurements.

• To determine the trigonometric parallax you measure the angle to a star, and observe how that changes as the position of the Earth changes.
• We know that in six months the Earth will be exactly on the opposite side of its orbit, and therefore will be two astronomical units from its location today.
• Using observations of the star which is to be measured against a background of much more distant stars, we can measure the angle between the star and the Earth in these two different positions in space, six months apart.
• As we know the size of the Earth’s orbit, geometry allows calculation of the distance to the star.
• Using observations of the star to be measured against a background of much more distant stars, we can measure the angle between the star and the Earth in these two different positions in space, six months apart.
• As we know the size of the Earth’s orbit, geometry allows calculation of the distance to the star
• By taking the picture of figure 7 and cutting it in half, we get a right-angled triangle formed as shown in figure 8, with the parallax angle,θ
• 
• Figure 8 The geometry of trigonometric parallax.
• [math]\begin{gather}
  \tan \theta = \frac{r}{d} \\
  d = \frac{r}{\tan \theta}
  \end{gather}[/math]
• At small angles, as we are usually using in astronomy, tan⁡θ ≈ θ. The distance, d, will come out in the same units used to measure r, so we also have:
• [math]d = \frac{r}{\theta}[/math]
• ⇒ The Parsec:
• A parsec (pc) is a measure of distance. It is an abbreviation of ‘parallax second’. It is the distance a star must be from the Sun in order for the angle Earth-star-Sun to be 1 arcsecond.
• [math]\begin{gather}
  1 \ \text{parsec} = \frac{1.5 \times 10^{11}}{\tan\left(\frac{1}{3600}^\circ\right)} \\
  1 \ \text{parsec} = 3.09 \times 10^{16} \ \text{m} \\
  1 \ \text{light year} = 3 \times 10^8 \times 365 \times 24 \times 60 \times 60 \\
  1 \ \text{light year} = 9.46 \times 10^{15} \ \text{m} \\
  1 \ \text{parsec} = 3.27 \ \text{light years}
  \end{gather}[/math]
• The accuracy of trigonometric parallax depends on the accuracy of the angle measurement. With atmospheric interference for Earth-based telescopes, this was for many years limited to stellar distances of about 100 light years. The European Space Agency’s Gaia mission is an orbiting telescope, which has an accuracy in the angle measurement of 24 micro arcseconds. This corresponds to allowing accurate distance measurement to stars as far away as 135 000 light years.
• We saw previously that the brightness of a star was linked to its size and its temperature. However, the Stefan-Boltzmann law only considers the power output of the star at its surface. The luminosity of a star is its overall power output over its entire surface.
• measurements of energy intensity are calculated as power per unit area. The general equation we had for intensity was [math]I = \frac{P}{A}[/math].
• The inverse square law A means that the energy emitted by a star will spread out in all directions over the surface of an increasing sphere (figure 9).
• As the surface area of a sphere is [math]4πr^2[/math], this gives us an equation for the radiant energy intensity at a certain distance, d, from a star:
• 
• Figure 9 The inverse square law, showing energy spread over greater and greater areas with increasing distance from a star.
• [math]I = \frac{L}{4 \pi l^2}[/math]
• Where L is the luminosity in watts.
• Some stars, including some variable stars and supernovae, have properties which mean their luminosity can be determined quite separately from other measurements. These are known as standard candles.
• If we have a figure for the luminosity, and measure the energy intensity (brightness) of the star reaching the Earth, we can then calculate how far away it is by comparing it with a standard candle with the same luminosity.
• Example:
• The luminosity of Betelgeuse is [math]5.4 × 10^{30}[/math]. Its radiant energy intensity at the Earth is [math]1.1 × 10^{-8} W m^{-2}[/math] How far away is Betelgeuse?
• [math]\begin{gather}
  I = \frac{L}{4 \pi d^2} \\
  d^2 = \frac{L}{4 \pi I} \\
  d = \sqrt{\frac{L}{4 \pi I}} \\
  d = \sqrt{\frac{5.4 \times 10^{30}}{4 \times 3.14 \times 1.1 \times 10^{-8}}} \\
  d = 6.2 \times 10^{18} \ \text{m}
  \end{gather}[/math]
• Or
• [math]d = 600ly[/math]
• Or
• [math]d = 200pc[/math]
• 
  

  167) Be able to sketch and interpret a simple Hertzsprung-Russell diagram that relates stellar luminosity to surface temperature

• 168) Understand how to relate the Hertzsprung-Russell diagram to the life cycle of stars

• Variable stars:
• Over a period of years at the beginning of the twentieth century, Henrietta Leavitt, working at the Harvard College Observatory, catalogued many stars in the nearby Magellanic Clouds.
• She monitored them over time and found that, for some stars, their brightness changed, changing in a repeating cycle. The time period of this oscillation in brightness was constant and, importantly, was in direct proportion to the luminosity of each star (figure 10).
• 
• Figure 10 The brightness variation typical of RR Lyrae variable stars. These are stars which have left the main sequence, but have not yet exhausted all of their nuclear fuels. They alternately expand and contract causing their light to change repeatedly.
• It was possible to calculate the intrinsic luminosity of these stars, as they were close enough to use trigonometric parallax to find their distance. Leavitt had discovered the period-luminosity relation.
• A longer time period for oscillation meant a brighter star. Astronomers then took this relationship and used it to determine the luminosity of variable stars at much greater distances.
• From the luminosity, the distance to these stars can be determined using our expression for the radiant energy flux (brightness) observed on Earth. RR Lyrae variable stars allow us to measure distances to about 760 000 parsecs.
• However, Leavitt particularly studied Cepheid variable stars. There are two types of these, and the more luminous Type I Cepheids give us the greatest distance measurements using the standard candle technique, out to about 40 million parsecs.
• ⇒ Temperature-Luminosity Relationship:
• One of the simplest methods of determining the luminosity of a star is to look at its spectrum.
• The peak wavelength gives the temperature from Wien’s law, and the width of spectral lines can determine whether or not it is a main sequence star.
• If it is, and you find its place on the main sequence of the H-R diagram as shown in fig G, you can read the luminosity from the y-axis. However, this is one of the least reliable standard candle measurements.
• 
• Figure 11 The relationship between stellar size and temperature gives luminosity, which in turn can give distance.

• 169) Understand how the movement of a source of waves relative to an observer/detector gives rise to a shift in frequency (Doppler effect)

• 170) Be able to use the equations for redshift

• [math]z = \frac{\Delta \lambda}{\lambda} \approx \frac{\Delta f}{f} \approx \frac{v}{c}[/math]

• for a source of electromagnetic radiation moving relative to an observer and [math]v = H_o d[/math]  for objects at cosmological distances

• 171) Understand the controversy over the age and ultimate fate of the universe associated with the value of the Hubble constant and the possible existence of dark matter.

• ⇒ The Doppler Effect:
• Astronomers first began to look at the spectra of stars in other galaxies during the 1920s.
• They noticed that the spectra looked very similar to the spectra from stars in our own galaxy but that all the features present were shifted by the same relative amount towards the red end of the spectrum. This phenomenon became known as the red shift.
• This shift is due to the relative motion of other galaxies with respect to ours, in an effect called the Doppler effect: an observer receiving waves emitted from a moving body observes that the wavelength of the waves has been altered to a new wavelength.
• 
• Figure 12 The Doppler effect causes a change in frequency and wavelength if there is relative motion between the wave source and the observer.
• A star or galaxy moving away emits light that appears to have a longer wavelength than expected.
• We also experience this effect when, for example, we hear a car coming towards us and driving past at a steady rate (figure 12).
• As it approaches, the note of its engine rises to a maximum pitch, and then falls as the car travels away.
• You could imagine the waves getting squashed closer together (shorter wavelength) as the car drives towards you, and then stretched further apart (longer wavelength) as it drives away.
• The amount of red shift a galaxy exhibits, z, allows us to calculate how fast it is moving.
• [mathz = \frac{\Delta \lambda}{\lambda} \approx \frac{\Delta f}{f} \approx \frac{v}{c}][/math]
• Example:
• In a laboratory sample, the hydrogen alpha spectral absorption line is at a wavelength of 656.285 nano-metres.
• In the spectrum from a nearby star, this line is observed at a wavelength of 656.315 nm. How fast is this star moving and in which direction?
• The wavelength from the star is longer than it should be, so the star is moving away.
• [math]\begin{gather}
  \Delta \lambda = 656.315 – 656.285 = 0.030 \ \text{nm} \\
  \frac{v}{c} = \frac{\Delta \lambda}{\lambda} \\
  v = c \times \frac{\Delta \lambda}{\lambda} \\
  v = 3 \times 10^8 \times \frac{0.030}{656.285} \\
  v = 13700 \ \text{m/s} \\
  v = 13.7 \ \text{km/s}
  \end{gather}[/math]
• Hubble’s Law:
• Astronomers quickly realised that red shift implied that galaxies surrounding us were travelling away from us. In 1929 the American astronomer Edwin Hubble published his finding that the value of a galaxy’s red shift is proportional to its distance from us – that is, the further away a galaxy is, the faster it is moving.
• Hubble’s paper had the same effect on the twentieth century view of the Universe as Galileo’s work on the solar system had some 300 years earlier. Instead of being static, the Universe was expanding.
• The philosophical implications continue to intrigue scientists and religious scholars.
• 
• Figure 13 As almost all galaxies show red shifts in their spectra, Hubble concluded that all the galaxies must be moving apart from each other and the Universe is expanding.
• o Considering the speeds of galaxies and their distances, there is a simple relationship between them. This is known as Hubble’s law:
• [math]v = H_o d[/math]
• So, the velocity of a galaxy is directly proportional to the distance to it. We can find the constant of proportionality, the Hubble constant, from the gradient of the graph in figure 13.
• This has had many values over the years, which demonstrates the immense difficulties involved in accurately determining astronomical distances.
• Since the launch of the ESA Planck Surveyor, the uncertainty in measurements and so the uncertainty in the Hubble constant has reduced significantly, and data from the Gaia survey will improve it even more. The current value is considered accurate to within 1% and is [math]H_0 =  70.9 km s^{-1} Mpc^{-1}[/math], although the most recent data give a slightly lower value.
• With an accurate value for Ho, astronomers can now also use Hubble’s law to determine distances to newly observed objects.
• Example:
• A supernova appears in the night sky, and astronomers find that it has a red shift of z = 0.45. How far away is the supernova?
• [math]\begin{gather}
  z = \frac{v}{c} \\
  v = z \cdot c \\
  v = 0.45 \times 3 \times 10^8 \\
  v = 1.35 \times 10^8 \ \text{m/s} \\
  v = 1.35 \times 10^5 \ \text{km/s}
  \end{gather}[/math]
• According to Hubble’s law:
• [math]\begin{gather}
  v = H_0 d \\
  d = \frac{v}{H_0} \\
  d = \frac{1.35 \times 10^5}{70.9} \\
  d = 1900 \ \text{Mpc}
  \end{gather}[/math]
• ⇒ How Old is the universe:
• All distant objects show a red shift and so they are all moving away from us.
• This implies that the Universe as a whole is expanding. If we imagine time running backwards from the present, then the known Universe would contract back to a point where everything is in the same place.
• This would be the time of the Big Bang, when everything first exploded outwards from that single point.
• Therefore, if we can find the Hubble constant, it will tell us how quickly the Universe is expanding. From this we can work out when it all started.
• For an object to travel a distance do from the beginning of time, at a speed of v_o, the time taken, T_o, can be calculated from the basic equation for speed:
• [math]\begin{gather}
  \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\
  v_0 = \frac{d_0}{T_0} \\
  T_0 = \frac{d_0}{v_0}
  \end{gather}[/math]
• If we consider the gradient of the Hubble Graph, [math]H_0 = \frav{v_0}/{d_0}[/math]
• [math]T_0 = \frac{1}{H_0}[/math]
• Note that in this calculation you should use the same units for the distance and for the length component of the units for recession velocity. Usually, though, the Hubble constant is quoted in units of [math]\text{km} \ \text{s}^{-1} \ \text{Mpc}^{-1}[/math].