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A-level Physics: Radio Telescopes Consolidation questions
For these questions, we will be considering galaxy NGC 660 (below), a rare polar-ring galaxy in the constellation of Pisces.
NGC 660 consists of an ordinary spiral disk, surrounded by a polar ring almost three times the size. Like all galaxies, NGC 660 has a supermassive black hole at its centre.
NGC 660 made the headlines in January 2013 because the central black hole threw out a huge eruption of gas, 10 times brighter than a supernova. It was caused by material being heated as it was drawn towards the black hole.
The following information may be useful in answering the questions: G = 6.67 x 10-11 m3 kg-1 s-2 1 light year = 1.5 x 1015 m Mass of the Sun = 2 x 1030 Kg 1 Angstrom = 1 x 10-10 m 1 parsec (1 pc) = 3.1 x 1016 m Speed of light c = 3 x 108 m s-1
Part 1 – Gravity and orbits
The force of gravity felt by a mass (m) orbiting around a second mass (M) is:
Where r is the distance between the two masses and G is the universal gravitational constant.
The centripetal force felt by an object undergoing circular motion about a central point is:
Where m is the mass of the object, vrot is the rotational velocity and r is the distance between the object and the central point.
The estimated visible mass of NGC 660 within the disc and polar-ring is 3 x 1040 Kg. The spiral disc contains approximately 75% of this visible mass. The estimated mass of NGC 660’s supermassive black hole is 4 x 1036 Kg. Jodrell Bank Discovery Centre www.jodrellbank.net
1. How much gravitational force would a sun-like star feel, orbiting very close to the central black hole of NGC 660, at distance of 0.1 parsecs?
2. How much gravitational force would an identical star feel, orbiting on the edge of the spiral disc of NGC 660, at a distance of 6,000 parsecs from the centre?
3. By considering circular motion, calculate the expected rotational velocities of the stars in questions 1 and 2.
4. What would be the rotational velocity of another identical star if it were orbiting on the edge of the polar-ring of NGC 660, at a distance of 15,000 parsecs?
5. Plot your three values for the rotational velocities of the stars on the graph below and then sketch the expected rotation curve for galaxy NGC 660.
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The graph below shows the actual measured rotation curve for NGC 660, using radio telescopes observing the 21-cm neutral hydrogen emission line.
The grey triangles represent data from the UK e-MERLIN network (centred at Jodrell Bank); the black line represents data from the Very Large Array of radio telescopes in New Mexico.
6. By using both your own graph and the observed data above a. describe the expected movement of matter in galaxies, based on your graph b. describe how the actual movement differs from expectation c. explain how this is evidence for the existence of dark matter in the universe
7. Using the observed data, estimate the total mass of NGC 660 (the polar-ring extends to approximately 15 kpc).
8. Approximately what percentage of the total mass of NGC 660 is visible? Jodrell Bank Discovery Centre www.jodrellbank.net
Part 2 – Cosmology
As light travels through the universe, it is stretched to longer wavelengths, due to the expansion of space. This “redshift” of light (z) can be used to calculate the apparent recessional velocity (vrec) of stars and galaxies by:
Where λobs is the observed wavelength of light from the
object, λ0 is the wavelength of light observed in a stationary frame and c is the speed of light.
Edwin Hubble (left) was the first to notice the redshift of distant galaxies in 1929. He concluded that the universe was expanding.
By plotting the recessional velocities of galaxies against distance (estimated by their apparent size) Hubble realised that recessional velocity was proportional to distance (see graph below). That is, the further away a galaxy is, the faster it is receding.
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This relationship is known as Hubble’s law. The constant of proportionality between recessional velocity and distance is Hubble’s constant; H0.
The relationship can therefore be written as:
-1 Where vrec is measured in km s and d in Mega parsecs.
-1 -1 The most recent estimate for H0 is 70 km s Mpc .
The graph below shows the spectra of galaxy NGC 660, taken from the Sloan Digital Sky Survey. The labelled peaks and troughs are identified emission and absorption lines. The Hα line originates from Hydrogen and it is often the brightest wavelength in visible light astronomy.
When measured in the lab, the Hα line has a wavelength of 6562.8 Angstroms. Jodrell Bank Discovery Centre www.jodrellbank.net
1. Using the spectrum, estimate the observed wavelength of the Hα line.
2. Describe what is meant by the term redshift.
3. Calculate the redshift of NGC 660.
4. Calculate the recessional velocity of NGC 660.
5. Explain how Hubble’s observations are evidence for an expanding universe.
6. Calculate the distance to NGC 660 in Mega parsecs.
NGC 660 is actually relatively close to the Milky Way. At shorter distances, Hubble’s law is not a good estimator of distance. By using other methods, the distance to NGC 660 has been estimated as 43 million light years, or 13.2 Mega parsecs.
7. Suggest one thing (other than inaccuracies in reading the Hα line from the spectrum) which might affect the accuracy of using Hubble’s law to calculate the distance to NGC 660.
8. Why is it important for astronomers to have several independent methods of determining the distance to galaxies? Jodrell Bank Discovery Centre www.jodrellbank.net
Part 3 – Resolving power
The resolving power of a telescope is described in terms of the smallest angle it can see in the sky. This is given by:
Where λ is the wavelength being observed and D is the diameter of the telescope.
The Lovell telescope (left) at Jodrell Bank has a diameter of 76.2 m.
The resolution of radio telescopes can be improved by combining them in interferometer networks. A network of telescopes acts as one giant telescope, the diameter of which is equal to the furthest possible distance between two telescopes in the network.
Jodrell Bank is the centre of the UK’s e-MERLIN network of seven radio telescopes. The furthest two points in this network are Jodrell Bank and Cambridge, 217 km apart.
1. The polar-ring of NGC 660 has a diameter of approximately 30,000 parsecs. Using the distance value of 13.2 Mega parsecs (or your previously calculated value) calculate the angular size of NGC 660 in the sky.
2. Would the Lovell telescope be able to resolve NGC 660, looking at the neutral hydrogen emission line (wavelength of 21cm)?
3. Would the e-MERLIN network be able to resolve NGC 660 at the same wavelength?
4. What would be the smallest possible diameter that a single telescope could be to resolve NGC 660 at this wavelength?
5. Give one advantage and one disadvantage of linking radio telescopes in interferometer networks, rather than building giant single dishes. Jodrell Bank Discovery Centre www.jodrellbank.net
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Answers Part 1 – Gravity and orbits The correct M must be used in these calculations. M should be the mass within the orbit radius. 1. 5.6 x 1025 N (M = mass of black hole only, 4 x 1036 Kg) 2. 8.7 x 1019 N (M = mass within spiral disc only, 75% of 3 x 1040 Kg = 2.25 x 1040 Kg) 3. a) 290 km s-1 b) 90 km s-1 4. 66 km s-1 (M = total visible mass within NGC 660, 3 x 1040 Kg) 5. Three points should be plotted relatively accurately, with a curve similar in shape to the one below.
6. a) Using Newtonian physics, the velocity of the galactic matter is inversely proportional to the square-root of the distance from the centre. The velocity should drop off rapidly as distance from the centre is increased. b) Measurements show that although there is a slight decrease in speed towards approximately 8 kpc, the rotation of matter is almost at a constant throughout the galaxy, rather than decreasing sharply towards the edge. c) Assuming that gravity is the only force acting on the matter within the galaxy, the constant rotational velocity suggests that matter is spread evenly throughout the galaxy. This matter cannot be observed, which suggests there is more matter in the universe than is currently known; i.e. Dark Matter. 7. Read an appropriate rotational velocity off the graph, e.g. 160 km s-1 Rearrange equation to get M = (v2 r) / G M = 1.8 x 1041 Kg (for v = 160 km s-1) Any value correctly calculated from a sensible reading of v is acceptable. 8. 17%
Part 2 – Cosmology 1. The Hα line on this spectrum is at 6,582.5 Angstroms. Any reading close to this, e.g. 6600 Angstroms is acceptable. Value should not be smaller than the lab value of 6,562.8 Angstroms. Jodrell Bank Discovery Centre www.jodrellbank.net
2. Redshift is the stretching of light to longer wavelengths as it travels through space, due to the expansion of the universe.
3. z = Δλ/λ0. Using Hα = 6,582.5 Angstroms gives z = 0.003 Any value correctly calculated from the Hα wavelength in question 1 is acceptable. For example, λ = 6600 Angstroms gives z = 0.006 4. v = z x c. Using z = 0.003 gives 900 km s-1 Any answer consistent with value of z from question 3 is acceptable. -1 For example, vrec = 1,800 km s (for z = 0.006) 5. Hubble observed that all distant galaxies show a redshift, which shows galaxies are moving further apart. Since the most distant galaxies appear to be travelling the quickest, this suggests that space is expanding between galaxies and that the universe is expanding.
6. d = v / H0. Any answer consistent with value of v from part 4 is acceptable. There is a large range of answers, depending on v. For example: 12.9 Mpc (for z = 0.003); 25.7 Mpc (for z = 0.006) 7. Any sensible reason is acceptable. For example: Nearby galaxies have small recessional velocities, so their own velocities are comparatively much larger – Hubble’s law applies less for nearby galaxies.
Hubble’s law is an empirical relationship; H0 is based on observations, which rely on an
independent estimate of the distance to distant galaxies. If these estimates are inaccurate, H0 will also be inaccurate. 8. To make the accepted value more reliable.
Part 3 – Resolving power 1. Using trigonometry, θ ≈ diameter/distance θ = 2 x 10-3 rads (distance = 13.2 Mpc) 2. Smallest angle Lovell can resolve at 21cm is 3 x 10-3 rads This angle is larger than angular size of NGC 660. Lovell cannot resolve NGC 660. 3. Smallest angle e-MERLIN can resolve at 21 cm is 1.2 x 10-6 rads This angle is smaller than angular size of NGC 660. e-MERLIN can resolve NGC 660. 4. 128 m 5. Examples of advantages: cheaper to build and maintain smaller dishes, single giant dishes require more engineering due to size and weight, difficult to maintain accurate parabolic shape for giant dishes Examples of disadvantages: engineering challenges required to interfere signals constructively, smaller dishes lack the collecting area of giant dishes