Solution for Homework# 3. Chapter 5 : Review & Discussion
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1 Solution for Homework# 3 Chapter 5 : Review & Discussion. The largest telescopes are reflecting telescopes, primarily because of 3 distinct disadvantages of the refracting telescope. When light passes through a lens, light of different wavelengths focus at slightly different places. This is known as chromatic aberration and can produce seriously out of focus images. It is not easy to correct when making large lenses. Note, however, that camera lenses are quite successful in correcting this aberration, otherwise all your color photos would be rather blurry. A second problem is the glass lens absorbs certain wavelengths of light that the astronomers need to observe. In the infrared, for instance, glass is not transparent like it is for visible light. In the infrared, the glass lens blocks light from entering the telescope. Lastly, it is difficult to keep a lens bigger than about one meter from bending in its support. Glass is flexible and lenses can only be supported around their edge. When a large glass lens bends due to its own weight, its curvature changes and so does the focus, thus ruining the image. 5. The Hubble Space Telescope is not affected by seeing because it orbits above Earth s atmosphere. It can also observe at wavelengths that are absorbed by Earth s atmosphere. Its disadvantages are several; it is a very complex telescope to operate and astronomers must use it remotely. If something goes wrong, they cannot easily fix it. Because the Hubble orbits close to Earth, half the sky is blocked by Earth. Because it orbits quickly around Earth, objects may be observable for only part of the time; the rest of the time they are blocked by Earth. 7. Image processing takes advantage of fast computers available today. Since the image is really nothing more than an array of picture elements stored in the computer s memory, it can be manipulated electronically/mathematically. The particular manipulation algorithm will be dependent on the desired enhancement or the particular defect to correct for. 11. The resolution of a telescope depends on the wavelength of the light observed; the longer the wavelength, the lower the resolution. Radio waves are very long relative to visible light. Since larger telescopes produce higher resolution, radio telescopes must be very large, compared to optical telescopes, in order to have a useful resolving power. 13. Even large radio telescopes have poor resolution when compared to optical telescopes. To improve their resolution would require radio telescopes of enormous size, at least kilometers in diameter. The technique of interferometry synthesizes a telescope of this size by separating several radio telescopes by this distance and simultaneously observing the same object. Using some rather complex computer processing, the individual images are combined to synthesize what would have been observed by a telescope the size of the separation between the telescopes. Radio interferometry can now reach resolutions that are far better than optical telescopes. 16. Because anything that is warm emits strongly in the infrared, the telescopes and instruments must be cooled to low temperatures to reduce the amount of interference from them. Some infrared observations must be conducted above much of Earth s atmosphere because the atmosphere absorbs certain wavelengths of the infrared radiation coming from space. 19. Many objects emit their peak amount of radiation at wavelengths other than visible. When the universe is observed at new wavelengths, these different objects suddenly become visible, and
2 astronomers are then able to study them. Generally, observing at many different wavelengths increases the total amount of information available to astronomers. Chapter 5 : Problems ) 6.9cm 1m 10μm λmax = 5.5K 100cm 1m, so λ max = 57μm, which is longer than the 3 to 00μm operating range. 4) The angular resolution of a telescope gets poorer as wavelength increases. The angular resolution in arc-seconds is proportional to the wavelength. (a) 3.5 µm = 3,500 nm. 3,500/700 = 5, so the wavelength is 5 times longer. The resolution should be 5 times poorer or 0.05" 5 = 0.5". (b) Similarly, for the ultraviolet, 140 nm/700 nm = 0. and 0.05" 0. = 0.01". 10) This is exactly the same problem as the previous problem but with a new distance,.5 million light years. Diameter =.5 million LY (3"/ 06,000") = 36 Ly; the other two answers are 0.6 Ly and 0.01 Ly ( = 760 A.U.) 1) Area is proportional to diameter squared: (10 m) = 00 m. Taking the square root of this to get the diameter of a single mirror of equivalent area gives 14.1 m. Similarly, 4 (8 m) = 56 m. Taking the square root of this to get the diameter of a single mirror of equivalent area gives 16 m. 14) Calculating the theoretical resolution of these telescopes will not help, since they do not image in the same way regular telescopes do. The text notes that the gamma-ray telescopes have about a one degree resolution, compared to one arc seconds for Chandra. The latter is the one to use. Chapter 6 : Review & Discussion ) Comparative planetology contrasts and compares the properties of objects throughout the solar system. Typical examples are the comparison of Mars and Venus to the Earth or Mercury to the Moon. By looking for similarities, new objects can be understood; by looking for differences, new processes are revealed. By understanding the similarities and differences among the bodies of the solar system, it will be possible to understand its origin and evolution. In the end, we might hope to understand the Earth much better than we do now. By seeing how other bodies formed and developed we may be able to understand how Earth developed an environment that has been so supportive of life. It is not clear at this time that other bodies have done this nor is it understood why they have or have not. 5) Mercury and Pluto have orbits tilted to the rest of the planets. Venus and Uranus have unusual rotations and Uranus is quite tilted over. Some moons have retrograde orbits. Comets from the Oort Cloud come in from all angles. Collisions by meteoroids and comets occur randomly. 11) The ices that make up a comet become gaseous as the comet nears the Sun. These gases emit light which can be analyzed. Comets also release dusty particles which reflect sunlight and reveal information about their size and composition.
3 14) All planets have been visited by spacecraft with the exception of Pluto. Spacecraft have landed only on Venus and Mars. The answer to this is rather speculative. Smaller, cheaper, faster is the new policy. One benefit is that a probe failure is not as devastating as with the big missions. Results come back faster, which helps in the planning of future missions. 16) The answer to this is rather speculative. Smaller, cheaper, faster is the new policy. One benefit is that a probe failure is not as devastating as with the big missions. Results come back faster, which helps in the planning of future missions. 17) Interstellar dust within the early solar nebula is the key ingredient for the modern condensation theory for the formation of the solar system. 19) The jovians are much more massive than the terrestrials since they have spent more time accreting mass. Additionally, they had more material available from which to collect their matter. Chapter 6 : Problems )(a) For Earth, assume your mass is 100 kg. Using Newton s law of gravity; F = (b) For Mars ( ) F = 978 N. With 4.45 N in a pound, this gives F = 0 lbs F = ( ) F = 383 N. With 4.45 N in a pound, this gives F = 86 lbs 6) 3000 kg/m 3 = kg / 4/3π(R) 3. R = 0,000 m = 0 km. Diameter = 40 km. 9) At perihelion, the spacecraft is 1 A.U. from the Sun. At aphelion, it is 1.5 A.U. from the Sun. So, 1 A.U. = a(1 e) and 1.5 A.U. = a(1 + e). There are two equations and two unknowns, so we can solve for both a and e. Using basic algebra, we get a = 1.6 A.U. and e = 0.1. Now, use Kepler s third law, P = a, to compute the orbital period, P. P = a = 1.6 = 1.4 years. The time required to travel from Earth to Mars is simply half of the orbital period or 0.7 years. 13) The initial rotation rate, ω 1, is years and the initial diameter, d 1, is 0. Ly or 1,900 AU. Since mass is constant, we have ω 1 d 1 = ω d where ω and d are the final rotation rate and diameter respectively. For Part a, given that d = 100 A.U. and solving for ω, we have ω = revolutions per year. This gives an orbital period of 600 years. For part b, we set d = AU. Using similar algebra as in part a, we get an orbital period of about 0.4 years.. For density to remain constant, the mass must increase with the cube of the radius. Therefore, if the radius is doubled, then the mass must increase by a factor of 8. Now, the force of gravity on the planet surface is proportional to mass/radius. If the radius is doubled, and recall that the mass increases by a factor of 8, then the gravitational force would increase by a factor of 8/, or.
4 Chapter 7 : Review & Discussion ) Rayleigh scattering is the effect by which particles selectively scatter waves, depending on the waves wavelengths. Particles tend to scatter waves whose wavelengths are about equal to or smaller than the particle size. Rayleigh scattering accounts for the blue sky of Earth and red sunrises and sunsets. 4) Both P-waves and S-waves are seismic waves that move outward from the site of an earthquake. P waves are the first to arrive; these pressure waves alternately expand and compress the material through which they move, just as sound waves do. They move through both liquids and solids. S- waves are shear waves and are like waves on the surface of water. They move more slowly than P- waves and move only through solids. These waves carry information about the earthquake that produces them and also about the material through which they have traveled. 6) Certain types of seismic waves cannot travel through liquid rock. It has been known for decades that these waves, produced by earthquakes, do not travel through certain parts of the Earth's interior. This region is now mapped out as being the outer core. The inner core appears to be solid. 8) Convection is the rising of hot material through cooler material. (a) In the Earth s atmosphere it transports heat from the surface into the atmosphere and is responsible for many of the weather patterns. (b) In the Earth s interior it helped for the crust, mantle, and core of the Earth. It currently is responsible for volcanism and plate tectonics. 1) Quasars are so far away that they never show any measurable motion on the sky because of their own motion in space. Thus, any apparent change in their position can be interpreted as being due to the motion of the telescope or of the tectonic plate on which it is located. 13) A dynamo needs both liquid metal and reasonably rapid rotation. The dynamo in Earth s interior is believed to create the planet s magnetic field. 18) The Moon would experience stronger tides than on Earth because of its weaker surface gravity; the tidal force is actually weaker due to the Moon s smaller size. Since the Moon rotates once during each orbit, the tidal bulges would not appear to move relative to the surface of the Moon. Chapter 7 : Problems ) Mass is equal to density times the volume. M = /3 π ( ) 3 M = kg Using this value for the Earth s mass in Problem 1 for the surface gravity and escape velocity gives a = 5.4 m/s and v escape = 8.3 km/s 4) Since the luminosity in Stefan s law is proportional to T 4, compare the Earth at 50 K to Earth at 90 K. (50/90) 4 = Subtracting this from 1 gives 0.45 or 45%. 6) The diameter of the Earth is 6378 km = 1,756 km. At 5 km/s the time it will take is 1,756 km/5 km/s = 551 s or 4.4 minutes.
5 8) 6000 km = cm. At a rate of 3 cm/yr, this will give cm / 3 cm/yr = 10 8 yr or 00 million years. DISCUSSION FOR PROBLEM 15: The tidal force is simply a differential gravitational force, i.e., it results from differences in the gravitational pull of the Moon from one side of the Earth to the other relative to the pull at the center of the Earth. The following simplified example may help. The numbers represent the relative strength of the Moon s gravitational pull on the far side, middle, and near side of the Earth, respectively =-1 4-4=0 5-4=1 The tidal pull can easily be shown using algebra and Newton s law of gravity. F M F 1 m r m r Calculation : GMm F = r - r ( ) 1 & F = GMm ( r + r ) Tidal force is the difference between these two forces. 1 1 F1 F = GMm ( r r ) ( r + r ) Simplifying and dropping small terms like r gives: 4 r F = F1 F = GMm r 3
6 The tidal force, df, is proportional to the mass of the body producing the tide and inversely proportional to the cube of its distance. It is easiest to answer this question by comparing Jupiter s tidal effect to that of the Moon. This allows lunar units to be used, i.e., Jupiter s mass in Moon masses and 4. A.U. in lunar distances. Jupiter s mass is 318 Earth masses and the Moon is about 80 Earth masses, so Jupiter is 5,400 Moon masses. The Moon s distance from the Earth is about 1/400 A.U. so 4. A.U. will be 1680 lunar distances. Jupiter's tidal effect is = 5,400 / (1680) 3 = or about 5 millionth that of the Moon's tidal effect on Earth.
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