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Earth as a Planet

Earth as a Planet. Mass M = 6 x 10 27 g . Radius R = 6371 km . Mean density = M/(4/3 p R 3 ) = 5.5 g/cm 3 Moment of inertia I of the Earth: I =  r 2 dm I/(MR 2 ) = 0.331. for a uniform sphere I/(MR 2 ) = 0.4. Differentiation early in Earth’s history. Interior of the Earth.

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Earth as a Planet

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  1. Earth as a Planet • Mass M = 6 x 1027 g. • Radius R = 6371 km. • Mean density = M/(4/3 p R3) = 5.5 g/cm3 • Moment of inertia I of the Earth: • I =  r2 dm • I/(MR2) = 0.331. • for a uniform sphere I/(MR2) = 0.4.

  2. Differentiation early in Earth’s history

  3. Interior of the Earth • Crust: variable thickness with an average value of 35 km in the continents and 7-8 km in the oceanic regions. Volume ~1019 m3 Mass 2.8 x 1022 kg. • Mantle: between the Moho discontinuity (crust-mantle) and the core-mantle boundary (R = 3480 km). Volume 9 x 1020 m3 Mass 4 x 1024 kg. • Core: from the center of the Earth to the core-mantle boundary. Volume 1.77 x 1020 m3 Mass 1.94 x 1024 kg.

  4. More Details about Layering…

  5. What is Earth made of? • Why do we need to look outside the Earth to learn about what is inside the Earth? • We know the Earth is layered and that what we can sample on the outside is not typical. • By studying members of the Solar System, it is possible to estimate its original composition and the physical and chemical processes that have led to its present state.

  6. The Solar System: A highly diverse zoo!

  7. Sun Mercury Venus Earth Moon Mars Overview of the Solar System • Asteroids • Jupiter • Saturn • Uranus • Neptune • Pluto

  8. The Origin of the Solar System Frank Crary, CU Boulder Here is a brief outline of the current theory of the events in the early history of the solar system: • A cloud of interstellar gas and/or dust (the "solar nebula") is disturbed and collapses under its own gravity. The disturbance could be, for example, the shock wave from a nearby supernova. • As the cloud collapses, it heats up and compresses in the center. It heats enough for the dust to vaporize. The initial collapse is supposed to take less than 100,000 years. • The center compresses enough to become a protostar and the rest of the gas orbits/flows around it. Most of that gas flows inward and adds to the mass of the forming star, but the gas is rotating. The centrifugal force from that prevents some of the gas from reaching the forming star. Instead, it forms an "accretion disk" around the star. The disk radiates away its energy and cools off.

  9. The Origin of the Solar System Frank Crary, CU Boulder • First break point. Depending on the details, the gas orbiting star/protostar may be unstable and start to compress under its own gravity. That produces a double star. If it doesn't ... • The gas cools off enough for the metal, rock and (far enough from the forming star) ice to condense out into tiny particles. (i.e. some of the gas turns back into dust). The metals condense almost as soon as the accretion disk forms (4.55-4.56 billion years ago according to isotope measurements of certain meteors); the rock condenses a bit later (between 4.4 and 4.55 billion years ago). • The dust particles collide with each other and form into larger particles. This goes on until the particles get to the size of boulders or small asteroids.

  10. The Origin of the Solar System Frank Crary, CU Boulder • Run away growth. Once the larger of these particles get big enough to have a nontrivial gravity, their growth accelerates. Their gravity (even if it's very small) gives them an edge over smaller particles; it pulls in more, smaller particles, and very quickly, the large objects have accumulated all of the solid matter close to their own orbit. How big they get depends on their distance from the star and the density and composition of the protoplanetary nebula. In the solar system, the theories say that this is large asteroid to lunar size in the inner solar system, and one to fifteen times the Earth's size in the outer solar system. There would have been a big jump in size somewhere between the current orbits of Mars and Jupiter: the energy from the Sun would have kept ice a vapor at closer distances, so the solid, accretable matter would become much more common beyond a critical distance from the Sun. The accretion of these "planetesimals" is believed to take a few hundred thousand to about twenty million years, with the outermost taking the longest to form.

  11. The Origin of the Solar System Frank Crary, CU Boulder • Two things and the second break point. How big were those protoplanets and how quickly did they form? At about this time, about 1 million years after the nebula cooled, the star would generate a very strong solar wind, which would sweep away all of the gas left in the protoplanetary nebula. If a protoplanet was large enough, soon enough, its gravity would pull in the nebular gas, and it would become a gas giant. If not, it would remain a rocky or icy body. • At this point, the solar system is composed only of solid, protoplanetary bodies and gas giants. The "planetesimals" would slowly collide with each other and become more massive.

  12. The Origin of the Solar System Frank Crary, CU Boulder • Eventually, after ten to a hundred million years, you end up with ten or so planets, in stable orbits, and that's a solar system. These planets and their surfaces may be heavily modified by the last, big collision they experience (e.g. the largely metal composition of Mercury or the Moon). • Note: this was the theory of planetary formation as it stood before the discovery of extrasolar planets. The discoveries don't match what the theory predicted. That could be an observational bias (odd solar systems may be easier to detect from Earth) or problems with the theory (probably with subtle points, not the basic outline.)

  13. The Big Questions • What is the origin of the solar system? It is generally agreed that it condensed from a nebula of dust and gas. But the details are far from clear. • How common are planetary systems around other stars? There is now good evidence of Jupiter-sized objects orbiting several nearby stars. What conditions allow the formation of terrestrial planets? It seems unlikely that the Earth is totally unique but we still have no direct evidence one way or the other. • Is there life elsewhere in the solar system? If not, why is Earth special? • Is there life beyond the solar system? Intelligent life? • Is life a rare and unusual or even unique event in the evolution of the universe or is it adaptable, widespread and common?

  14. Solar abundance of elements • Determined from spectral absorption lines • Light from visible surface of sun passing through cooler gases above the surface • This is thought to represent total solar abundance because nuclear reactions powering the star take place deep inside and there is little convection there to mix modified material up with original material.

  15. Meteorites

  16. Meteorites: Summary • The fabric of chondrites is quite unlike that of any terrestrial rock and required very different conditions in which to form. These are identified with early stages in the development of the Solar nebula. • The carbonaceous chondrites are a close approximation to the material of the Solar Nebula, having lost only the most volatile elements. It is, therefore, plausible to regard them as a starting point from which the composition of the Earth has evolved. This leads to the Chondritic Earth Model. • The meteorites derive from the asteroids by collision. • The differentiated meteorites were formed within minor planets, or asteroids, which heated sufficiently to segregate into layers, forming an iron core, silicate mantle and transitional region between. Subsequent break-up due to collisions produced iron, achondrite, and stony-iron meteorites.

  17. Next Question: What is the origin of the distribution of elements in the Solar System? Hydrogen, the simplest element, is the basic building block.

  18. Hydrogen burning – 4 protons become alpha particle (helium nucleus) • Helium burning - 3 alpha particles become 12C (which can absorb another to become 16O • Carbon burning and oxygen burning produce 28Si (very stable), 24Mg , 32S, and other elements • Each of these requires more heat than the fusion reaction before it.

  19. Silicon burning involves breaking of pieces of other nuclei and adding them to others. This produces many stable nuclei heavier than Si. • As temperature rises, this becomes the equilibrium e-process which is like shaking and breaking up all the existing nuclei and recombining them randomly to make all possible stable nuclei up to the iron group elements. • Everything bigger than the iron group is less stable and the e-process would rearrange them into iron group elements.

  20. Neutron capture becomes the method that builds larger nuclei. • Slow-neutron or the s-process. Add a neutron to a nucleus. If nucleus becomes unstable because its neutron/proton ratio is too high, it has time to “fix” itself before another neutron arrives. It “fixes” things by a beta decay. A neutron converts to a proton and an electron is emitted. The nucleus has moved one element up the periodic table. • The s-process can build elements up to 209Bi. At that point there is no neutron/proton ratio stable enough to allow the one by one conversion of neutrons to protons.

  21. The rapid-neutron or r-process involves adding neutrons to a nucleus faster than things can be “fixed.” Much heavier nuclei can be built up. Once the bombardment is over, the neutron-rich nucleus will undergo repeated beta decays to produce nuclei that are relatively more stable but which, in turn, are unstable to alpha decay and so break down into lighter nuclei. These include 238U, 235U, and 232Th which have half-lives comparable to the age of the Earth, and so have not yet decayed to negligible amounts.

  22. The s-process also helps fill in gaps between some of the lighter elements such as between 12C and 16O. This mechanism can only produce neutron-rich nuclei, so other processes must account for known nuclei with lower than average neutron/proton ratios. • The p-process resolves this by adding protons to nuclei. • Light elements Li, Be, and B are not produced by any of the above processes. In fact, they are destroyed at the temperatures required for hydrogen burning. They are probably formed as fragments when heavy nuclei in interstellar dust are struck by cosmic rays. This is a very slow process, but interstellar dust spends a lot of time in space!

  23. The most likely place for these reactions to take place is in the interior of a large star. • The Sun is not large enough to ever get beyond hydrogen burning, and therefore will not generate the distribution of elements found in the Sun or meteorites. • One or many larger stars were needed to produce these elements that formed the nebula that became the Solar System. • The matter from these stars would have been disseminated by supernova explosions at the end of their existence as stars. • There is time between the formation of our galaxy, 15x109 years ago and the formation of the Solar System 4.6x109 years ago for many generations of stars to form, explode and slowly enrich the interstellar medium. This heavy material is the 2% of the Solar System.

  24. One additional observation suggests that the last of these supernovae must have occurred just 2-3 million years before the initiation of the formation of the Solar System. It must have occurred close to the dust cloud that became the Solar System. • There is evidence that certain elements like 26Al with very short half-lives were present in the material that formed the Solar System. If these had been formed gradually by many stars, these elements would have decayed away. They could only have been formed and disseminated by a very recent supernova. • It is possible that this supernova not only contributed material to the cloud, but also initiated its collapse.

  25. Differentiation of the Earth • Differentiation is the process by which random chunks of primordial matter were transformed into a body whose interior is divided into concentric layers that differ from one another both physically and chemically. • This occurred early in Earth’s history, when the planet got hot enough to melt.

  26. What was the starting point for differentiation? • Heterogeneous/Hot starting model • Initial layering as Earth solidified from gas • Homogeneous/Cold starting model • Little or no initial layering because Earth formed from the agglutination of cold, uniform particles • Neither model seems to work completely

  27. When did differentiation happen? • About 4.5 billion years ago • After beginning of Earth’s accretion at 4.56 billion years ago • Before the formation of the Moon’s oldest known rocks, 4.47 billion years ago

  28. Sources of heat to melt Earth • Frequent and violent impacts • There was likely one particularly large impact • Moon aggregated from the ejected debris • Earth’s spin axis was tilted • Decay of radioactive elements • This heat generation was greater in the past than today

  29. Basic processes of differentiation • In a liquid or soft solid sphere, denser material sinks to the center and less dense material floats to the top. • When rock is partially melted, the melt and the remaining solid generally have different chemistry and density. The melt is usually less dense than the “residue.” The melt is enriched in “incompatible” elements. The residue is enriched in “compatible” elements.

  30. Earth’s Core • Iron, nickel, and other heavy elements were the densest material and formed the core. Core radius is 2900 km. • They are about 1/3 of the planet’s mass • Inner core is solid. Inner core radius=1200 km. Inner core is solid because pressure is too great for iron to melt at Earth’s current temperature. • Outer core is liquid. Some of the iron in the outer core is iron sulfide.

  31. The Iron, Oxygen, Sulfur, Magnesium, and Silicon story • There were large amounts of these five elements in the early Earth • The fate of the iron was controlled by its affinity for bonding with oxygen and sulfur. • Iron bonds preferentially with sulfur. All available sulfur is consumed. Iron remains. • Oxygen bonds preferentially with magnesium and silicon. This uses up the magnesium and silicon. Oxygen remains. • Iron then combines with oxygen. Oxygen is now used up. Iron remains as elemental iron. • The iron, magnesium, and silicon oxides are light and form the Earth’s crust and mantle. • The iron sufide is dense, but less dense than iron, so it forms the outer part of the core of Earth. • The elemental iron is densest of all, so it forms the inner core of the Earth. • Note: The amount of oxygen in the starting material plays a key role in determining the size of the core of a planet. What does adding oxygen do to the core radius? What does adding sulfur do to the core radius?

  32. Earth’s Crust • Lighter rocks floated to the surface of the magma ocean. • The crust is formed of light materials with low melting temperature and is up to 40 km thick. • These are generally compounds of silicon, aluminum, iron, calcium, magnesium, sodium, and potassium, mixed with oxygen. • Fragments of crustal rocks (zircons) of age 4.3-4.4 billion years were found recently in western Australia. If this is confirmed, we can conclude that Earth cooled enough for a solid crust to form only 100 million years after the large impact.

  33. Earth’s Mantle • Lies between the crust and the core. • Depth range is 40 km to 2900 km. • The mantle consists of rocks of intermediate density, mostly compounds of oxygen with magnesium, iron, and silicon • New continental crust may be produced during partial melting of mantle material.

  34. Radiometric Dating: General Theory • The radioactive decay of any radioactive atom is an entirely random event, independent of neighboring atoms, physical conditions, and the chemical state of the atom. • It depends only on the structure of the nucleus. • λ, the decay constant, is the probability of an atom decaying in unit time. It is different for each isotope.

  35. Suppose that at time t there are N atoms and that at time t+δt, δN of those have decayed, then δN can be expressed as δN = -λ N δt In the limit as δN and δt go to 0, this becomes dN/dt = -λ N Thus, the rate of decay is proportional to the number of atoms present. Rearrangement and integration gives: loge N = -λ t + c If at t=0 there are N0 atoms present, then c = loge N0 N = N0 e-λt

  36. The half-life, T½, is the length of time required for half of the original atoms to decay. N0/2 = N0 e-λT½ or T½ = (loge 2) / λ Consider the case of a radioactive Parent atom decaying to an atom called the Daughter. After time t, N = N0 – D parent atoms remain and N0 – D = N0 e-λt Where D is the number of daughter atoms (all of which have come from decay of the parent) present at time t. Thus D = N0 (1 – e-λt)

  37. However, it is not possible to measure N0, but only N Use the previous equation and N = N0 e –λt yields D = N (eλt – 1) This equation expresses the number of daughter atoms in terms of the number of parent atoms, both measured at time t, and it means that t can be calculated by taking the natural log t = loge (1 + D/N) / λ In practice, measurements of D/N are made using a mass spectrometer. http://www.chemguide.co.uk/analysis/masspec/howitworks.html

  38. Major radioactive elements used in radiometric dating

  39. Radiometric dating is not always that simple! • There may have been an initial concentration of the daughter in the sample • Not all systems are closed. There may have been exchange of parent and/or daughter with surrounding material. • If dates from different isotope systems match within analytical error, we say the ages are concordant. If they are not, then we say they are discordant. • When discordant, we suspect problems like those above with one or all of the systems. • The date t obtained is not always the date of formation of the rock. It may be the date the rock crystallized, or the date of a metamorphic event which heated the rock to the degree that chemical changes took place. • Radioactive decay schemes are not all as simple as a parent and exactly one daughter. 87Rb to 87Sr is a simple one step decay. The two U to Pb series have a number of intermediate daughter products.

  40. Fission Track Dating • As well as decaying to 206Pb as described before, 238U is also subject to spontaneous fission. • It disintegrates into two large pieces and several neutrons. This is a very rare event, occurring just once per 2 million α decays. • Each event is recorded as a trail of destruction about 10 m long through the mineral structure. • These “fission tracks” can be observed by etching the polished surface of certain minerals. The tracks become visible under a microscope.

  41. Spontaneous Fission Tracks

  42. Consider a small polished sample of a mineral. Assume that it has [238U]now atoms of 238U distributed throughout its volume • The number of decays of 238U, Dr, during time t is: • The number of decays of 238U by spontaneous fission, Ds, which occur in time t is: • Where s is the decay constant for spontaneous fission of 238U. • To determine an age, we must count the visible fission tracks, estimate the proportion of the tracks visible (crossing) the surface, and measure [238U]now.

  43. Fortunately, we do not need to do this in an absolute manner, because another isotope of Uranium, 235U, can be made to fission artificially. This is done by putting our sample in a nuclear reactor and bombarding it with slow neutrons for a specified time (hours). This provides us with a standard against which to calibrate the number of tracks per unit area (track density). The number of induced fissions is: Where σ is the known neutron capture cross-section and n is the neutron dose in the reactor. We assume that if the two isotopes of U are equally distributed in the sample, then the proportion of tracks that cross the surface will be the same. We can combine equations to get: Where Ns and NI are the numbers of spontaneous and induced fission tracks counted in an area.

  44. The equation can be rearranged and the known present ratio of the two isotopes of Uranium, [238U]now/[235U]now=137.88,can be inserted to give: In practice, after the number of spontaneous fission tracks Ns has been counted, the sample is placed in the reactor and then etched again. The spontaneous tracks are enlarged and the induced tracks are exposed. The number of induced tracks NI are counted and the age calculated. Spontaneous Fission Tracks Induced Fission Tracks

  45. There is an additional (and very powerful) way to use fission tracks. Fission tracks in a mineral crystal are stable at room temperature, but can “heal” if the temperature of the crystal is high enough. At very high temperature, the tracks heal completely very quickly. This means that the “age” of a rock can be completely “reset” by heating. The rate at which tracks are healed varies with temperature and mineral type. Therefore there is a “closure” temperature that is a function of mineral type and rate of cooling.

  46. For example, fission track ages determined from sphene are always greater than ages determined from apatite. This is because healing tracks in sphene (~300C) requires much greater temperatures than healing tracks in apatite (~90C). Imagine that rocks are being uplifted and eroded during the creation of a mountain range. The individual rocks are cooling as they are brought closer to the surface. A progression of fission track ages in different minerals record the uplift/cooling history of the rock. There are newer, even more sophisticated methods, that use the rate at which tracks heal, they actually shorten before disappearing, to determine more complicated temperature history curves from each mineral. http://www.geotrack.com.au/ttinterp.htm

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