ASTC25. PROBLEM SET#4
express代做 | 代写angular – 这是一个关于express的题目, 主要考察了关于express的内容,是一个比较经典的题目, 涉及了express/angular等代写方面
1 [30p] A day at the extraterrestrial beach
https://en.wikipedia.org/wiki/Ontario_Lacusis theotherLake Ontario: a similarly shaped and sized lake on Titan, on of many lakes on Saturns moon Titan, the largest of the 82 known moons of Saturn, and second-largest moon in the Solar System after Jupiters Ganymede. It is the first extraterrestrial lake ever found and the first river on the western shore of the lake. The northern shoreline features1 km height hills, and wave-swept shoreline. The lake was discovered in 2008 by the Cassini spacecraft. It is filled with methane, ethane and propane (more or less a cigarette lighter fluid). We want to go to the beach at Lake Ontario on Titan. Since Titan is about 9 AU from the sun, we need to dress well for the occasion: it could be fairly colder there than here. You will compute the atmospheric temperature in an indirect way, from remote observations of density profile in the atmosphere. (A) To establish the surface temperature, you will utilize the known structure of the dense, N 2 -dominated, atmosphere of Titan (molecular mass=28, pressure 1.45 atm is significantly larger than in the N 2 -dominated air in your room). Function(z)describing the vertical density profile of the atmosphere was obtained from remote observations and can be approximated near the ground by
(z) = 0 exp(z/H)
whereH= 18 .5 km is the exponential scale height, and 0 a surface density which, by the way, exceeds the atmospheric density on Earth. Assume for simplicity that the atmosphere is isothermal, i.e. p= (kT/mH)=c^2 s, whereT=const. is temperature,mHis the mass of the dominant molecule, andk= Boltzmann constant,cs=constis the isothermal soundspeed. Write and solve by variable separation an equation of the vertical hydrostatic force balance in the atmosphere. Gravitational acceleration is in local balance with gas pressure gradient acceleration, equal to( 1 /)d p/dz. In computation of gravitational accelerationgTnear the surface of Titan, you can assume that it is constant in the whole atmosphere. Titans mass isM=1.35e23 kg = 0.0225 Earths masses = 3.3 Moon masses, and its radiusR=2576 km, which is 2576/6371 of Earths radius, only 25% smaller than Mercurys radius. Prove that the solution indeed is an exponential function of heightz, and write the relation connectingHto the constant temperatureTnear the surface. Then check the units and evaluate the temperature. [You should be concerned, if that temperature is not somewhere between 70 and 100 K, since these are the typical temperatures of satellites in the outer solar system.] KnowingT, you now know how to dress for the beach-going. But more can be obtained from the observa- tion of the atmospheric scale height. (B) Write the equation for the temperatureTe f fof a spherical body placed at distancerfrom the sun, having an ideal absorption and emissivity ofQIR=1 in infrared, but non-zero scattering coefficient or Bond albedoA in the visible. [Cf. problem set # 3]. Caused only partly by the reflecting surface, the albedo is mostly due to Titans atmosphere, which is always hazy and scatters a significant fractionAof solar radiation back into space. Scattering dominates over retention of infrared radiation, so much so the atmosphere is called anti-greenhouse for its surface being colder than blackbody.
UsingTcomputed in (A), estimate the mean albedoAof Titan. Sun-Saturn distance during the observations wasr= 9 .12 AU. Any other data like solar luminosity can be found in wikipedia.
2 [20p] Janus and Epimetheus – corotating satellites
Two satellites, Janus and Epimethus were found in 1980 to orbit Saturn on almost identical, nearly circular orbits. Before the discovery of Epimetheus, only one satellite was thought to occupy Janus orbit (discovered in 1966). On Dec 31, 2003, Epimetheus semi-major axis wasaE= 151410 10 km and its orbital period was P = 0.694333517 days. Find the mass of Saturn, assuming the masses of both satellites are vanishingly small relative to the planets mass. Janus at the same time had the following orbit:aJ= 151460 10 km, P = 0.694660342 days. Approaching very slowly, at small distance Janus and Epimetheus exert non-negligible mutual gravitational force, which modifies their orbits. The more distant satellite loses energy and angular momentum, which the less distant gains. Tis decreases the semi-major axis of the outer moon and increases the semi-major axis of the inner moon. Before they achieve conjunction, the moons cross the same value of semi-major axis, at which point they stop approaching each other along the orbit. The gravitational interaction continues, with the slower moon becoming faster and vice versa. After the swap, the corotational satellites start drifting apart, never being physically closer thans=10000 km apart. Sketch the motion of the satellites in one of the possible rotating frames of reference, for instance the frame of reference of the more massive Janus. Compute thesynodic period Psynof Janus and Epimetheus pair, defined as the period at which the two bodies and the planet repeat their configuration (this could be opposition, i.e., when the moons are in one line with the planet, on opposite sides). Based on it, predict which of the moons is closer to Saturn right now. NOTE:https://www.planetary.org/articles/janus-epimetheus-swap. I mentioned opposition and not conjunction (moons in one line with the planet on the same side) which is normally used to define synodic period, because corotational satellites do not have conjunctions. Even thoughsis a small distance compared with the circumference of their orbits, please take it into account in the calculation of synodic period: the moons relative angular position changes by less that the full 360 degrees in one synodic period, because of the minimum distances. HINT: Use Keplers relationship for angular speedas a function of radius, and assume eccentricities are negligible. You may use expansion of any quantity like orbital frequencyor orbital periodPwith respect to radius. For instance,(a+x) =(a)+(d/dr)x, where(d/dr)is evaluated atr=a, and equals(d/dr) = ( 3 / 2 )(/a), in view ofa^3 /^2 ). Herexis the difference of two similaras. Alternatively, you can use the quantitiesaJandaEas they are given, and do arithmetic in double precision.
3 [30p] Derivation of greenhouse effect
The subject of this problem is to model the greenhouse effect in a planetary atmosphere by assuming that it has optical thickness zero (is transparent) to starlight, but large in infrared thermal radiation (IR). Starlight thus heats the surface, which re-emits the entire obtained energy in IR. Before eventually leaving the planet, IR
radiation is temporarily trapped in opaque atmosphere, which affects the temperature profile in the atmosphere and makes the ground warmer than in the absence of greenhouse gases. The atmosphere will be assumed to transfer radiation only in vertical direction. It will be conceptually divided into an integer numberof plane, horizontal layers, each able to efficiently absorb and re-emit at IR wavelengths.0 is called the IR optical depth of the atmosphere. (Physically, absorption and re-emission of IR is done by greenhouse gases.) By this assumption, each layer is heating only two neighboring layers (one of which can be ground, or space, in case of the lowest/highest layer). The ground and all the layers are in thermal equilibrium, i.e. at constant temperatureTn, wheren= 0 , 1 ,…,; index 0 describes the ground. The layers neither store energy nor cool down faster than they are heated by immediately neighboring layers (by layer 1 and the starlight, in the casen=0). Each atmosphere layer absorbs IR photons on upper and lower side and re-emits them in up and down directions. The ground only has one side, and layer numberonly absorbs on lower side but emits from both up and down-facing side equally. The one-sided fluxes of infrared energy emitted by a surface at temperatureTncan be written as blackbody flux Fn=Tn^4
(power per unit surface area). Consider a unit area of surface, so that you do not have to multiply all the fluxes by the same area (how large the area is does not matter for temperatures that we are seeking). The ground receives some amount of starlight which for convenience will be written as fluxF=T^4. The Tcan be considered a known quantity, although a particular value is not important here. Your task is to find how the surface temperatureT 0 ()under an atmosphere of IR optical thicknessdepends on. How much higher will it be for= 1 , 2 ,…, i.e. what areT 0 ( 1 )/T 0 ( 0 ),T 0 ( 2 )/T 0 ( 0 ), andT 0 ( 3 )/T 0 ( 0 )? To find this, build a series of models for the atmospheres with optical depths= 0 , 1 , 2 ,3. The first model has 1 equation, describing energy gain and loss by the ground only, as=0 corresponds to no atmosphere. Case=1 is modeled by 2 equations (one for the ground and one for a single opaque layer), and so on. A pattern will become obvious after you do the first few models, allowing you to write an expression for the temperature valid for any. In fact, your result will be a good approximation for any real-valuedas well. HINTS Make a sketch depicting the ground and all the layers of a given model, indicating by arrows energy transfer routes. It is tedious to write the associated fluxesTn^4 , so why dont you label the arrows by the equivalent name,Fn, and also useFnwhen you write the set of+1 energy balance equations. After you find the values of allFnby algebra, you can express the final result as the ground temperatureT 0 = (F 0 /)^1 /^4.
4 [30p] Migration type III criterion
Derive the criterion of fast migration of protoplanets in disks (type III) using the following simplified model: A giant planet with massmp=M on a circular orbit opens a gap in a disk of uniform surface density. It depletes the CR (corotational) region, creating an empty gap between radiiarL, whereais the orbital radius of the planet,rL=a(/ 3 )^1 /^3 its Roche lobe radius, andis a non-dimensional multiplier which is equal = 2 .5 in a gaseous disk. Disk outside of the gap has density=const..
Initially the planet is located in the middle of the gap, but the symmetry is perturbed by something shifting the planet radially by a virtual displacement 0 toward one of the disk edges. It does not matter which, but to avoid ambiguity lets say thataincreases by 0 >0. This causes a narrow ring of gas of width 0 to be peeled off the disk, enter a horseshoe trajectory and flow across the gap switching sides from outer to inner disk (w.r.t. planet). The flow across the gap results in the decrease of gas angular momentum, corresponding to initial and final orbits spaced by 2rL. Compute the amountmof disk material transferred across the gap, and the amount of angular momentummL, transferred to the planet as a result of gas flow, using angular momentum conservation principle. (Denote asL=
GMathe specific angular momentum, and byLits jump across the gap.) Compute the resulting change 1 of the semi-major axis of the planet, and compare it with the virtual displacement assumed. If| 0 |<| 1 |then the migration will be accelerating (self-sustaining); if| 0 || 1 | then migration will die down. Formulate the criterion for fast migration, show that it has the form: mp(?)const.M, where (?) mean either