PTYS Geology and Geophysics of the Solar System Solutions for homework #4

Transcription

1 Y - Geology and Geohysics of the olar ystem olutions for homeork # If the core volume decreases by a factor F, then sho that the surface area of the lanet decreases by a factor ( c F Assume F is 0.995, ho many square kilometers did Mercury loose? he volume of the lanet s core changes from core to F core, causing the surface area of the hole lanet to change from to X, e ould like to find X. As all the volume change is caused by the change in core size. he change in the lanets volumes (-F core. We can relate volume and surface area of a shere ( rearrange so We ll make the st order aroximation that δ δ or then ce sin δ δ he change in surface area,, is (-X so ( ( ( ( core core core F X F X F X If F0.995, (and core / 0.75 for Mercury then X o Mercury lost 0.% of its surface area, hich is ( mercury 0km roughly 05,00 km.

2 Mercury s lithoshere broke along many thrust faults during this eisode. If each fault is about 500km long and has a dislacement of km, ho many faults does Mercury need to accommodate this shrinkage? If e assume a tyical fault di of about 5 degrees then a dislacement of km causes ~.km of surface to be overridden by the thrust sheet. If the faults are 500km long then that corresonds to ~700 km to be lost. As Mercury lost 05,00 km in total that corresonds to about 50 faults. his system of global thrust faults is unique to Mercury, yet the other terrestrial lanets also ossess cooling cores. Why don t e see this haen on the Earth, enus or Mars? he anser is different for each body. Earth he surface is already slit into lates that can slide under each other to accommodate any global shrinkage. enus enus as resurfaced ~700 Myr ago, there hasn t been much core cooling in the meantime so F (and by extension X is small. Moon he lunar core is very small or erhas non-existent. ( core / ill be very small for the Moon and by extension so ill X. Mars Mars has a very thick lithoshere so it is harder to generate the thrust faults. It has also been recently argued that the sulfur in the Martian core ill ensure that it stays comletely molten, even until today, so F is very small.

3 Imacts on enus ho that the ram ressure equals ram v atmoshere he rojectile sees u atmosheric articles in its ath and changes the momentum of those articles so that their velocity equals the imactor velocity. he force needed to change their momentum leads to a reactionary force on the imactor. he resulting ressure is the force er unit area and is knon as the ram ressure. he change in the velocity of the atmosheric articles is from 0 v ms -. he change in momentum is the mass of this material times the velocity change. A unit area ( m on the front of the imactor moves through the atmoshere at a velocity v, so the volume it sees out in one second is v x m, and the mass it sees out is v. o the momentum change every second is unit area, and so the ressure. atmoshere atmoshere atmoshere v v v. his is the force er he ressure variation ith height is given as z ( z k e here. Convert the atmosheric ressure equation gµ to density. What is the atmosheric surface density and scale height for enus, Earth and Mars. Use temeratures of 750, 70 and 00K and surface ressures of 00,, 0.0 bars resectively. tart from the ideal gas la n, here n is number of moles, is the universal gas constant, is volume and is temerature. his can be reritten in terms of the number of articles (N rather than number of moles N k here k is the Boltzmann constant. After some rearrangement Nµ k µ here µ is the mass of one molecule. o k or µ µ k so µ z z s e e k he molecular mass for atmosheres on Mars and enus is that of CO and on Earth is that of N. Given values for the surface ressure ( s and atmosheric temerature, e can calculate the scale height ( and surface atmosheric density ( s to be ressure(z0 Atm. k s s gµ k g Mars 0.0 bars 00 K 0.0 km 0.07 kg m - Earth bar 70 K 8. km.7 kg m - enus 00 bars 750 K 5.88 km 7.98 kg m -

4 If an imactor barely makes it to the surface ithout fragmentation on Mars, at hat altitude ill it break u if it had hit enus. If the imactor falls from rest at infinity it ill reach the surface of Mars at 5.0 kms -. he ram ressure at the surface given the surface atmosheric density (calculated above times the imact velocity squared and is 6.8x0 5 a. If the body breaks u at his oint (and not just because it slammed into the ground at 5 kms - then this is the strength of the body. If the same body again falls from rest but instead imacts enus then the velocity ill be 0.6 kms -. he ram ressure at the surface given the density calculated above and this velocity is given by z 9 ( a e z ram v v e When the body breaks u the ram ressure is equal to the strength e already calculated so a 9 ( a a z ln km a Alternatively, e could assume the same imact velocity for both lanets so it ould effectively cancel out of the exression for breaku-elevation. In this case, e need to find out hat elevation the atmoshere on enus has an equivalent density to that at the surface of Mars. _ MA z ln _ ENU _ ENU _ MA e z e z km One ay to recognize meteors is by their fusion crust i.e. the exterior if the rock is melted during its assage through the atmoshere. o hot do the gases at the leading edge of the meteor get, just before imact into the martian surface? (assume they re adiabatically comressed. Is this hot enough to melt rock? o dee does this thermal disturbance enetrate into the meteorite? We can calculate the temerature of the atmosheric gasses at the leading edge of the meteor by assuming they have been adiabatically comressed to the ram ressure. For adiabatic comression of an ideal gas const., here gamma is the ratio of secific heats (~7/5 for air. Combining this ith the ideal gas la

5 const const and so Nk and are 00 K and 0.0 bars and is the ram ressure (calculated above as 6.8x0 5 a. Both the ressure and ram ressure fall off exonentially ith the same scale height so the ratio of these surface values should aly to the rest of the atmoshere as ell. Inutting these values gives a heated air temerature of 60 K. Enough to barely melt the front face of the meteor; but, there isn t enough time to conduct much heat into the interior so this molten layer is thin. he thermal disturbance is conducted to a deth d κt k c t yically (for basalt k.5 W/m/K, c 800 J/Kg/K and 000 kg/m. he time is that taken to traverse the atmoshere, a fe seconds. o d ~ mm i.e. retty thin.

6 Oceanic lithoshere subduction. At some distance from the sreading ridge, the lithoshere is 00km thick and the seafloor has subsided by km. Assuming the uncooled mantle has a density of 00 Kg m - and that the late is isostatically suorted everyhere, is this late ready to be subducted? he figure belo shos a cartoon of this situation (ith scale very distorted. Isostatic equilibrium imlies that the total eight of vertical columns should be the same everyhere. ince everything above the sreading ridge is seaater and everything belo the loest ortion of the lithoshere is mantle material e need only consider material beteen these to vertical limits hen adding u the eight of column A (at the sreading center and column B (at some distance aay here the lithoshere has subsided km. he eight of the column A material mg ( 000m + L he eight of the column B material g + g ( m L 000 Where m, and are the densities of the mantle, slab and ater resectively. Equating these gives g + g 000m g 000m + L ( m ( L ( 000m + ( 000m m L L Using the values rovided, e find to be 69 kg m -, hich is slightly denser than the underlying mantle material. o yes, the slab can be subducted. What's the density ratio beteen the cooled mantle material no art of the lithoshere and the uncooled mantle material? Assume a 5km thick crust ith a density of 000 kg m -. he slab density is 69 kg m -, hich is made u of 5km of crust at 000 kg m - and 95km of cooled mantle material (together forming the 00km thick lithoshere. he eighted average can be ritten as 5 c he density of the cooled mantle material is therefore cool mantle cool mantle 00 5c 88 kg m 95 he density ratio beteen the cooled and uncooled mantle material is.07.

7 At the Mariana trench the lithoshere basically takes a right angle turn and lunges vertically into the mantle at least 650 km. ight above the subducting slab the Marianas rench has formed. Calculate ho dee you'd exect this trench to be. As before e assume isostatic equilibrium and add u mass in to equivalent columns. Column A is far from the trench and column B runs through the trench and subducting lithosheric slab. he arrangement is shon in the cartoon on the left. he eight of the column A material mg( 650 km + h L + s g( L he eight of the column B material g( 650 km + gh Where m, and are the densities of the mantle, slab and ater resectively. Equating these and rearranging gives g 650km + h + g g 650km + gh h h m ( L s ( L ( ( m ( 650km m ( 650km L s ( L ( m ( 650km L 6.5km ( he trench is redicted to be 6.5km dee. m he actual trench-bottom is about.5km belo the surrounding seafloor hat's the reason for the discreancy beteen your anser and this number? his rediction far exceeds the observed trench deth, the reason for this is that the density of the mantle increases ith deth so there is in actuality a larger buoyancy force hich acts to reduce the trench deth.

8 and dunes on riton? (From Cha 9 of Melosh 0 riton, Netune s largest moon, ossesses a very thin atmoshere that is comosed mainly of N gas at a chilly 8 K. Nevertheless, geysers sout lumes 8 km high into the atmoshere. uose that loose sand grains of ice (erhas from imact ejecta lie on the surface. o fast do the inds of riton have to blo to just entrain such ice grains? Comute both the minimum friction velocity needed to loft these grains and the minimum ind seed meter above the surface. Comare this velocity to the seed of sound in riton s atmoshere. What can you conclude about the robability of finding sand dunes on riton hen it is be visited by a sacecraft ith an imaging system caable of resolving such features? Facts that you may find useful he viscosity of nitrogen gas at 8 K is about. x 0-6 a-s and its density at riton s atmosheric ressure of.5 a is. x 0 - kg/m. he acceleration of gravity at the surface of riton is 0.78 m/sec. he adjusted eight of a article is F don 6 d ( s a g he uard force is roortional to the drag force F!"#$!! d!! C!!!! u Equating these at the motion threshold ith a constant of roortionality $ u * A s a ' & gd % a ( Exeriments sho that A ~0. for a fully turbulent flo. Using sand sized grains of ater ice (d00 microns, s 90 kg/m on riton gives u *. ms - he seed m above the surface can be found from the la of the all u u κ ln # z & % ( * $ z o ' he arameter z o is usually d/0 for closely acked grains and κ (on Karman s constant is ~0.. o, for a sandy bed on riton, hen zm then u is 98. ms -. ound seed in an ideal gas is given by c Where is the ratio of secific heats and is 7/5 for a diatomic gas. lugging in the numbers reveals than c on riton is 7ms -. o the ind (at zm ould need to be bloing at a substantial fraction of the seed of sound to mobilize sandy material (incidentally, such inds ould be bloing at mach- m above the surface. his is an unlikely situation so, sadly, e ill robably never see dunes on riton unless it had a denser atmoshere in the ast.