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Final

EARTH 135 Lecture 1: Lecture1.1Exam


Department
Earth Science
Course Code
EARTH 135
Professor
Tanimoto
Study Guide
Final

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Earth&135&Lecture&1:&Pressure&in&a&planet&(TS&section&2.5,&p105)&
1.&Pressure&at&MOHO&
!What!is!the!pressure!at!the!MOHO!?!MOHO!is!the!boundary!between!the!crust!
and!the!mantle!and!is!about!35!km!under!the!surface!(under!continents).!Density!of!
crustal!rocks!are!typically!r=3000!kg/m3.!Since!gravitational!acceleration!g!is!9.8!
m/s2,!the!pressure!is!
!
P
MOHO =
ρ
gh =3000 ×9.8 ×35000 =1.029 ×109
!(Pa)=1.029!(GPa)!
!
2.&Pressure&at&the&center&of&the&Earth&
A!planet!actually!is!a!sphere!and!to!first!order!has!spherically!symmetric!
structure.!Then!density!r!is!a!function!of!radius!(!r).!In!such!a!case,!the!pressure!at!
radius!r!follows!the!equation!
!
dP
dr =
ρ
g
!!!!!!!!(1)!
where!
ρ
!is!density!and!g!is!the!gravitational!acceleration!at!radius!r.!An!important!
difference!from!the!above!case!(MOHO)!is!that!g!is!not!constant!in!the!planet!any!
more.!
For!simplicity,!we!take!an!average!density!
ρ
!and!assume!the!sphere!is!
uniform!in!density.!Gravitational!acceleration!g!is!not!uniform,!however.!At!radius!r,!
it!is!given!by!
!
g=GM (r)
r2
!!!!!!!!(2)!
where!M(r)!is!the!mass!within!the!radius!r.!For!a!uniform!planet,!it!is!given!by!
!
M(r)=4
π
3
r3
ρ
!!!!!!!(3)!
thus!
!
g=4
π
3G
ρ
r
!!!!!!!!(4)!
!
Using!(4)!in!(1)!and!integrating!from!r=0!to!the!surface!r=R,!we!get!
!
P(R)P(0) =2
π
3
G
ρ
2R2
! ! ! ! ! ! (5)!
!
Since!P(R)!is!very!small,!we!put!P(R)=0.!Then!we!have!
!
!!!!!!!(6)!
In!the!case!of!the!Earth,!
ρ
=5517
!(kg/m3)!and!R=6371!km.!So!
!
P(0) 2
π
36.67 ×1011 ×55172×637100021.72 ×1011
!Pa!=!172!GPa!!(7)!
This!is!actually!about!a!factor!of!two!off!from!the!correct!value!(about!360!GPa)!
because!we!simply!used!the!average!density!instead!of!Earths!layered!structure.!!
&
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