Announcements•Homework: Chapter 8 # 46, 47, 50 & 51 plus Estimate the size of the meteor that formed Barringer’s Crater in Arizona. Assume it was a pure iron meteor and the rock in the area was silicon.

•Exam 2 is three weeks away

Cratering is an energy transformation problem

The impactor has kinetic energy and it is converted into heat, kinetic energy of ejecta and vaporizing the ground and impactor

212KE mv Q mL Q mC T

Energy TransformationsTotal Initial Energy = Total Final Energy

Before an impact, the initial energy is only the kinetic energy of the impactor: ½mv². After the impact the energy takes six forms:1. Heat some of the impactor and ground rock to

their melting point2. Melt some of the impactor and ground rock3. Heat some of the molten impactor and ground

rock to their vaporization temperature4. Vaporize some of the impactor and ground rock5. What isn’t melted or vaporized is hurled out with

some kinetic energy6. The ground rock around the edge of the crater is

deformed

ExampleA spherical iron meteorite with a radius of 100m impacts the Earth at a speed of 44,000 km

/hr. If all the meteorite is vaporized on impact, what volume of ground rock will also be vaporized? How large a hole will this make? For simplicity, assume there is no ejecta. Constants needed:

Silicon Iron

3 3

1,799 247

12,800 6,090

705 449

700 400

1414 1538

3265 4200

2.33 7.87

Si Fe

Si Fe

solid solid

liq liq

Si Fe

Si Fe

KJ KJKg Kgf f

KJ KJKg Kgv v

J JKg C Kg CSi Fe

J JKg C Kg CSi Fe

m m

v v

g gSi Fecm cm

L L

L L

C C

C C

T C T C

T C T C

Example SolutionInitial Energy

The initial energy is just the kinetic energy of the impactor: ½mv2. To find the mass, find the volume of the sphere and multiply by the density of iron.

3

334 43 3

10

7870 100

3.2966 10

kgFe m

Fe

mm V r m

V

m kg

Example Solution 2After the impact, there will be a number of energies:• Heat to raise temperature of impactor to its

melting temperature• Heat to raise the ground rock to its melting

temperature• Heat to melt the impactor• Heat to melt the ground rock• Heat to raise the temperature of the molten

impactor to its vaporization temperature• Heat to raise the temperature of the molten

ground rock to its vaporization temperature• Heat to vaporize the impactor• Heat to vaporize the ground rock

Q mC T

Q mC T

fQ mL

fQ mL

Q mC T

Q mC T

vQ mL

vQ mL

Example Solution 3Sum up all the energies after, equate to the initial energy.

212 ( 20 ) 20

( )

solid Fe solid Si Fe Si

liq Fe Fe liq Si Si Fe Si

Fe Fe Fe m Si Si m Fe f Si f

Fe Fe v m Si Si v m Fe v Si v

m v m C T m C T m L m L

m C T T m C T T m L m L

Now solve for the mass of the ground rock

212

212

( 20 ) ( )

20

( 20 ) ( )

20

solid Fe Fe liq Fe Fe Fe

solid Si Si liq Si Si Si

solid Fe Fe liq Fe Fe Fe

solid Si Si liq

Fe Fe m f Fe v m v

Si Si m f Si v m v

Fe m f Fe v m v

Si Fe

Si m f Si v

m v C T L C T T L

m C T L C T T L

v C T L C T T Lm m

C T L C T

Si Si Sim vT L

Example Solution 4Using the constants given with the problem we find the mass of ground rock to be 1.301 x 1011kg.This mass was what filled the hemispherical hole that the impact creates so find the radius of the hole.

3

323

11

333 1.301 103

298.7 3002 2 2330

(half a sphere)Si

Sikgm

mmV r

V

kgmr m m

The hole will be about 300 m deep and 600 m wide.

Rock dating can be done by radioactive decay analysis

0

1

2

nfP

P

Pf is the final amount (moles) of an isotope, P0 is the initial amount (moles) of the isotope and n is the number of half-lives that have elapsed.

Example

A rock sample is found to contain 2.35 grams of U235 and 6.21 grams of Pb207? What is the age of the rock?

The atomic mass of an isotope is the subscript in the symbol for the element.

Example SolutionFirst find the number of moles of U235 and Pb207. The atomic mass of U235 is 235 grams so there are 0.01 moles of it. The atomic mass of Pb207 is 207 grams so there are 0.03 moles of it. That means there was initially 0.04 moles of U235.

14

12

0.01 1 log( )2

0.04 2 log( )

n

n

Thus, two half-lives have elapsed. The half-life of U235 is 713,000,000 years so the rock is 1,426,000,000 year old