Planets & LifePlanets & LifePHYS 214PHYS 214

Dr Rob ThackerDr Rob ThackerDept of Physics (308A)Dept of Physics (308A)

thacker@astro.queensu.cathacker@astro.queensu.caPlease start all class related emails Please start all class related emails

with “214:”with “214:”

Pop Quiz 4Pop Quiz 4 On lectures 16-20 (I’m not including On lectures 16-20 (I’m not including

the guest lecture)the guest lecture) 10 minutes10 minutes Solutions for assignment 2 are up Solutions for assignment 2 are up

the in solutions cabinet on the 3the in solutions cabinet on the 3rdrd floor of Stirlingfloor of Stirling

WeekWeek MonMon WedWed FriFri

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99 Mid term reviewMid term review Earth History Earth History (geological issues)(geological issues)

Rare Earth hypothesisRare Earth hypothesis

1010 MarsMars(Book: 85)(Book: 85)

TitanTitan(Book: 171)(Book: 171)

Icy bodiesIcy bodies(Book: 127)(Book: 127)

1111 Broadcasts & ETIBroadcasts & ETI(Book 281)(Book 281)

Drake Equation in Drake Equation in retrospectiveretrospective(Book: 199)(Book: 199)

SETI ISETI I(Book: 281)(Book: 281)

1212 SETI IISETI II(Book: 281)(Book: 281)

Kardyshev classification Kardyshev classification & Dyson spheres& Dyson spheres

ReviewReview

Today’s LectureToday’s Lecture Review of midterm short answer Review of midterm short answer

questionsquestions

(a)(i) 8 marks(a)(i) 8 marks (a) (i) (a) (i) CarefullyCarefully draw a diagram of draw a diagram of

the Milky Way galaxy and indicate the Milky Way galaxy and indicate approximately where the Sun lies. approximately where the Sun lies. Explain the concept of a galactic Explain the concept of a galactic habitable zone, what processes habitable zone, what processes contribute to it, and use your contribute to it, and use your diagram to help illustrate this idea.diagram to help illustrate this idea.1 mark for spiral structure

1 mark for nucleus1 mark for Sun positions (approx2/3 of the way to the edge)

Galactic Habitable Zone(1 mark)

Key issue: need several generations ofStars to develop a high enough metallicity to form terrestrial planets. (1 mark)Star formation in the outer regionsis very slow, therefore we cannothave had enough generations ofStars. (1 mark) Key issue: strong radiation

can prevent life from developing (1 mark)Sources in the inner regions:High SN rate (1 mark)Gamma-ray bursters (1 mark)SM black hole (1 mark)(2 marks available here)

GHZ: A regionin a galaxy in whichconditions are favourablefor the formation of lifeas we know it.

(a)(ii) 7 marks(a)(ii) 7 marks The hydrogen The hydrogen line emission line of the quasar line emission line of the quasar

3c273 is measured on the Earth at a wavelength 3c273 is measured on the Earth at a wavelength of 565.7 nm, while it's wavelength when emitted, of 565.7 nm, while it's wavelength when emitted, is 486.1 nm. Calculate the speed at which 3c273 is 486.1 nm. Calculate the speed at which 3c273 is moving away from us. If Hubble's constant is 70 is moving away from us. If Hubble's constant is 70 km skm s-1-1 Mpc Mpc-1-1 how far away is 3c273? how far away is 3c273?

Doppler shift equation Doppler shift equation v/c where v/c where observedobservedemittedemitted, , so so =565.7-486.1=79.6 nm (1 =565.7-486.1=79.6 nm (1 mark) mark)

Calculate ratio: Calculate ratio: mark), mark), Rearrange to give v= Rearrange to give v= ccmark), mark), v=0.163*3.0*105 km sv=0.163*3.0*105 km s-1-1=49 000 km s=49 000 km s-1-1 (1 mark) (1 mark)

Common errors: not calculating properly or not taking the correct choice of

(a)(ii) cont(a)(ii) cont Hubble’s Law (not given on formula Hubble’s Law (not given on formula

sheet) v=Hsheet) v=H00d (1 mark) d (1 mark) rearrange for d: d=v/H rearrange for d: d=v/H00 (1 (1 mark) d=49 000/70=702 mark) d=49 000/70=702 Mpc (1 mark)Mpc (1 mark)Most common error: not remembering Hubble’s Law

(b)(i) 8 marks(b)(i) 8 marks Using the Hertzsprung-Russell diagram of Using the Hertzsprung-Russell diagram of

temperature versus luminosity, explain the temperature versus luminosity, explain the evolutionary stages of a star like the Sun. Ensure evolutionary stages of a star like the Sun. Ensure you discuss its formation through to its final state.you discuss its formation through to its final state.

Main sequence

Log L

Decreasing Log TO B A F….Spectral classes

Correct axes 1 markMain sequence 1 mark

Formation & protostarEvolution track1 markMain sequence track

1 mark

Giants

Red giant phase1 mark

Red supergiant phase1 mark

Supergiants

White dwarfs

White Dwarfend state1 mark

Planetary nebula1 mark

Marks also givenfor labelling variousparts of the HRdiagram.Marks also availablefor mentioning whichfuel is burnt at a givenstage (e.g. H or He).

AGB1 mark

Horizontal branch1 mark (Yellow giant)

(b)(ii) (7 marks)(b)(ii) (7 marks) Suppose a planet of radius rSuppose a planet of radius rpp, and , and

temperature Ttemperature Tpp, orbits a star, of luminosity , orbits a star, of luminosity L*, at a distance dL*, at a distance dpp. Write an equation for . Write an equation for the fraction of the stars luminosity, L* that the fraction of the stars luminosity, L* that arrives at the planets surface. If the albedo arrives at the planets surface. If the albedo is represented by is represented by aa, write down the total , write down the total amount of radiation arriving at the planet. amount of radiation arriving at the planet. Equate this to the luminosity of the planet Equate this to the luminosity of the planet to derive the equation behind the radiation to derive the equation behind the radiation balance model.balance model.

(b)(ii) (7 marks)(b)(ii) (7 marks)

Fraction of radiation arriving: To get this we just Fraction of radiation arriving: To get this we just divide the area of the planet, by the total area of the divide the area of the planet, by the total area of the sphere, and multiply by the stars luminosity (2 marks)sphere, and multiply by the stars luminosity (2 marks)

Total surface area ofthe sphere, of radius dp, that the star radiatesinto is 4dp

2Star

Planet

Radiation

Distance to planet is dp

dp

Planet’s radius is rp, surface area on the sphere ittakes up is rp

2

2

2

2

2

4*

4*luminosity offraction

p

p

p

p

dr

Ldr

L

(b)(ii) (7 marks)(b)(ii) (7 marks) Fraction arriving at planet after albedo (1 Fraction arriving at planet after albedo (1

mark)mark)

Luminosity of planet (2 marks)Luminosity of planet (2 marks)

Equate (1 mark)Equate (1 mark)

)1(4

*luminosity offraction 2

2 adr

Lp

p

424 TrL pp

)1(4

*r4 2

2 42

p adr

LTp

p

(b)(ii) (7 marks)(b)(ii) (7 marks) Simplify by cancelling like factors & Simplify by cancelling like factors &

rearrange (1 mark)rearrange (1 mark))1(*16 2

4 adLTp

(c)(i) (9 marks)(c)(i) (9 marks) Explain the key features of the solar Explain the key features of the solar

nebula theory: where does the nebula theory: where does the material in the solar nebula come material in the solar nebula come from, what does it explain in relation from, what does it explain in relation to the structure of the solar system. to the structure of the solar system. Ensure you mention differentiation Ensure you mention differentiation and how it affects the formation of and how it affects the formation of planets.planets.

(c)(i) (9 marks)(c)(i) (9 marks) 1 mark is available for each of the following1 mark is available for each of the following

Material comes from the interstellar medium which is Material comes from the interstellar medium which is enriched with heavy elements from previous SN eventsenriched with heavy elements from previous SN events

Cloud collapses under mutual gravitation and under Cloud collapses under mutual gravitation and under conservation of angular momentum it speeds up its conservation of angular momentum it speeds up its rotationrotation

Centrifugal forces prevent material in the equatorial Centrifugal forces prevent material in the equatorial plane from falling in and a disk is formedplane from falling in and a disk is formed

Radiation from the protostar keeps the interior regions Radiation from the protostar keeps the interior regions of the disk hotter than the outer regionsof the disk hotter than the outer regions

In the interior only materials with a high melting point In the interior only materials with a high melting point such as silicates and metals can condense to form such as silicates and metals can condense to form solidssolids

At larger distances ices (both water and ammonia) can At larger distances ices (both water and ammonia) can condense due to the lower temperaturescondense due to the lower temperatures

Diff

eren

tiatio

n of

the

Sola

r sy

stem

(c)(i) (9 marks)(c)(i) (9 marks) Formation of planets begins from dust grains which merge to Formation of planets begins from dust grains which merge to

form ever larger systems and so on (up to plantesimals)form ever larger systems and so on (up to plantesimals) Planets form in “disks within disks” and gain satellites in this Planets form in “disks within disks” and gain satellites in this

processprocess Planetary differentiation means that they should have rocky Planetary differentiation means that they should have rocky

cores with volatile gases being outgassedcores with volatile gases being outgassed Protostars T-Tauri phase blows out remaining gas as star Protostars T-Tauri phase blows out remaining gas as star

begins nuclear burningbegins nuclear burning What does the theory explain:What does the theory explain:

Why planets orbit in a plane around the SunWhy planets orbit in a plane around the Sun Also why planets tend rotate in the plane of the solar systemAlso why planets tend rotate in the plane of the solar system Differentiation ensures rocky planets are found in the inner Differentiation ensures rocky planets are found in the inner

regions while outer planets are gas giantsregions while outer planets are gas giants Asteriod belt is left over planetesimalsAsteriod belt is left over planetesimals Expect icy bodies in the outer solar systemExpect icy bodies in the outer solar system Can also have mentioned how exceptions are explained within Can also have mentioned how exceptions are explained within

the theorythe theory

(c)(ii) (6 marks)(c)(ii) (6 marks) A protostellar nebula has a mass of 3 solar A protostellar nebula has a mass of 3 solar

masses and a diameter of 0.30 light years. masses and a diameter of 0.30 light years. What is the density of this nebula in g cmWhat is the density of this nebula in g cm-3-3? If ? If the nebula rotate once every two million years the nebula rotate once every two million years what is the speed of the outer edge of the what is the speed of the outer edge of the nebula in km snebula in km s-1-1??

Volume of sphere = 4r3/3 (not given on formula sheet)

r=D/2=0.15 ly=0.15*63240*1.5*1011*100 cm=1.42*1017 cm (1 mark)

V=4 * 3.141 * (1.42*1017)3 = 1.21*1052 cm (1 mark)

Density=mass/volume so find mass in g

m=3*1.99*1030*1000 g = 5.97*1033 g

Density = m/V = 5.97*1033/1.21*1052 g cm-3 = 4.95*10-19 g cm-3 (1 mark)

Common errors: not remembering what density is, or volume of sphere

(c)(ii) (6 marks)(c)(ii) (6 marks) For the speed of the outer edge (two For the speed of the outer edge (two

possible ways to do this, I’ll use the possible ways to do this, I’ll use the simple one, see solutions to simple one, see solutions to assignment 2 for alternative)assignment 2 for alternative)Circumference = 2r = D

D = 3.141*0.3*63240*1.5*1011/1000 km = 8.94*1012 km (1 mark)

Period in seconds, T T = 2*106 yr =2*106*365*86400= 6.31*1013 s (1 mark)

Speed = D/T = 8.94*1012 / 6.31*1013 km s-1 = 0.14 km s-1 (1 mark)

Alternative solutions where v=r is used are also acceptable.

(d)(i) (9 marks)(d)(i) (9 marks) The greenhouse effect may well be The greenhouse effect may well be

necessary for life to develop on the necessary for life to develop on the Earth. Explain the mechanism behind the Earth. Explain the mechanism behind the greenhouse effect. Be sure to mention greenhouse effect. Be sure to mention the underlying key physical concepts the underlying key physical concepts (such as the parts of the electromagnetic (such as the parts of the electromagnetic spectrum that are relevant), the gases spectrum that are relevant), the gases which do and don't contribute, and also which do and don't contribute, and also the net flow of energy in and out of the the net flow of energy in and out of the planet.planet.

(d)(i) (9 marks)(d)(i) (9 marks) Again 1 mark for any of the following factors in the Again 1 mark for any of the following factors in the

greenhouse effectgreenhouse effect Incoming radiation is largely at Incoming radiation is largely at visible wavelengthsvisible wavelengths (peak of (peak of

Sun’s emission from Wien’s Law) which is transmitted Sun’s emission from Wien’s Law) which is transmitted wellwell by by the atmospherethe atmosphere

Black-body temperature of the Earth corresponds to Black-body temperature of the Earth corresponds to infrared infrared wavelengthwavelength which are which are strongly absorbedstrongly absorbed and effectively and effectively reflected by the greenhouse gases in the atmospherereflected by the greenhouse gases in the atmosphere

HH22O and COO and CO22 are the dominant ghg’s although CH are the dominant ghg’s although CH44 and O and O33 also also play smaller roles along with more exotic man made moleculesplay smaller roles along with more exotic man made molecules

NN22 and O and O22 are not ghg’s which is important since they make up are not ghg’s which is important since they make up the bulk of the atmospherethe bulk of the atmosphere

Overall the system is in equilibrium with the net emission from Overall the system is in equilibrium with the net emission from the Earth balancing the net incoming radiation, which includes the Earth balancing the net incoming radiation, which includes contributions from both the atmosphere and the Sun (after contributions from both the atmosphere and the Sun (after albedo)albedo)

(d)(i) (9 marks)(d)(i) (9 marks) In terms of the mechanism, 1 mark is available for In terms of the mechanism, 1 mark is available for

(some of these are slight repeats)(some of these are slight repeats) Incoming radiation reaching the Earth’s atmosphere is Incoming radiation reaching the Earth’s atmosphere is

partial reflected (albedo)partial reflected (albedo) A small fraction goes into direct heating of the atmosphere A small fraction goes into direct heating of the atmosphere

itself rather than the planetitself rather than the planet Remainder reaches Earth’s surface and heats it upRemainder reaches Earth’s surface and heats it up Earth reradiates at infrared wavelengths which is strongly Earth reradiates at infrared wavelengths which is strongly

absorbed and reradiated back towards the Earthabsorbed and reradiated back towards the Earth A small fraction of the IR emission from the Earth goes A small fraction of the IR emission from the Earth goes

directly into spacedirectly into space When the atmosphere radiates some of this radiation goes When the atmosphere radiates some of this radiation goes

directly out into space and balances the incoming net solar directly out into space and balances the incoming net solar radiation after albedo lossesradiation after albedo losses

Earth’s temperature is thus maintained above the expected Earth’s temperature is thus maintained above the expected value from the incoming radiation by the Sun combined value from the incoming radiation by the Sun combined with the atmospherewith the atmosphere

Between 35-40 K overall temperature riseBetween 35-40 K overall temperature rise

(d)(i) (9 marks)(d)(i) (9 marks) Or you could have drawn the energy Or you could have drawn the energy

flow diagramflow diagram

Atmosphere

Albedo loss

Greenhouse effect

Outgoing (~235 Wm-2)

Net incomingafter albedo(~235 Wm-2)

Visible incoming

Atmosphere

Earth

Emission at IR

(d)(ii) (6 marks)(d)(ii) (6 marks) The star Rigel shows a parallax angle of 4 milli arc The star Rigel shows a parallax angle of 4 milli arc

seconds. Calculate the distance to Rigel in parsecs. If the seconds. Calculate the distance to Rigel in parsecs. If the diameter of Rigel is 0.56 AU, calculate the angular size of diameter of Rigel is 0.56 AU, calculate the angular size of its disc on the sky in arc seconds using the distance you its disc on the sky in arc seconds using the distance you estimated from the parallax measurement.estimated from the parallax measurement.Parallax of Rigel = 4 mas = 4 * 10-3 arcseconds (1 mark)

d=1/p=1/4*10-3 pc=250 pc (1 mark)

To use small angle formula need to convert units to be the same. In this case we consider converting 250 pc to AU (but you can do it the other way)

250 pc = 250 * 3.26 * 63240 = 51.5 * 106 AU (1 mark)

=D/d=0.56/51.5*106 = 1.09 * 10-8 rad (1 mark)

Convert to milli arc seconds

1.09*10-8 * 60 * 60 *360 / (2) = 2.2 * 10-3 arc seconds = 2.2 milli arc seconds (2 marks)

Common errors: not using parallax formula

Next lectureNext lecture History of EarthHistory of Earth

Early formation issuesEarly formation issues WaterWater Plate tectonicsPlate tectonics