1.##Cleaned#up#image#of#the#Sunspots#(noon#2/14)####

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Soho#data#for#noon#2/14##

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The#sun#is#1943.4#arc#seconds#in#angular#diameter#at#noon#on#February#4#2014.#In#the#SOHO#image,#the#angular#diameter#is#481#pixels.#This#means#that#each#pixel#in#the#SOHO#image#is#4.04#arc#seconds#wide.###Measuring#the#distance#between#sunspot#group#B#and#C,#we#see#that#they#are#54.32#pixels#apart.#Looking#at#our#image,#we#see#that#the#same#sunspot#groups#are#228.27#pixels#apart.#This#means#that#there#are#more#pixels#in#our#image.#Therefore,#there#are#228.27/54.32#times#more#pixels#between#the#sunspots#in#our#image#compared#to#the#SOHO#image.#Now#we#can#calculate#the#angular#size#of#each#pixel#in#our#image.#For#our#image,#this#corresponds#to#0.96#arc#seconds#per#pixel.###The#angular#size#of#the#largest#sunspot,#the#dominant#one#in#sunspot#group#D#is#46#pixels,#or#using#our#conversion,#44.2#arc#seconds.#To#calculate#the#size#in#kilometers,#we#must#also#find#the#distance#of#the#sun#when#we#took#our#photo.#This#distance#d#is#1.477373E8#km.#The#diameter#of#our#sunspot#w#is#then#given#by#this#equation:#w#=#d*tan#(44.2’’).#The#diameter#turns#out#to#be#about#31,660#km.#This#is#about#two#and#half#times#the#diameter#of#the#earth!################

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Soho$Images$(12:00$pm;$2/1442/18)$!

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2/14!

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2/15!!

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2/16!

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2/17!

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2/18!

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We!can!now!track!the!position!of!the!sunspots!by!pixel.!Since!we!determined!that!each!

pixel!of!the!Soho!image!was!4.04!arc!seconds!wide,!we!can!also!convert!pixels!to!kilometers!since!we!know!the!distance!from!the!earth!to!the!sun!using!the!parallax.!!Using!this!method,!

we!find!that!each!pixel!is!2894!km!wide.!!!

Now,!we!can!select!a!sunspot!and!record!its!coordinate!for!each!of!the!5!days.!!I!will!track!

sunspot!A!here.!Now,!I!can!plot!the!x!coordinate!of!pixel!A!as!a!function!of!time:!!!

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I!can!fit!these!points!to!a!line!using!a!least!squares!fit.!This!gives!a!line!with!a!slope!of!2.229!pixels/hour.!This!is!the!average!tangential!speed!of!the!sunspot!across!the!sun.!Converting!

to!units!of!kilometers!per!second,!we!find!that!the!speed!of!the!sunspot!is!1.79!km/s.!Now,!

assuming!that!the!sunspot!is!fixed!to!the!reference!frame!of!the!sun,!we!can!find!the!rotation!rate!of!the!sun!if!we!know!its!radius,!which!is!about!695,500!km.!Since!w=v/r,!we!

find!that!the!angular!frequency!of!the!rotation!of!the!sun!is!2.574EQ6!radians/second.!The!

period!is!then!2.44E6!seconds,!or!about!28!days.!This!means!that!the!sun!takes!28!earth!days!to!make!a!complete!rotation.!!

!Using!this!method,!the!period!found!is!probably!slightly!longer!than!the!true!period!since!I!

have!not!taken!into!account!the!radius!of!curvature!of!the!sun.!I!recorded!the!pixel!position!

as!if!the!sunspot!just!moved!in!the!plane!of!the!image.!However,!in!reality,!the!sunspot!has!

moved!a!little!further!as!it!also!moves!into!the!plane!of!the!image!as!well.!Therefore,!I!

would!have!underestimated!its!velocity!and!overestimated!its!period.!!

Spectrum!of!the!Sun!!!1. First,#we#find#k,#the#unit#conversion#between#the#spectrometer#units#and#watts/m^2.#We#find#

this#by#plotting#k#as#a#function#of#wavelength#and#taking#the#median#value#of#k#for#wavelengths#between#800#and#900#nm,#since#those#wavelengths#are#affected#least#by#scattering.###

##By#taking#the#median#values#around#800^900#nm,#we#find#k#to#be#around#2.439E^33#counts*m^2/watt.###2.#Now,#with#our#value#of#k,#we#can#convert#the#units#of#the#spectrum#we#took#from#counts#to#watts/m^2,#and#plot#this#against#the#reference#spectrum.#This#is#the#graph#obtained:##

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Qualitatively,#we#can#see#that#light#at#shorter#wavelengths#is#scattered#more,#as#the#intensity#difference#between#the#two#spectra#is#greater,#whereas#the#intensity#at#longer#wavelengths#is#about#the#same.#We#can#also#divide#the#two#spectra#to#find#the#“response#function”#of#the#atmosphere,#or#how#much#the#atmosphere#scatters#as#a#function#of#wavelength.#This#is#the#graph#obtained:##############

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Although#Rayleigh#scattering#is#proportional#to#the#inverse#4th#power#of#the#wavelength,#the#atmospheric#scattering#displayed#in#this#graph#behaves#only#like#an#inverse#square#law.#This#occurs#because#we#flattened#the#longer^wavelength#part#of#the#graph#by#normalizing#with#k,#the#conversion#factor#we#previously#determined.#To#remove#this#error,#we#only#fit#the#graph#at#shorter#wavelengths:###

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#Here,#we#can#see#the#inverse^fourth#power^law#more#clearly.##################

3.#Now,#we#compare#the#lines#in#our#spectrum#to#a#table#of#Fraunhofer#lines#for#longer#wavelengths#

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##759#nm#O2#absorption#line###

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# 687#nm#absorption#line#also#can#be#seen###

###Possibly#898#nm#O2#absorption####################

Elliptical!Effect!#This#is#the#difference#(in#radians)#between#the#angle#traveled#by#the#earth#in#an#ideal#circular#orbit#and#the#angle#traveled#by#the#earth#in#the#true#elliptical#orbit:#

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#It#takes#the#earth#229.18#minutes#to#rotate#one#radian.#With#this,#we#can#now#convert#the#above#graph#to#a#graph#of#time#difference#versus#day.#We#simply#multiply#the#angular#difference#by#229.18#min/radian,#the#conversion#factor.#This#yields#the#following#graph:###########

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#The#physical#interpretation#of#this#graph#is#that#the#sun#will#be#“behind#schedule”#up#to#about#the#170th#day#of#the#year#and#“ahead#of#schedule”#after#that.##

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Inclination!effect!!#

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