HW4: TO ANSWER QUESTIONS WITH EASE, SKIP TO ENLARGED
PURPLE FONT!!!!
Ch. 2 pg. 67 #2,4,8,10
Ch. 3 pg. 97 #3,4,9
Ch.10 pg. 289; #3,5,8
Ch. 2 pg. 67 #2,4,8,10
#2
c=fλ
so λ=c/f=
(3*10^8)/(100*10^6)
λ=?
-#4
object_a is 200nm (UV)
object_b is 650nm (red)
Wein’s law:
λnm_max = (3 *10^6)/ TK
where λnm_max is the peak wavelength of a blackbody in nanometers
and likewise
TK = (3 *10^6)/ λnm_max for the temperature in Kelvins
corresponding to this peak wavelength of a blackbody
TK_A = (3 *10^6)/ λnm_max_A = (3 *10^6)/200
TK_B = (3 *10^6)/ λnm_max_B = (3 *10^6)/650
TK_A/ TK_B = λnm_max_B/ λnm_max_A
λnm_max_B/ λnm_max_A = 650/200 > 1 so TK_A > TK_B
brightness has units of W/m2
Flux ,which is brightness at the radius of the sphere in question, is F = σT4
This also has the same units of brightness. This is what they are asking in the problem.
If you combine these, you get B= 4π R2σT4 / (4πd2) = R2σT4 / d2
If d=R, B= σT4 which is brightness at R, which is defined as flux,F
(TK_A)4/( TK_B)4 =(TK_big)4/( TK_small)4 = Fbig/Fsmall = (650/200)4
On google enter: (650/200)^4
Fbig/Fsmall =?
-8. what are all possibilities?
2nd excited is n=3
ground state is n=1;
List all possibilities.
also look
up wavelengths from my astro-notes and write them down!
3rd excited is n=4
ground state is n=1;
List all possibilities.
also look
--
up wavelengths from my astro-notes and write them down!
10. Doppler Shift
vorb = √(GM/r) =( IvRI+IvBI)/2 = IvRI = IvBI
vR is –ve (receding) and vB is +ve (approaching);
vR= (1-λR/λ0)*c; λR > λ0 so vR is –ve; (1-λR/λ0) = (λ0/λ0)-(λR/λ0)= (λ0-λR)/λ0 =Δλ/λ0 and is -ve
vB= (1-λB/λ0 )*c; λB < λ0 so vB is +ve; (1-λB/λ0) = (λ0/λ0)-(λB/λ0)= (λ0-λB)/λ0 =Δλ/λ0 and is +ve
-The following approach will spare us the torture of keeping track of –ve numbers:
where λR/λ0 = 1 + IvRI/c where vR is RECEDING so λR is Red-Shifted from λ0, and λR > λ0 (increased
wavelength); IvRI= (λR/λ0 – 1)*c
and
where λB/λ0 = 1 - IvBI/c where vB is APPROACHING so λB is Blue-Shifted from λ0, and λB < λ0 (decreased
wavelength); IvBI= (1-λB/λ0 )*c
λ0 = (λR+λB)/2
-Hint: Since 3.00036m > 2.99964m,
λR is 3.00036m
λB is 2.99964m
λ0 = (λR+λB)/2 = (3.00036 + 2.99964)/2 = 3
λ0 =3
-IvRI= ((λR/λ0) – 1)*c =
((3.00036/3)-1)*(3*10^8) =36000
IvBI= (1-(λB/λ0) )*c =
(1-(2.99964/3))*(3*10^8)=36000; wow!!!, it’s the same, and it should be.
( IvRI+IvBI)/2=(36000+36000)/2=36000 m/s
Given: radius = 100,000 km = 10^8 meters
. vorb = √(GM/r) =( IvRI+IvBI)/2
√((6.67*10^-11)*M/(10^8)) = 36000
so (6.67*10^-11)*M/(10^8) = (36000)2
M = (vorb2)*(r)/(6.67*10^-11) =
M = (36000^2)*(10^8)/(6.67*10^-11) = ?
OR
ALTERNATIVE ON GOOGLE:
M = (((λR/((λR+λB)/2))-1)*c)^2 *(r)/(6.67*10^-11) =
(((3.00036/((3.00036+2.99964)/2))-1)*(3*10^8))^2*(10^8)/(6.67*10^11) = ?
Ch. 3 pg. 97 #3,4,9
3.
t1*(D1)^2 = t2*(D2)^2
or (D1)^2/(D2)^2 = t2/t1
or (D2)^2/(D1)^2 = t1/t2
Given:
D1=2 m, t1=1 hr
D2=6 m, t2=?
SOLVE!
t2= (t1*(D1)^2)/(D2)^2 in hours=
60*(t1*(D1)^2)/(D2)^2 in minutes
-D1=2, t1=1
D2=6, t2=?
t1>t2 because D2>D1
First find t1/t2 = (D2)^2/(D1)^2 =
(6)^2/(2)^2)= ? on Google
t1/t2 =?
So, it takes ?X as much time for a 2m sensor to collect the same amount as light as the 6m sensor
Second, find t2 in minutes.
On Google, substitute the data:
60*(t1*(D1)^2)/(D2)^2 on Google
60*(1*(2)^2)/(12)^2 = ??
t2=??
-D1=2, t1=1
D2=12,t2=?
First find t1/t2 = (D2)^2/(D1)^2 =
(12)^2/(2)^2)= ? on Google
t1/t2 = ?
So, it takes ?X as much time for a 2m sensor to collect the same amount as light as the 12m sensor
Second, t2=
60*(1*(2)^2)/(12)^2 = ??
4.
arcsec >.25m/ Dm where arcsec is the minimum resolution angular distance in arcseconds for
accurately resolving 2 objects using a given diameter mirror and a given imaging wavelength;
the smaller the better, and we can’t get any smaller than arcsec …..
.05" > .25*(.7)/Dm, where.05" is the minimum resolution needed
so
Dm>.25μm/arcsec is needed for the minimum diameter mirror needed for accurately resolving 2
objects at a distance of a given number of arcseconds
Dm= .25*(.7)/.05 = ? meters on Google;
We use this Dm for a) and b)
Now that you have gotten the hang of this, what are the answers in arcseconds?
Use this Dm from above (Dm= .25*(.7)/.05)
a) .25*(3.5)/ Dm = ??
b) .25*(.140)/ Dm = ??
9.
(N(Dhex)2) = Deq
N= the number of hexagonal mirrors which make the total, bigger, equivalent hexagonal mirror
Dhex =diameter of each hexagonal mirror
Deq =diameter of equivalent mirror(total )= your answer
(N(Dhex)2) = Deq
For N = 36 hexagonal mirrors, Dhex=1.8 meters in diameter each….
On Google, enter:
sqrt(36*(1.8)^2)
Answer is 10.8 meters
Now do problem 9.
sqrt(N*( Dhex)^2) on Google
a) 2 separate 10 meter mirrors
b) 4 separate 8 meter mirrors
Ch.10 pg.289 3,5,8
L= 4π R2σT4 for luminosity in watts
σ = 5.67*10^-8 W/m2K4
Rsun= 6.96*10^8 m
Tsun= 5800K
Lsun = 4*pi* (6.96*10^8)^2*(5.67*10^-8)*(5800)^4
= 3.90592029*10^26 W
3. a) L=?? when T= 10000K and R=3*Rsun
JUST ENTER CRAP INTO GOOGLE:
L= 4*pi* R^2*σ*T^4
=
4*pi* (3*6.96*10^8)^2*(5.67*10^-8)*(10000)^4
= ??
(??) /( Lsun)= ??? x Lsun
Lsun is from above.
b)T= 2*Tsun; L=64*Lsun; R=?
NO GOOGLE HERE; (2*Tsun);
I’ll do this for you!!!
nice:
L= 4π R2σT4
and Lsun = 4π Rsun2σ Tsun4 , so Rsun2 = Lsun/4π σ Tsun4
4
Since L= 64Lsun and T= 2Tsun from the givens, and knowing 2 = 16
64Lsun = 4π R2σ(2Tsun)4
Lsun =( 4π R2σ16Tsun 4)/64 = ¼ R2(4π σ Tsun4 )
¼ R2 = (Lsun/4π σ Tsun4 ) = Rsun2
¼ R2 = Rsun2
R2 = 4Rsun2
R = 2 Rsun
You can also use solar units, where 4π σ=1, Rsun =1, Tsun =1, and this makes Lsun =1;
This simplifies to L= R2T4
64(1)= R2 *(2(1))^4
R2=64/16=4
R=2=2(1) 2 Rsun
alternate ugly:
L= 4π R2σT4
so R2= L/4π σT4 so R= sqrt(L/4π σT4) =
R= sqrt(1/4π σ)* sqrt(L/T4)
and likewise
Lsun = 4π Rsun2σ Tsun4
so
***Rsun = sqrt(1/4π σ)* sqrt(Lsun / Tsun 4)*** This is a great format!!!
R= sqrt(1/4π σ)* sqrt(L/T4)
Since T= 2*Tsun and L=64*Lsun from the givens,
R = sqrt(1/4π σ )* sqrt(64*Lsun /(2*Tsun)4)=
sqrt(1/4π σ) )* sqrt(64 /(2)4)* sqrt(Lsun /(Tsun)4),
SO
R= sqrt(64 /(2)4)*sqrt(1/4π σ) )* sqrt(Lsun /(Tsun)4)
Remember, Rsun= sqrt(1/4π σ) )* sqrt(Lsun /(Tsun)4)
SO
R = sqrt(64 /(2)4)* Rsun= 2 Rsun
5.
b2 /b1 = [ (100)1/5]m1-m2 = [2.512]m1-m2
Given: Absolute magnitudes, 3 and 8, but same apparent magnitude, mapp
Let mapp_a = ma = m1 = mapp
and mapp_b = mb = m2 =mapp
and mabs_a = 3
and mabs_b = 8
This means m1 – m2 = 0 because same apparent magnitudes;
mapp – mapp = 0
b2 /b1 = [ (100)1/5]m1-m2 = [ (100)1/5]0 = 1, so b1 =b2
(100^.2)^(m1-m2) is Google notation
On Google: (100^.2)^(0)
Answer = 1
--
b= L/ (4πd2) and L=b*(4πd2)
L and d are different for stars A and B, though because different absolute magnitudes
Let’s find d’s
mabs = (mapp) – (5*log(dpc /10))
so,
mapp = mabs +(5*log(dpc /10))
Since mapp are the same, mabs_a +(5*log(dpc_a /10)) = mabs_b +(5*log(dpc_b /10))
and remember, mabs_a = 3 and mabs_b = 8
so
3 +(5*log(dpc_a /10)) = 8 +(5*log(dpc_b /10))
5 = (5*log(dpc_a /10)) - (5*log(dpc_b /10))
5 = 5*(log(dpc_a /10) - log(dpc_b /10))
1= log(dpc_a /10) - log(dpc_b /10))
using algebra, logx - logy=log(x/y) and 10logx= x
so 1 = log((dpc_a /10)/ (dpc_b /10))= log(dpc_a /dpc_b)
log(dpc_a /dpc_b) = 1
10log(dpc_a/ dpc_b) =101
dpc_a /dpc_b=10
answer: dpc_a = 10*dpc_b
-L=b*(4πd2); La/Lb = ba/bb *da2 / db2
ba/bb =1, so this means La/Lb = da2 / db2 = 100
so, furthermore, the answer is: La =100*Lb
In order for brightness to remain the same, as d increases, L must increase also
Google alternate for same mapp , but different mabs:
Since mapp are the same, mabs_a +(5*log(dpc_a /10)) = mabs_b +(5*log(dpc_b /10))
dpc_a /dpc_b = 10^((mabs_b-mabs_a)/5)
10^((8-3)/5) = ?
8.
mapp = mabs +(5*log(dpc /10))
10 = 2.5 +(5*log(dpc /10))
7.5 = (5*log(dpc /10))
7.5/5 = log(dpc /10))
7.5/5 =(15/2)/5 = 15/10 = 3/2 = 1.5
1.5 = log(dpc /10))
using 10logx= x
10(1.5)
= 10log(dpc /10)) = dpc /10
dpc = 10*101.5
dpc=10*10^1.5 on google;
dpc=?
-Google alternate:
dpc=10*10^((mapp-mabs)/5)
so
dpc=10*(10^((10-2.5)/5))
=?