the sun, blackbodies, & earth’s moon problem set #1 due

9/2/21

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THE SUN, BLACKBODIES, & EARTH’S MOON

Workshops start this afternoon!

Problem Set #1 due Tuesday

First observations

September 2, 2021 ASTRONOMY 111 | FALL 2021 1

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THE SUN, BLACKBODIES, & AN INTRODUCTION TO THE MOONNucleosynthesisBlackbody radiation and temperatures of starsThe spectrum of blackbodies, and solid angleWien’s LawPhysical aspects of the Lunar surfaceSidereal timeCratering and the history of the Lunar surfaceThe Moon’s temperatureRocks and minerals


Close-up of the Sea of Tranquility, with Buzz Aldrin shown for scale (N. Armstrong/NASA)

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https://www.hq.nasa.gov/office/pao/History/ap11ann/

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NUCLEOSYNTHESIS

Models of the Sun’s interior indicate that the composition at the very center is 33.6% H and 64.3% He by mass. Much of the H there has been burned into He, mostly due to the proton-proton reaction chains. The interior of the Sun is so hot that enough collisions occur to ionize H, resulting in a plasma Individual protons collide, fusing together to form deuterium; deuterium continues to fuse with other

protons in a chain reaction to form He. The alpha particle (He) is less massive than the four protons that combined to make it, so the proton-

proton chain releases energy equal to the mass difference via 𝐸 = 𝑚𝑐!

In hotter, denser stellar cores, fusion of He (by the “triple alpha” reaction), C, O, Ne, and even heavier elements can take place at respectable rates…

…and elements heavier than Fe can be made in small quantities in such stars by the endothermic (opposite of exothermic) s-process reactions.


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NUCLEOSYNTHESIS

The Universe was born (in a fireball we can see, which we call the Big Bang) with hardly any elements heavier than helium. This is because fusion of two normal helium nuclei (!"He) produces an isotope of beryllium ("#Be) that is

very unstable, dissolving back into helium within 10$%& sec.

Fusion of three normal helium nuclei produces a stable result:3 !

"He → #$!C + 𝛾

But very high density and temperature are necessary in order to make three He nuclei collide simultaneously, energetically enough to fuse. And by the time the Universe cooled enough for nuclei to condense, the density was too low.


Triple-α process

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NUCLEOSYNTHESIS

In the cores of some stars, temperatures and densities can both be large enough for the triple-α process to proceed. And the carbon produced thereby can react to form other heavier nuclei. Thus, the heavy elements that the planets – and you – are made of were synthesized by nuclear

reactions in the cores of stars that lived and died billions of years ago.

Stars die before they burn more than about 10% of their total H into He, so the Universe will take a very long time to run out of hydrogen.


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THE NUCLEAR-CHEMICAL EVOLUTION OF THE MILKY WAY

New stars

Younger stars

Older stars


1.0E-11

1.0E-10

1.0E-09

1.0E-08

1.0E-07

1.0E-06

1.0E-05

1.0E-04

1.0E-03

1.0E-02

1.0E-01

1.0E+00

0 10 20 30 40

Num

ber o

f ato

ms p

er h

ydro

gen

atom

Atomic number, Z

Interstellar gasSunPop I starsPop II stars

C ONe

Mg Si SCa

Ti

Fe

Ni

Na AlP

N Ar

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THE TEMPERATURE OF THE SUN’S SURFACE

The Sun is a ball of hot gas; it has no sharp, solid surface. So what do we mean by the “surface” of the Sun? Usually, we mean the deepest place we can see in the Sun’s atmosphere; this is called the photosphere. The absorption of light in the Sun varies with wavelength, though, so the position of the photosphere is different

for different parts of the spectrum. The temperature varies with depth within the Sun’s atmosphere and interior, so the physical temperature of the

surface that we see is different for different wavelengths.

It is more convenient to characterize the Sun’s “surface” temperature in a way that depends upon the luminosity (total power output), which of course is wavelength-dependent. Effective temperature of the Sun: the temperature that would produce the Sun’s observed luminosity, if the Sun

were a spherical blackbody with the same radius. You will learn all about blackbodies in PHYS 143/123 and PHYS 227. We will only list the properties and

equations pertaining to blackbodies that we will need to use this semester and defer the detailed explanations to those courses.


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BLACKBODIES AND BLACKBODY RADIATION

A blackbody is a body that is perfectly opaque: it absorbs all the light incident on it.

If such a body has a constant temperature, it must therefore also emit light at exactly the same rate as it absorbs light. This is called blackbody radiation.

The flux – total power emitted in all directions by a blackbody at temperature 𝑇 per unit area – of a blackbody is given by

𝑓 = 𝜎𝑇"

𝜎 = 5.67051×10%& erg sec%$ cm%! K%"

= 5.67051×10%' J sec%$ m%! K%"


Stefan’s Law

Stefan-Boltzmann constant

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DEFINITIONS: LUMINOSITY AND FLUX

Luminosity is the total power output (in the form of light) emitted in all directions by an object. Units = erg/s The Sun’s luminosity, as we listed before, is

𝐿⊙ = 3.83×10(( erg/s

Flux is the total power (in the form of light) per unit area that passes through a (sometimes imaginary) surface. Units = erg s-1 cm-2

The flux of sunlight at the Sun’s surface is

𝑓 = 4𝐿⊙

4𝜋𝑅⊙!= 6.29×10%) erg s$% cm$!

The flux of sunlight at Earth’s orbit is

𝑓 = 4𝐿⊙4𝜋 1 AU ! = 1.36×10& erg s$% cm$!


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THE SUN’S EFFECTIVE TEMPERATURE

If the Sun were a blackbody with radius equal to that of its visible photosphere (𝑅⊙ = 6.96×10$) cm) and luminosity as observed (𝐿⊙ = 3.83×10** erg/s), then its (effective) temperature 𝑇+ would be given by

𝐿⊙ = 𝑓𝐴 = 𝜎𝑇+"4𝜋𝑅⊙! or

𝑇+ =,⊙

"-./⊙*

⁄+ ,= *.'*×$)-- erg sec.+

"- &.#3×$)./ erg sec.+ cm.* K., #.4#×$)+0 cm *

⁄+ ,= 5770 K

Hence the number given earlier for the surface temperature. This turns out not to be far off from the physical temperature typical of the photospheric region.


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THE SPECTRUM OF A BLACKBODY

One reason the effective temperature comes so close to the relevant physical temperature is that the spectrum of the Sun resembles the spectrum of a blackbody.


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BLACKBODY SPECTRUM

The spectrum of blackbody radiation is given by the Planck function:

𝑢1 =2ℎ𝑐!

𝜆21

𝑒 345167 − 1

where

ℎ = 6.626×10$!8 erg sec

𝑘 = 1.381×10$%& erg K$%

The dimensions of the Planck function may seem confusing at first: they are power per unit area, per unit solid angle, per unit wavelength interval; for example, erg sec-1 cm-2 ster-1 μm-1


Planck’s constant

Boltzmann’s constant

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SOLID ANGLE?

Solid angle is to area what angle is to length. It is usually defined by a differential element in spherical coordinates:

𝑑Ω = sin 𝜃 𝑑𝜃 𝑑𝜙

For small angles, the solid angle is calculated from the angular widths of the “patch” in the same manner as plane-geometrical areas. For a rectangle,

ΔΩ = sin 𝜃 Δ𝜃 Δ𝜙September 2, 2021 ASTRONOMY 111 | FALL 2021 13

𝑑𝜃𝑑𝜙 Angles in radians

Unit of solid angle = steradian

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REMINDER: THE SPHERICAL COORDINATES 𝑟, 𝜃, 𝜙


𝑧

𝑥

𝑦

𝑥, 𝑦, 𝑧 𝜃

𝜙

𝑟

𝑟, 𝜃, 𝜙

𝑥 = 𝑟 sin 𝜃 cos𝜙𝑦 = 𝑟 sin 𝜃 sin𝜙𝑧 = 𝑟 cos 𝜃

𝑟 = 𝑥! + 𝑦! + 𝑧!

𝜃 = tan$%𝑥! + 𝑦!

𝑧𝜙 = tan$%

𝑦𝑥

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SIMPLE EXAMPLE

What is the solid angle of a square patch of sky 15.4 arcminutes on a side? (This is the solid angle covered by the CCD camera you will use on the Mees 24-inch telescope.)

15.4 arcminutes is a small angle (≪ 1 radian). Since I did not specify where the z-axis is, we are free to point it perpendicular to the square’s center so that sin 𝜃 = 1:

ΔΩ = Δ𝜃 Δ𝜙 = 15.4′×1°

60′×2𝜋 radians360°

!

∆Ω = 2.01×10%& steradians


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COMPUTING SOLID ANGLESWhen the angles are not small, you must integrate the differential element over the range of 𝜃and 𝜙. We will not be doing this in ASTR 111, but just so you can see how it works…

The solid angle of the entre sky:

Ω = E)

!9𝑑𝜙E

)

9sin 𝜃 𝑑𝜃 = −2𝜋 Hcos 𝜃

)

9= −2𝜋 −2

Ω = 4𝜋 steradians

The solid angle of a cone with an angular radius of π/8 radians:

Ω = E)

!9𝑑𝜙E

)

39 #sin 𝜃 𝑑𝜃 = −2𝜋 Hcos 𝜃

)

39 # = 2𝜋 1 − 0.924

Ω = 0.478 steradians


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BLACKBODY SPECTRUM

The reason for the funny dimensions of the Planck function is that it has to be integrated over wavelength, solid angle, and area in order to produce a luminosity:

𝑓 = O6

𝑑ΩO)

7𝑑𝜆 𝑢8 𝐿 = O

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𝑑𝐴 O6

𝑑ΩO)

7𝑑𝜆 𝑢8

In PHYS 227 or ASTR 241, you will learn how to prove that the power per unit area emitted in all directions by a blackbody is

𝑓 = O)

!-𝑑𝜙O

)

:- !𝑑𝜃 cos 𝜃O

)

7𝑑𝜆 𝑢8 =

2𝜋&𝑘"

15𝑐!ℎ*𝑇" ≡ 𝜎𝑇"

that is, Stefan’s Law follows from the Planck function.


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THE PLANCK FUNCTION

Note that the higher the temperature, the shorter the wavelength at which the peak occurs.

Note also that visible wavelengths do not include much of the luminosity, whatever the temperature.


0.01 0.1 1 10 1001 .108

1 .109

1 .1010

1 .1011

1 .1012

1 .1013

1 .1014

1 .1015

1 .1016

10000 K5000 K2000 K1000 K

Wavelength (µm)

u l(T

) (er

g se

c-1

cm-3

ster

-1)

Visible light

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WIEN’S LAW

We will not be using the Planck function much, and we will certainly not be integrating it. It appears now because of a useful consequence of the shape of the function.

It is fairly easy to convince yourself – as you will, in Problem Set 1 – that there is a simple relation between the temperature of a blackbody and the wavelength at which it is brightest:

𝜆:;<𝑇 = 0.2897756 cm K

= 2.897756×10$( mK

The shape of the spectrum dictates what the effective temperature is. This shape is easier to accurately measure than luminosity.


Wien’s law

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EXAMPLES

At what wavelength do you (𝑇 = 98.6℉ = 37℃ = 310 K) emit the most blackbody radiation?

𝜆:;< =0.29 cm K

𝑇=0.29 cm K310 K

= 9.4×10$" cm = 9.4 𝜇m

Gamma-ray bursters emit most of their light at a wavelength of about 10$%) cm. If this emission were blackbody radiation, what would be the temperature of the burster?

𝑇 =0.29 cm K𝜆:;<

=0.29 cm K10$%) cm

= 2.9×10= K


Infrared light

Pretty hot

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THE MOON’S VITAL STATISTICS

Mass 7.349×10!2 g (0.012𝑀⊕)Equatorial radius 1.7381×10# cm (0.273𝑅⊕)Average density 3.350 g/cm3

Albedo 0.12

Average surface temperature 274.5 K

Orbital semimajor axis 3.844×10%) cmSidereal revolution period 27.3217 days

Synodic revolution period 29.53 days

Recession rate from Earth 3.8 cm/yr


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HOW DO WE DEFINE TIME?


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SIDEREAL TIME V. SOLAR TIME


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SIDEREAL TIME (ST)

Sidereal time at some location is defined to be the hour angle of the vernal equinox at that location; it has the same value as the R.A. of any body that is crossing the local meridian.

When the vernal equinox crosses the local meridian, LST = 00h00m00.0s .

An object will be highest in the sky when the ST is equal to the object’s R.A.


Office of Naval Research

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https://www.onr.navy.mil/

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WHY DO WE CARE ABOUT (L)ST?

RA - LST = HA

Hour angle (HA): object’s position relative to observer 0h occurs when the object is at its highest

position in the sky

At a given (L)ST, the sky will always look the same.

So, if you know the R.A. of the object, and you know the (L)ST, then you can figure out where to look in the sky for the object! (Our telescope does this math for us.)


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the sun, blackbodies, & earth’s moon problem set #1 due

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