E.1.1 Outline the general structure of the solar system. Students should know that the planets orbit the Sun in ellipses and moons orbit planets. (Details of Kepler’s laws are not required.) Students should also know the names of the planets, their approximate comparative sizes and comparative distances from the Sun, the nature of comets, and the nature and position of the asteroid belt.
http://solarsystem.nasa.gov/planets Tons of interesting stuff about our solar system
Aug. 2006 the International Astronomical Union declared the official definition of a planet: A “planet” is a celestial body that: a) is in orbit around the Sun b) has sufficient mass for its self-gravity to overcome rigid body forces so that is assumes a hydrostatic equilibrium (nearly round) shape c) has cleared the neighborhood around its orbit.
8 Planets – name them…. Elliptical orbits – to have a circular orbit an object must have a very specific velocity. Any variations create and elliptical or hyperbolic shape Moons Period - 27.3 days Which planets have them????
Asteroids belt between Mars and Jupiter size – dust to hundreds kilometeres. Comets Similar to asteroids but made up of loose particles of ice and rock. Tail is blown off by solar winds and melted by radiation. Some orbit, others only pass the sun once Planetoids??? Pluto….
Light year (ly) – the distance that a beam of light will travel in one year. How far is that? (3 x 10 8 m/s = c) Used to measure distances outside of our solar system Other important units. Astronomical unit (AU) – the average distance between the Sun and Earth 1AU = 1.5 x 10 11 m Used to measure distances inside our solar system Parsec (pc) – 1parsec = 3.26 ly Defined by making a triangle between the Earth, the Sun and a distant object. If the angle at the distant object is 1 arcsec then it would be 1 parsec away. (more later)
E.1.2 Distinguish between a stellar cluster and a constellation E.1.4 Compare the relative distances between stars within a galaxy and between galaxies, in terms of order of magnitude.
Stars are not evenly distributed. Stellar cluster – small groups of stars that gravitationally interact with one another. Physically close to each other Closest star, besides the sun is Proxima Centauri - 4.25ly
Galaxy – a very large number of stars bound together by gravity Trillions of stars 10 3 – 10 5 light years across Each star is approx. 1 ly apart Andromeda is about 2.5x10 6 ly away Galaxy cluster – small group of galaxies that gravitationally interact with one another There are about 20 other galaxies we are clustered with. Supercluster – bigger than a cluster
The simplest explanation is that if all the gas is made into stars before the gas has time to form a disk, then you get an elliptical galaxy. if the gas has time to stabalize into a disk before it is all used up, then you get a spiral galaxy. Or perhaps some of the elliptical galaxies are made from merging of other types of galaxies. Observations of distant galaxies indicates that spiral galaxies were more common in the past than they are today. So maybe yesterday's spirals are todays ellipticals. This is an active research area. One problem is that if most of the mass in galaxies is unaccounted for, we have a hard time understanding the dynamics of galaxy formation.
Constellations – groups of stars that are “linked” visually Ancient civilizations played “connect the dots” Located in the same general direction from Earth Not necessarily close to each other 88 total Different ones are visible at different times during the year.
E.1.5 Describe the apparent motion of the stars/constellations over a period of a night and over a period of a year, and explain these observations in terms of the rotation and revolution of the Earth. This is the basic background for stellar parallax. Other observations, for example, seasons and the motion of planets, are not expected
Why do the stars move through the night sky? Because the rotation of the Earth It also matters where you are located on Earth Ex. North Pole http://www.yorku.ca/ns1745b/figs-ch1.html This rotation takes 23h and 56min every time. The effect is that it seams that the stars position at 12:00 changes each night. This means that the Earth rotates 360º in 23h and 56min. Which means 4min it will rotate 1º. Which means it only takes 360 DAYS for the constellations to make one compete rotation.
The Sun doesn’t make the same path through sky every day. For us, the summers are high in the sky, winters are low on the horizon. This is because the axis of rotation for the Earth and the axis in which we orbit around the sun aren’t the same angle.
The Earth is not a perfect sphere. This means that depending on it’s location in it’s orbit, it will feel more or less pull from the Sun This pulls on the Earths axis of rotation and makes it wooble. This is technically called presession. This means that the “North Star” won’t always be the north star. Period – 26,000 years
The word planet comes from the Greek word for wanderer. Planets will shift back and forth in the night sky relative to the constellation background. Apparent east/west motion comes from the Earth’s orbit around the sun. See Diagram on board Apparent north/south motion come from the other planet’s orbital plane being at a different angle from ours.
E.2.1 State that fusion is the main energy source of stars Students should know that the basic process is one in which hydrogen is converted into helium. They do not need to know about the fusion of elements with higher proton numbers. E.2.2 Explain that, in a stable star (for example, our Sun), there is an equilibrium between radiation pressure and gravitational pressure.
How does our sun work? Fusion of hydrogen into helium that provides the energy, for our sun Happens on the inside of the sun (Yes, there are different layers) Produces neutrinos that leave the sun and travel to Earth
This is the same reaction discussed in Topic 7. Each complete chain reaction produces 26.7MeV.
Remember you need 4 H to end up with one He See simplified equation:
Gravity pulls inward So much the sun should collapse. Nuclear explosions push outward These two have to balance out to be at pressure equilibrium Ex. Balloon. Rubber is like gravity Air is like the explosions If the temp changes the inside pressure will change and won’t be stable
E.2.3 Define the luminosity of a star. E.2.4 Define apparent brightness and state how it is measured.
Light measurements give us information about the temperature, size and chemical composition of a star. Luminosity(L) is the total amount of energy emitted by the star per second. Unit is watt (same as power) Depends on the temp. Ex. Two stars have same temp, the bigger one will give out more energy Sun’s luminosity of 3.839 x 10 26 W
Some stars appear brighter than others. Brightness depends on: How much energy is radiated (luminosity) How far away it is located Apparent brightness is the amount of energy per second received per unit area. Unit is W/m 2 b = (L) / 4πd 2 d is distant to the star
E.2.5 Apply the Stefan–Boltzmann law to compare the luminosities of different stars. E.2.6 State Wien’s (displacement) law and apply it to explain the connection between the color and temperature of stars.
Black bodies absorbs all wavelengths of light and reflects none. It also is a perfect emitter of radiation. If temp is increased the energy available is increased. Means the electrons can gain more energy and move into higher energy levels Means more photons released, and their average energy is greater. E = hf, Higher energy means higher frequency/shorter wavelength
Each peek represents the intensity(apparent brightness) of radiation at different wavelengths. Total intensity is the area under the curve. Power per unit area = σ T 4 σ = 5.6 x 10 -8 W/m 2 K 4 (Stefan-Boltzmann constant)
If a star has a surface area A and temperature T then the total power emitted (luminosity), L is given by: L = σAT 4
At the temperature increases, the peak wavelength is shorter Relationship between peak wavelength and temp is Wien displacement law: λ max = (2.90 x 10 -3 km) / T
Example The maximum in the black body spectrum of the light emitted from the sun is at 480 nm. Given that the Sun’s radius is 7.0 x 10 8 m, calculate the temperature of the sun, the power emitted per square meter, and the luminosity. Answers: 6000K, 7.3 x 10 7 W/m 2, 4.5 x 10 26 W
E.2.7 Explain how atomic spectra may be used to deduce chemical and physical data for stars. Students must have a qualitative appreciation of the Doppler effect as applied to light, including the terms red-shift and blue-shift. E.2.8 Describe the overall classification system of spectral classes. Students need to refer only to the principal spectra classes (OBAFGKM).
Remember: Electrons only exist in certain energy levels When excited only produce specific wavelengths. (Emission Spectrum) When white light passes through same gas these wavelengths are absorbed. (Absorption spectrum)
Stars emit a continuous spectrum of EM Peak intensity depends on the temp. As this EM pass through the outer layer of the star, some is absorbed. The absorption spectrum of a star tells us what elements are present because of the missing lines.
The absorption spectra also helps us to calculate the temperature of the gas. Hot gas Most electrons are already in higher energy levels Meaning they can’t make the biggest jump See “Energy Levels” diagram Means the higher energy photons will not be absorbed. Means a weak absorption line Which can let us find the temp
E.2.8 Describe the overall classification system of spectral classes. Students need to refer only to the principal spectra classes (OBAFGKM).
The spectrum of a star is related to it’s temp and chemical composition. Also the color. The peak points to it’s color Oh Be A Fine Girl Kiss Me ClassTemperatrueColor O30k - 60kBlue B10k - 30kBlue-White A7.5k - 10kWhite F6k - 7.5kYellow-White G5k - 6kYellow K3.5k - 5kOrange M2k - 3kRed
As objects move, the wave lengths they produce is either pushed together or spread apart. Called doppler effect. Applies to all waves including light from stars. Red shift – longer λ – star moving away Blue shift – shorter λ – star moving closer
E.2.11 - Identify the general regions of star types on a Hertzsprung–Russell (HR) diagram. Main sequence, red giant, red supergiant, white dwarf and Cepheid stars should be shown, with scales of luminosity and/or absolute magnitude, spectral class and/or surface temperature indicated. Students should be aware that the scale is not linear. Students should know that the mass of main sequence stars is dependent on position on the HR diagram.
E.2.9 - Describe the different types of star. Students need to refer only to single and binary stars, Cepheids, red giants, red supergiants and white dwarfs. Knowledge of different types of Cepheids is not required. E.2.10 - Discuss the characteristics of spectroscopic and eclipsing binary stars.
E.3.9 - State that the luminosity of a star may be estimated from its spectrum E.3.10 - Explain how stellar distance may be determined using apparent brightness and luminosity
E.3.11 - State that the method of spectroscopic parallax is limited to measuring stellar distances less than about 10 Mpc. E.3.12 - Solve problems involving stellar distances, apparent brightness and luminosity
Limited to 0.01 arc sec on Earth due to bending of light in atmosphere. – 100pc Using satellite telescopes, limited to about 0.001 arc sec – 1000pc
Measure the spectra of a star. Determine the strongest wavelength (see graph) Use Wien’s law to calculate temp λ max = (2.90 x 10 -3 ) / T
Assuming it’s in the main sequence, use HR diagram to find its luminosity. Measure the apparent brightness. Using luminosity(L) and apparent brightness(b) we can solve for distance(d) b = L / (4πd 2 )
The maximum wavelength of a distant star is measured to be 600nm and its apparent trightness is 1.0 x 10 -12 Wm -2. What is it’s distance from the Earth? First Wien’s Law for temp: λ max = (2.90 x 10 -3 ) / T T = (2.90 x 10 -3 ) / (600 x 10 -9 ) T = 4800K
Next use the HR diagram to determine it’s luminosity is about 1 L(special dot **ask**) Remember 1L = 3.84 x 10 26 W Use apparent brightness formula to solve for distance b = L / (4πd 2 ) Answer: 584ly
Exercise 15 Hamper The spectrum of a main sequence star has a maximu intensity at 400nm and an apparent brightness of 0.5 x 10 - 12 Wm -2. Use Wien’s law to find the temp. Use the HR diagram to find the luminosity Calculate the distance from the star to the Earth.
E.3.13 - Outline the nature of a Cepheid variable Students should know that a Cepheid variable is a star in which the outer layers undergo a periodic expansion and contraction, which produces a periodic variation in its luminosity. E.3.14 - State the relationship between period and absolute magnitude for Cepheid variables.
E.3.15 - Explain how Cepheid variables may be used as “standard candles”. It is sufficient for students to know that, if a Cepheid variable is located in a particular galaxy, then the distance to the galaxy may be determined. E.3.16 - Determine the distance to a Cepheid variable using the luminosity–period relationship
Stars further than 10Mpc are to weak to determine their temp. Cepheid variable is unstable. Expands and contracts on a regular period. Time vs Apparent brightness graph
There are many Cepheid variable close enough to the Earth for us to use the stellar parallax method to find their distance. If we measure their apparent brightness (b) we can then calculate their luminosity (L) b = (L) / (4πd 2 )
If the luminosity and period are plotted on a graph, we find that they are directly related. Notice this is only evident on a logarithmic scale.
Using the graph below and knowing that the period of a Cephied variable is 10 days and its brightness is 1.0 x 10 - 10 Wm -2, calculate how far is it from the Earth? Answer: 3.0 x 10 19 m or 3200 ly
A Cepeid variable has period 20 days and brightness of 8.0 x 10 -10 Wm -2. Calculate a) its luminosity b) its distance from the Earth.
E.4.1 - Describe Newton’s model of the universe. Students should know that Newton assumed an infinite (in space and time), uniform and static universe. E.4.2 - Explain Olbers’ paradox. Students should be able to show quantitatively, using the inverse square law of luminosity, that Newton’s model of the universe leads to a sky that should never be dark.
Infinitely big Gravitational force on each star is the same, giving static equilibrium Stars don’t move Stars haven’t ever moved Stars have always been Universe must be uniform, if not then there would be movement of stars. (And we all know that’s true!! LOL!!)
Infinitely big, infinitely old, static and uniform
If universe is infinite, then must be infinite number of stars, then why is the sky dark? Answer: the universe is not infinite. Also time is not infinite, meaning that som of the light just hasn’t reached us yet. See Minute Physics Vdeo https://www.youtube.com/watch?v=gxJ4M7tyLRE
Light from distance galaxies is red-shifted. Remember that a shift in spectral lines can tell use if a star is moving toward or away. Red-shifted implies that the galaxies are receding. Galaxies that are the furthest away are shifted more than the close ones. Implying that the expansion is accelerating. It’s like a big explosion. The outer parts flew off the fastest.
What was there before the Big Bang? What is it expanding into? Length and time are created to define objects and events. With out a universe, these questions can’t be answers. There was no “space” to exist in. Don’t know when because there was no time.
Where was the big bang? Was there a center? https://www.youtube.com/watch?v=W4c-gX9MT1Q Are we expanding with the universe? https://www.youtube.com/watch?v=th_9ZR2I0_w
Time and space grew out of the Big Bang. It didn’t expand into a void because there was not void. Don’t think expanding, spreading into nothingness. Do think growing This also helps to explain why everything is red-shifted.
Calculating red shift (∆λ / λ ) = (v / c) ∆λ = change in wavelength λ = original wavelength v = relative velocity c = speed of light