Notes

Great job! Atoms are only allowed to absorb specific wavelengths of light, not all of them. This results in "gaps" in the spectrum. All other wavelengths are unaffected. You can think of this as the atom's "shadow". By looking at this shadow, or absorption spectrum, we can identify which atom is causing the shadows.
Good job! Electrons, the negatively charged particle, are storing the missing photon. Electrons can occupy a variety of energy levels. When they are storing a photon, they will jump to a higher energy level. How high? Depends on the energy of the photon that is absorbed.
box-black.jpgbox-black.jpgelectronlow.gifelectronhigh.gif
Violet = high energy
Electron = high orbital
Red = low energy
Electron = low orbital

Electrons in atoms can absorb and emit light. The configuration of the atom determines exactly which energy levels electrons are allowed to occupy and, hence, which energies of light (colors) they are allowed to absorb and emit.

We observe this microscopic behavior as absorption ...

A rainbow with four black lines indicating missing colors (two violet, one cyan, one red), tagged to the star.
... and emission spectra.

A black band with four colored lines (two violet, one cyan, one red), tagged to interstellar space.
Because each element has a unique configuration of electrons, each element will create its own unique absorption/emission spectrum.
This "barcode" allows us to identify which elements are present!

RaINBOW Hydrogen absorption lines indicate hydrogen in this star's atmosphere
BLACK BAR WITH LINES Hydrogen emission lines indicate energized hydrogen in interstellar space.

Recall that a variety of elements are required for constructing technologically advanced habitable worlds.

Rocks
Aluminum, Magnesium, Oxygen, Carbon, Calcium, Silicon, Sodium, Potassium, Iron ...
Plate Tectonics
Iron, Uranium (radioactive), Potassium (radioactive) ...
Atmosphere
Carbon, Oxygen, Nitrogen, Argon, Neon, Hydrogen ...
Oceans
Hydrogen, Oxygen, Sodium, Chlorine, Magnesium, Sulfur, Calcium, Potassium ...
Life
Carbon, Hydrogen, Nitrogen, Oxygen, Phosphorus, Sulfur, Iron, Magnesium ...
Technology
Aluminum, Copper, Lithium, Gold, Plutonium, Uranium, Palladium, Silver, Platinum, Nickel, Iron ..

Good job. Metallicity is a simple comparison of iron to hydrogen in a star. Whether this ratio is higher or lower than that same ratio in the Sun (defined as 0) tells us whether that star is element-poor or element-rich. Our exploration of exoplanet systems has revealed that planets (especially large ones) are more likely to be found around element-rich stars.
MINIMUM METALLICITY
Z~-0.52
That's correct! Planets are not impossible to form around low-metallicity stars ... they're simply less likely to form around stars with a metallicity less than -0.5.
All of the following stars can host planets
Red Dwarfs
Sun-like Stars
Blue Giants
Observations suggest that the current star formation rate in the Milky Way is about 7 stars per year. However, estimates range from 1 to 10.

There are also some observations that suggest that the star formation rate may not be uniform through time or space. Star formation rates may be higher in parts of the galaxy where there is a higher concentration of gas and dust, like in the arms or core. Star formation may have also been higher early in the galaxy's history because of a high abundance of gas and dust in a newly formed galaxy.
ne way that we can rotate a system is to change its inclination, or how much it is tilted towards or away from us.

An animation showing the star-planet system at 0° where the orbit appears as a circle around the star and the planet is located at the bottom of this circle, transitioning to a 90° view where the orbit of the planet appears as a horizontal line and the planet appears directly in front of the star, and finally to a 180° view which is identical to the 0° view, but the planet is now located at the top of the orbit circle.
TRANSITS
For example, if the planet passes in front of its host star and blocks some of the light, the star will appear to dim.
Good job! We can observe the whole transit or even just part of it as long as the inclination of the system is between about 88 and 92 degrees. Tip the system any more than that, and the planet's orbit does not pass in front of its star, and we would not observe any changes in brightness as the planet orbits its host star
We describe a system's inclination using angles, ranging from 0° to 180°.
Good job! You discovered that an almost completely edge-on inclination (90°) is necessary to see a transit. But there's a bit of a wiggle room. You can tilt the system a few degrees in either direction and still see a transit as the planet nicks the edge of the star.

What fraction of system inclinations results in a visible transit for each type of star-planet system?

Planet Close to Star

(92.5° - 88°) / 180° =0.025

Planet Far from Star

(90.5° - 89.6°) / 180° =
0.005
Fantastic! What we have discovered is that for large planets close to their stars, only 2.2% of all system inclinations will ever result in a transit. For planets at Earth-like distances, this value drops to well below even half a percent. This means that the vast majority of planets will never transit their host stars! Using transit photometry, we can only ever observe the existence of a very tiny percentage of all planets out there.

EX)1
If only 5% (0.05) of all planet transits are visible, and we observe 100 in a field of 100,000 stars, how many planets are out there?
0.05 (5%) = 100 planets
1 (100%) = ? planets
100 / 0.05 = ? / 1
100 / 0.05 = ?
2000 = ?
There are 2,000 planets in our field of stars.

100,000 stars = 2,000 planets

100,000,000,000 stars = ? planets
2,000 planets / 100,000 stars = 0.02 planets per star
100 billion stars x 0.02 = 2 billion planets
There are 2 billion planets in our galaxy.
EX)2
Only a small percentage of alien solar systems are optimally aligned for us to see a transit. But because we know this, we can still estimate how many alien worlds are actually out there!

A field of 100,000 stars has been observed for transits. The following observations have been made:


Planets Observed

445


Fraction of Transits Visible

(assuming random orientation)

0.02

Calculate how many planets are present in this field of view and extrapolate to the entire galaxy (100 billion stars).

Number of Planets

Field of View
22250
Galaxy
22250000000
Great job! Assuming that each system has an equally probable random orientation, and knowing that only 2 percent (0.02) of all orientations are visible as transits, we can calculate that we have tens of thousands of planets in our field of view. Assuming again that our field of view is representative of the rest of the galaxy, then we can estimate billions of planets in our galaxy!
But wait. That's a lot of assumptions isn't it? What if these assumptions aren't correct?

We know, of course, that stars have various brightnesses. The brightness we observe is dependent on the star's luminosity and its distance from us.
Good job. Remember that brightness is related to the square of the distance. So if the star is twice as far away, its brightness would be reduced by 1/22 = 1/4 = 0.25 times
Because brightness can vary so much, we normalize star brightnesses to 100. That means we set the "normal" brightness of the star (whether 0.01x or 1000x that of our Sun) to 100 and scale everything else accordingly. This makes it easier to compare stars to each other.

Here is a sample light curve. The number of days (time) observed is on the x-axis. The normalized flux is on the y-axis. The line is noisy because a star's luminosity naturally varies due to sunspots and solar flares.

Days on x-axis (0-400), normalized flux on y-axis (99.9990 to 100.0002). Light curve dips from 100 to 99.9991 at days 20, 120, 220, 320.21
Here was can see a brightness "dip". When the planet passed in front of this star, it blocked part of the light from that star.


Normally, 100% of the star's light reaches us. Estimate how much light reached us during the transit.

99.9991% of normal
Good work. If we read off the graph, the brightness dip falls exactly between 99.9992% and 99.9990%. That means that the star appeared only 99.9991% as bright as when no planet is blocking the light.
Therefore, how much light was blocked by the planet?

0.0009% blocked
That's correct. If 99.9991% of the light of the star reached us, then 100% - 99.9991% = 0.0009% of the light was blocked by the planet.

Brightness drop: 0.1%

Star radius: 2.4 RS


To compare to the size of Earth:

RE = (0.1% / 100%)1/2 x (2.4 RS) x 109 = 8.27 RE


To compare to the size of Jupiter:

RJ = 8.27 RE x 0.0892 = 0.738 RJ
The square root is the same as raising something to the 1/2 or 0.5 power.

Good job! Our observations seem to indicate that the mathematical relationship between period and distance is as follows:
period = distance1.5
Another way of putting it is like this:
period = distance3/2
period2 = distance3
This relationship is known as Kepler's 3rd Law. Kepler's laws mathematically explain the properties of orbits. The third one show the relationship between how far the planet is from the star and the amount of time it takes for the planet to complete an orbit.
So if we know the amount of time it takes to complete an orbit (time between transits) we can very easily calculate how far away the planet is from its star.
You have discovered the period for a planet is 45 days. How far is it from its star, which has a mass of 3.7 solar masses (MS)?
Days to Years:


45 days to years


(45 days) / (365 days/year) = 0.123 years


Period to Distance:


period2 x star mass = distance3


(0.123 years)2 x 3.7 MS = distance3

0.0562 = distance3

(0.0562)1/3 = (distance3)1/3

0.383 AU = distance