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37

That's right. The inclination of the orbital plane around stars is considered to be random throughout the galaxy, thus the planets we can detect by the transit method is just a tiny fraction of the planets that we should expect in our stellar neighbourhood. The transit method allows for planetary detection only when the line of sight from Earth to the ...


11

One thing to keep in mind is that the Kepler instrument is not a telescope like Hubble. It is a photometer and though it uses CCDs to look at the sky, it doesn't return a picture in the usual sense. The way it operates is that you only look at the pixels around the object you're interested in because otherwise you'd never get all those pixels transmitted ...


9

Kepler's third law is that $R^3/P^2$ is a constant. However it is not a universal constant; it depends on the mass of the body that is being orbited. $$\frac{R^3}{P^2} \simeq \frac{GM}{4\pi^2},$$ where $M$ is the mass of the orbiting body (assuming that $M\gg$ the mass of the moons). Depending on your level of sophistication, you could try plotting $R^3$ ...


8

Yes. The probability of a transit taking place is something like $r/a$, where $r$ is the stellar radius and $a$ is the radius of the planetary orbit. If you assume that planet orbits are randomly inclined to our line of sight, then each detected planet corresponds to $a/r$ planets in actual fact. Note that this approximation is ok for circular orbits ...


6

There are two separate points of interest you're looking at so I'll separate this into sections. Sudden Drop at Day 1559 As near as I can tell, this is the result of a quarterly roll of the satellite, specifically the end of quarter 17. Every 90 days, which NASA calls a quarter, the space craft does a 90 degree roll to optimize the solar panel efficiency. ...


6

I'm going to try to take a stab at answering this. With our current technologies, detecting exomoons can prove hard however there are various techniques being used today such as: Analyzing data from the Kepler Spacecraft Dynamic effects – the exomoon tugs the planet, which causes deviations in the times and durations of the host planet’s transits. This is ...


6

Changing fields breaks the timeline of the observations. If you have variation with a period similar to when you switch, you might completely miss it. Even if the periods are different, it degrades your observations and makes it harder to draw conclusions. The more uninterrupted viewing you get, the better. This mission document covers some of the ...


5

Starting from the index you mentioned, I clicked through the links for some individual planets, which in turn link to discovery papers or other relevant observations. For planets around Kepler-23, -24, -25, -26, -27, and -28, the relevant papers are Ford et al. (2012) and Steffen et al. (2012), two out of a series of papers. Both papers used transit timing ...


5

There are many planets known which have orbits longer than the longest exoplanet orbital periods found by Kepler. These planets were discovered using the "doppler wobble" or radial velocity technique. The plot below (a few months out of date now) shows many planets orbiting with similar periods to Mars and Jupiter. The red points we discovered by transits (...


5

The prime objective of the Kepler mission was to attempt to find "Earth-like" planets using the transit technique. To establish that you definitely have a transiting planet requires, at a minimum, that you see three regularly spaced transits. The Kepler mission (originally) was planned for 4 years. Thus to ensure the detection of 3 transits for planets in ...


5

Planet "candidates" are Kepler Objects of Interest (KOIs)that have a transit-like light curve and have passed a number of observational tests. They are candidates, because although they do show a transit in the light curve of the star in question, there is no independent confirmation of a planetary mass. One problem to overcome is that of "false positives"....


5

There is an observational bias and it is taken into account when you see inferences about planet frequency. The methods to find planets are inherently biased towards finding large, close-in planets. Both the transit and doppler-based methods suffer from this bias. The paper you reference takes into account this bias to arrive at the statistics you ...


5

I believe you are right, this is extracted from the "Explantory supplement to the Astronomical Almanac (2006?)" (although it might date back to the 1960s...) For example a similar PDF document gives that citation. The supplement is published in a separate binding from the main astronomical almanc and a new version is published each year. Finding ...


5

If you plot the log of the period against the log of the semi-major axis then it is obvious that $P^2 \propto a^3$. Any other power law relationship simply wouldn't fit. The following passage (from https://www.mathpages.com/rr/s8-01/8-01.htm ) seems relevant: Is it just coincidental that John Napier's "Mirifici Logarithmorum Canonis Descripto" (...


4

They don't stop. The Kepler CCDs read out the signals collected during the accumulated time of 6.02 seconds. The fixed read out time is 0.52 seconds. So each CCD gets one frame every (6.02 + 0.52) = 6.54 seconds. Then Kepler sums up every 9 frames (short cadence) and 270 frames (long cadence). The time between two short cadences is (6.54 x 9) = 58.9 seconds,...


4

That is not correct. The area is the total area between the two radius lines, so there is a curved side. Imagine you have two points almost 180 degrees from each other. Using just a triangle, the area is close to zero. Now, two points placed closely together can have the same area in between them. Then you have an equal are, but not equal time, so your ...


4

This can best be described by two slides from this presentation. "Forward-facing" implies looking towards Earth in the spacecraft's orbit, in the direction of the spacecraft's velocity vector: "Backward-facing" implies looking in the opposite direction, away from Earth and in the opposite direction to that of the spacecraft's velocity vector: These are ...


4

The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits. There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time. These give quite different results - qualitatively different. Average distance on ...


4

Well "if no known astrophysical model can explain it" then nobody told Wright & Sigurdsson (2016) who, cognisant of Montet & Simon's results, explore a number of astrophysical models. They conclude by saying that the most "plausible model" is that of small scale intervening material between us and the star that may be responsible for the short-term ...


4

Go to the Kepler/K2 MAST database https://archive.stsci.edu/kepler/data_search/search.php or https://archive.stsci.edu/k2/epic/search.php Search on a temperature ($T_{\rm eff}$) range: e.g. for K-stars 4500 .. 5200 A (big) list of objects will be found and presented to you in a table. Mark the ones you want (all the Kepler objects should have available ...


4

Just to provide an analytical formula for @uhoh's correct time-averaged distance, here the derivation of $\langle r\rangle_t=1+\epsilon^2/2$: $$a=1 \qquad c=e\qquad b=\sqrt{1-e^2}\\ \vec{r}=(\cos \beta-e,\sqrt{1-e^2}\sin \beta)\\ \vec{r'}=(-\sin \beta,\sqrt{1-e^2}\cos \beta)\\ |r|^2=\cos^2 \beta -2e\cos \beta+e^2+\sin^2\beta-e^2\sin^2\beta=1-2e\cos \beta+e^...


4

Great project! and welcome to Stack Exchange. I'll post a short answer but I think someone can add a more detailed, thorough and insightful answer. I think that website is not well suited, so I'll answer based on you switching to Horizons. If you like Python then it's more fun to use Skyfield. If you want apply an equation based on a Kepler orbit model, you'...


3

It's not proof that they've ejected other inner planets, because there are plenty of other explanations for why we haven't observed companions. Steffen et al. (2012) analyzed Kepler data - likely some of the same examples you've looked at - and came up with several explanations besides the no-companions-because-of-planet-planet scattering hypothesis: Inner ...


3

Gyrochronology uses the rotation periods of stars, caused by rotational modulation by starspots to estimate a stellar age. In the absence of differential rotation, the rotation axis inclination has no effect on this measurement. The doppler broadening of spectral lines plays no role in gyrochronology. The rotation period of a star is just that. It is the ...


3

I do not know whether complimentary observations of the light curve are being done, but I will try to answer the rest of the question. Is it easy to observe the light curve with ground-based telescopes? No, it's quite difficult. From the Kepler homepage: Since transits only last a fraction of a day, Kepler must monitor all target stars continuously. ...


3

What you are looking for is here. http://archive.stsci.edu/kepler/search_retrieve.html This one is good for exoplanet light-curves https://exo.mast.stsci.edu/


3

How can I do this by not knowing the gravitational constant? Sarcasm: You could do what Henry Cavendish did, which was to measure how the horizontal deflection rather heavy balls separated horizontally by a small distance and held in place by torsion rods. The concept of the Newtonian gravitational constant did not exist in Cavendish's time, and would not ...


3

Kepler's first law is that a planet moves in an ellipse with the sun at one focus. Your equation is that of an ellipse about the focus, so, you have proven Kepler's first law. The $\varphi$ is what astronomers call true anomaly. To put your equation in the usual form, $a/b^2$ is $1/p$ so $$ r = \frac{p}{1+\epsilon cos\varphi}$$ With this equation, the ...


2

This is a well known, well researched phenomena. Yes, there certainly is a correlation between metallicity and the likelihood of observing a hot Jupiter. There are two classes of explanation. (1) The correlation is real and due to the fact that it is easier to form planetary embryos from metal-rich material in the core-accretion model of giant planet ...


2

Hohmannfan's answer is correct, but as I understand your question, you do understand the general idea correctly and the answer to your question is yes, no matter how eccentric the orbit, the planet spans equal areas over equal time, at least nearly perfectly. (more on that later) You're mistake is in calling the sectors "triangles". They're ...


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