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I think what you have established here is just that $\rho$ tends to increase with mass. The density of planets isn't constant. Let $\rho = \rho_0 (M/M_{earth})^{\alpha}$, so that $M = (4/3)\pi R^{3} \rho_0 (M/M_{earth})^{\alpha}$ Then $$g = \frac{GM}{R^2} = \frac{4\pi G}{3} R \rho$$ Replace $R$ with $(3M/4\pi \rho)^{1/3}$ so that $$ g = \frac{4\pi G}{3} ...


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Data from observatory archives is a good way to go. Here is another one with tons of imaging datasets: http://archive.eso.org


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For planets of constant mean density you have: $$ M=\rho \times 4\pi r^3 $$ and the surface value of $g$ is: $$ g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r $$ So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ ...


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In addition to the capture mechanisms mentioned by Andy, you also need to take into account the stability of any orbit due to perturbations from tidal effects. E.g. in case of our Moon, there are no stable orbits possible. Every satellite put in orbit around the Moon had had to implement course corrections to prevent it from prematurely crashing into the ...


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This is due to the mechanics of capture. For one object (moon) to be captured by a another (planet), some energy has to be removed from the system. If the incoming moon has an existing satellite then it would be ejected, carrying a lot of kinetic energy. If a small body were to be captured by the planet/moon combination, it would usually be captured by ...


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It's pretty easy to create a dataset for yourself using remotely controlled telescopes. I've used iTelescope in the past and had a good experience. Their "starter" account costs $20 and should be good to get you going.


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Try the NASA data catalog, https://data.nasa.gov/data. Also Try the Keck Observatory Archive, KOA, http://nexsci.caltech.edu/archives/koa/index.shtml. Look for raw images. of clusters or galaxies.


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You could use archival data, e.g., from the WISE mission: http://irsa.ipac.caltech.edu/applications/wise/ The data are reduced as well as photometrically and astrometrically calibrated, which should make the image combination a lot easier. Just type in random coordinates and download data for a field with the desired number of frames. You can also try ...


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Read the fine print under the Horizons output when you select an OBSERVER table. It says (emphasis mine) R.A._(ICRF/J2000.0)_DEC = J2000.0 astrometric right ascension and declination of target center. Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f) You could do the same, but it will take some extra work. The Mars you see ...


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The 7 days of the week certainly fit somewhat as a quarter of a month. But there was another argument for the ancients to use the number seven, because they knew about seven "planets", seven objects which wander across the sky relative to the thousands of fixed stars. The Sun, the Moon, Mercury, Venus, Mars, Jupiter, Saturn. Earth wasn't a planet according ...


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The synodic period of the moon is $29.53$ days, a little shorter than a calendar month, which is on average about $30.4$ days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision. ...


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The Kaggle galaxy zoo challenge is an example of a problem begging for ideas from outside the field. Sander Dieleman, with a background in deep learning and feature learning, bravely stepped forward, creating an image classifier utilising convolutional neural networks; his full solution is described fluently here. These kinds of techniques could be applied ...


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If you have good knowledge of software development and pattern recognition, there are several problems that you could assist in solving. Much of observational astronomy requires long time series data, and removing the noise from this data. I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques ...


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I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star? The answer is, you can sort of estimate this, by using something called the Rossiter-McLaughlin effect (see also Rossiter 1924; McLaughlin 1924). You can find plenty of information on the web - I'll add a couple of links when I have a moment - ...


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Is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star? It's unclear what precisely you're asking. The planet will be orbiting (to very good approximation) on a Keplerian orbit (closed ellipse co-focal with the parent star). Such an orbit has two relevant angles. One is the angle between the orbital plane and the ...


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It depends a bit on how precise you would want to be. A very good discussion on how to calculate the orbits of solar system objects is given in the book by Jean Meeus, Astronomical Algorithms (1999), which is at an advanced amateur level. At professional level you have the Explanatory Supplement to the Astronomical Almanac by Urban and Siedelmann. For ...


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Look up Gaia hypothesis, James Lovelock, and Lynn Margulis.



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