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do you guys think this would be able to focus tightly on any planets No. The focal length of your telescope is too short to effectively see planets. I have a William Optics GT71 and, with its focal reducer, it is approximately the same focal length as yours. I also use an APS-C format camera and when I attempt to photograph planets, they are tiny white dots....


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A planetary mass object (also callled a planemo) is an astronomical object large enough to be pulled into a roughly spherical shape by its gravity compressing its matter. A planetary mass object must also have less than about 13 times the mass of Jupiter or about 4,131.4 times the mass of Earth. If a planetary mass object orbits around the Sun in our solar ...


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They would be called exoplanets (or just planets) if large enough, small bodies if not. There's a good reason for calling them exoplanets (or just planets) if they are large enough rather than moons. Non-stellar objects orbiting a multi-star system must either be well removed from the stars that comprise the star system, in which case they would be orbiting ...


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Consider a particle with mass m, orbiting in a circle a body with mass M. The gravitational force must be the centripetal force causing circular motion, so $$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ Cancelling the $m$ and $r$ and square rooting gives: $$ \frac{\sqrt{GM}}{\sqrt{r}} = v$$ So you see the velocity is inversely proportional to the square root of ...


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The brightness of a Solar System object, seen in reflected light, depends on how far it is from the Sun, $d_s$, and how far away it is from the observer, $d_o$, (and the angles between them). Both dependencies are "inverse square laws": $${\rm brightness} \propto \left(\frac{1}{d_s^2}\right)\left(\frac{1}{d_o^2}\right)\ . $$ Both Uranus and Neptune ...


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There is nothing specifically preventing it, but there is no mechanism that can create or maintain this situation. Tidal locking occurs as a transfer of energy between rotation and orbit. This transfer of energy stops when the rotation rate of the planet equals its orbital time. So planets will tend to evolve towards this state of tidal locking. Now the ...


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If you go to Wikipedia article of conjunction, there is a formula given for the average time between two conjunctions between a planet pair in siderial years. $$\mathrm{p_{conjunction} = \frac{1}{\frac{1}{p_{1}}-\frac{1}{p_{2}}}}$$ where p1 and p2 are the orbital time periods of two planets respectively. The orbital time periods of Uranus and Neptune are 84....


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If I have calculated it right then the apparent magnitude of Neptune as seen from Uranus is approximately 4. The absolute magnitudes of Neptune (-7.11) and Uranus (-7.00) are nearly identical. The apparent magnitude of Uranus is 5.38. The distance between Neptune and Uranus is roughly half of the distance between Earth and Uranus. At half the distance a ...


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I'm going to break this up into two parts. Supernovae The effect a supernova will have on a planet is, as one would expect, dependent on a lot of factors. Like how close a planet is to the supernova if it's far away then how long will it take for the remnant to pass by it, etc. But a couple of guiding principles boil down to the following: Once the mass ...


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I note that the planet which appears to have the largest angular diameter as seen from Earth is Venus. Venus, our sister planet, comes closest to Earth, ranging from 9.7" to a whopping 66.0". https://medium.com/starts-with-a-bang/which-planet-appears-the-largest-from-earth-d95f2f72a8f8 In astronomy " is the symbol for an arc second of ...


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Here is a table with planets, their minimum/maximum distances from the Sun, their minimum distance (assuming the planet's aphelion and its neighbour's perihelion coincide, which they don't, so the actual minimum distance is a bit higher), the size of the largest of the two (indicated with a *), and finally the maximum apparent size $s$, which is given by $s =...


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Supplementary answer supporting @PierrePaquette thorough and well-source answer: I tried the nice new JPL Horizons interface and fired up Excel which I haven't used in a long time. For years 1800 to 2100 in Observer mode it calculates apparent magnitude using all the bells and whistles (albedo model, phase angle, illumination, etc.) and gives the following ...


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Possibly but "Habitable" doesn't automatically mean "Suitable for Earth-creatures" Mars will be heated significantly, enough to boil off any water. The same will happen to the moons of Jupiter, though there will be a period during which the ice on Europa melts and evaporates. Titan is a more interesting case. It would be in the habitable ...


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According to https://arxiv.org/pdf/1808.01973.pdf, the magnitude of Neptune follows the relationship (formula 17, page 25): $ V = 5 \log_{10} (rd) - 7.00 + 7.944 \times 10^{-3} α + 9.617 \times 10^{-5} α^2 $ Where r is the distance of Neptune to the Sun, d is the distance of Neptune to the observer, and α is “the arc between the Sun and the sensor with its ...


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I don't know what you mean by a descrete number of months in a year. Do you mean an integer number of months in a year? A non integer number of months in a year would be 11.319, 19.8, 35.42, 429.1082, etc. An integer number of months in a year would be 11.0000, 19.0000, 35.0000, 429.000, etc. And it is perfectly possible that as the orbit of a moon slowly ...


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