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Outside Earth's atmosphere, the Solar flux is S0 = 1370 S 0 = 1370 Watts per square meter, for a surface placed perpendicular to the Sun rays. Since Earth is 1 AU from the Sun, the received flux for a planet at a distance of x x AU from the Sun is then given by. Fplanet = S0 ∗(1 x)2 F p l a n e t = S 0 ∗ (1 x) 2.
As far as I see, the separation between Earth and Pluto this year is always bigger than 34.045586 AU (5,093,147,172 km). This minimum separation is reached on 20th July 2024 around 21h UTC checked to about 1 hour accuracy. In the graph the horizontal axis reads "days from today" (21 May 2024), the vertical "distance in AU", the time in the ...
Lookback time also describes the distance the quasar's light travelled through an expanding universe before reaching the observer. This lookback time distance, D, determines by how much the object's light intensity falls off due to distance and due to intergalactic medium extinction. D = C x T, where C = 1, describes D in light years,
5. 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. pconjunction = 1 1 p1 − 1 p2 p c o n j u n c t i o n = 1 1 p 1 − 1 p 2. where p1 and p2 are the orbital time periods of two planets respectively. The orbital time periods of Uranus ...
So, should be fine. From Pluto, the Sun is point-like to human eyes, but still very bright, giving roughly 150 to 450 times the light of the full Moon from Earth (the variability being due to the fact that Pluto's orbit is highly elliptical, stretching from just 4.4 billion km to over 7.3 billion km from the Sun). https://www.discovermagazine ...
At these times L = mvr L = m v r. For Pluto the periapsis speed is v = 6.10km/s v = 6.10 k m / s the distance is 4.44 billion km and the mass is 1.31 ×1022 1.31 × 10 22 kg. To get the angular momentum you've got to multiply them together. If you want SI units, convert those km to m first.
Alternatively, software like Stellarium will be able to calculate the position of a planet on any given date in the near future or past. When you have the position of the planets as RA and Dec, and you have converted the units to decimal degrees, you can calculate the angle A A between them using. cos(A) = sin(Dec1) sin(Dec2) + cos(Dec1) cos ...
Another issue is time. JPL uses its own time scale, JPL ephemeris time (T eph). This is a relativistic time scale. It differs from Terrestrial Time (TT) by at most a few milliseconds, so for most purposes you can use TT in lieu of T eph in your homebrew ephemeris calculator. TT currently differs from UTC by 67.184 seconds, and beginning in July ...
(too long for a comment, this answers why barycenter data is available for a much longer time span than planet center data) When NASA computes planetary positions, they treat each planet and its moons as a single point mass (the barycenter of the planet and its moons). They compute these positions +-15000 years from now.
Now let's plug in some numbers. If you want all 8 planets to be aligned to within 1 degree of longitude, then the average time between two such alignments is roughly equal to P = 3606 = 2.2 × 1015 orbits of the fastest planet. For the Solar System, Mercury is the fastest planet, with a period of about 0.241 years, so then the average time ...