2024年2月13日 · What you can do is take the angular size for a from the table, treat it as if it were perpendicular to your line of sight, and use the small-angle approximation (look at the 'geometric' section): ##a=\theta d##, where a is the size of the semi-major axis in AUs, d is the distance to the black hole in AUs, and ##\theta## is the value for a in ...

2012年6月8日 · semi-major axis: 39.264 AU semi-minor axis: 34.031 AU orbital eccentricity: 0.249 approx orbital circumference: 230.85 AU orbital inclination is 17.142° (I'm thinking this number isn't relevant to this question) So I guess what I'm asking is, given those figures, how do I determine the "section" of the circumference that is above/below the ...

2016年12月12日 · O type star, mass ~90M☉ and a type G star with mass 1M☉. It is likely to have an orbit , described simply -- the smaller star orbits the bigger one. One semi-major axis. When companions are closely matched mass-wise a simple one-axis description does not work well. Two semi-major axes, one for each orbital component more closely matches the ...

2013年3月10日 · From geometry there are various measures of the ellipse which, in appropriate combination, are related to its eccentricity. These include the semi-major axis a, the minor axis b, and the distance between the foci c. Extrapolating to the jargon of orbits, the perihelion (periapsis) and aphelion (apoapsis) distances are also related.

13.29. The dwarf planet Pluto has an elliptical orbit with a semi- major axis of 5.91 X 102 m and eccentricity 0.249. (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, …

2009年11月23日 · I was thinking to much of your problem in term of converting cartesian position and velocity to orbital elements (six of them). The specific angular momentum and the eccentricity vector are the keys to unlocking that puzzle. However, all you wanted was the semi-major axis. The vis-viva is the key to answering that particular problem.

2011年7月1日 · Getting a semi-major axis of [itex]a = 1/2 r[/itex] is correct since the orbiting object will be "wipped around" the primary body and head out the same way it came in. If you draw the orbit it will be a line (flat ellipse) from the distance r down to the primary, with the center of the ellipse being half-way between r and the primary body.

2016年6月6日 · I'm doing my physics coursework on kepler's third law and I'm finding the minimum mass and semi-major axis of a unknown planet. I have the following data: Stellar mass Mstar = 1.31 ± 0.05 Msun Orbital period P = 2.243752 ± 0.00005 days Radial velocity semi-amplitude: V = 993.0 ± 3 m/s Inclination i = 84.32º ± 0.67 We assume eccentricity e=0

2010年2月18日 · Why is the total energy of an elliptical orbit given by: E_{tot}=\\frac{-GMm}{2a} Where a=semi major axis. I agree for a circular orbit I can do the following: F_c=F_g ma_c=\\frac{GMm}{r^2} \\frac{v^2}{r}=\\frac{GM}{r^2} v^2=\\frac{GM}{r} Since the total energy also equal to the kinetic plus...

2009年12月22日 · It's major axis would therefore be 1.6, half of that is a semi-major axis of 0.8AU. Now that this is an elliptical orbit, we can use Keplers relationship T^2 = a^3 to find the time for 1 cycle in years, half of which should be when the …