43 votes

Are the stars distributed in uniform distribution, on the celestial dome, with respect to brightness?

Naked eye stars are not distributed uniformly in the sky. That is because the median naked eye star is at a distance of 440 light years, and this is far enough away that some of the details of ...
ProfRob's user avatar
  • 146k
21 votes

Why does the Fourier transform of this CMB image have a hole in it?

Having now looked at the paper by Aiola et al. (2020), it emerges that for that map, they filtered the data to exclude low frequency multipoles with $|l|<150$, corresponding to about 1 degree. This ...
ProfRob's user avatar
  • 146k
20 votes
Accepted

At what distance from the Sun can planetary moons exist?

There are several factors determining the inner limit to moons. Perhaps the simplest is that it needs to stay inside the Hill sphere, the region around the planet where the planet's gravity dominates ...
Anders Sandberg's user avatar
19 votes
Accepted

Why does the Fourier transform of this CMB image have a hole in it?

For that specific E-mode map we have applied a Wiener filter to highlight the high SN modes (those "rings"). I also further apply the following filter: $((1 + (kx/5)^{-4})^{-1}) * ((1 + (k/...
Simone Aiola's user avatar
18 votes
Accepted

What is the difference between the two terms named "Eccentricity" and "Ellipticity"?

Both ellipticity $f$ (also called flattening) and eccentricity $e$ are measures of how elongated an ellipse is, based on the semi-major axis $a$ and the semi-minor axis $b$ (figure from wikipedia). ...
Connor Garcia's user avatar
  • 16.2k
16 votes

14,000 square degrees

The whole sphere has approximately 41,253 square degrees of solid angle. $$4\pi\left(\frac{180}{\pi}\right)^{2}\approx 41,253$$ so for a hemisphere there should be half this number or about 20,627 ...
GrapefruitIsAwesome's user avatar
14 votes
Accepted

Why does the Earth's orbital eccentricity oscillate with a period of about 100,000 years?

Saying why gets tricky beyond "because of Jupiter", but to clarify on the quote, the statement "Earth's eccentricity follows a 100,000 year cycle" is loosely true but it's also an oversimplification. ...
userLTK's user avatar
  • 23.9k
14 votes

Why can't we use the semi-minor axis in Kepler's third law?

The relationship between $a$, $b$ and the eccentricity of an orbit $e$ is $$ a = b\left(\frac{1+e}{1-e}\right)\ .$$ It is clear therefore from a mathematical point of view that you cannot get a ...
ProfRob's user avatar
  • 146k
13 votes
Accepted

Role of power laws in astronomy?

Scale invariance and self-similarity Power laws basically mean that there is no preferred scale, i.e. that a physical property is scale invariant. Any deviation from a power law means that the ...
pela's user avatar
  • 37.1k
13 votes

What if Earth gained 1 km/s orbital velocity?

I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum. This is a relatively small ...
ProfRob's user avatar
  • 146k
13 votes
Accepted

N-body simulation still losing precision despite using arbitrary-precision arithmetic and symplectic integrator

There are various possible causes for an orbit simulator to behave poorly, even when using an excellent arbitrary precision package like mpmath. When supplying numerical values, you need to pass them ...
PM 2Ring's user avatar
  • 13.1k
13 votes

Confusion over habitable zone

The reason you get a too cold temperature for Earth is of course because you ignored the greenhouse effect. The atmospheric composition strongly affects the actual temperature of planets, which ...
Anders Sandberg's user avatar
11 votes

Are the stars distributed in uniform distribution, on the celestial dome, with respect to brightness?

Alright I finally finished this program so I could take a look at each tier individually and see for myself. First of all, the projection type does indeed matter, so I will explain it here. It needs ...
DrZ214's user avatar
  • 1,940
10 votes

What is the difference between the two terms named "Eccentricity" and "Ellipticity"?

Ellipses have a "long radius" called the "semi-major-axis" which is the length from the centre to the ellipse measured along the long axis. And a "semi-minor-axis" which ...
James K's user avatar
  • 116k
10 votes
Accepted

At what point are orbital resonances no longer "ordered" but "chaotic?"

Consider a child on a stationary swing. The fastest way to get them going is to push once every time they swing (a 1:1 resonance). If you push 581 times for every 137 swings, the pushes will mostly ...
Connor Garcia's user avatar
  • 16.2k
9 votes
Accepted

How to mathematically describe a planet's eccentric orbit?

The formula is Kepler's equation, but to understand it you need to know three values: $M$ is the "Mean Anomaly". It increases linearly from 0 to 360 over the period of one orbit, measured from ...
James K's user avatar
  • 116k
9 votes
Accepted

What is the RGB curve for blackbodies?

Ok, here's my take on calculating the color of a blackbody, or any spectrum in fact: Disclaimer: I'm not a color theorist, and there may be more accurate methods. But the result, shown in the bottom, ...
pela's user avatar
  • 37.1k
9 votes

Why can't we use the semi-minor axis in Kepler's third law?

Consider this gif from wikipedia. All the orbits in the animation have the same orbital period $T$ and the same semi-major axis $a$, but different semi-minor axes $b_1,b_2...b_5$. This shows that ...
Connor Garcia's user avatar
  • 16.2k
9 votes
Accepted

Why is there a deviation between the ratio of $a^3$ and $T^2$ for the outer planets?

Kepler's third law is not that $a^3/T^2$ is a constant. It is that $$ \frac{a^3}{T^2} \propto (M_\odot + M_{\rm planet}) $$ and so the left hand side depends on which planet you are looking at and ...
ProfRob's user avatar
  • 146k
9 votes
Accepted

What kind of perspective projection is shown in this image?

The iphone camera, in common with most other lenses, produces a rectilinear projection - also known as a gnomonic or perspective projection. The main feature of this projection is that straight lines (...
James K's user avatar
  • 116k
8 votes

Role of power laws in astronomy?

I have to admit that power-laws (in general) used to be my shtick so I am happy to shed some light on their general importance in physics which obviously also hold for astronomy. The main idea of a ...
B--rian's user avatar
  • 5,526
8 votes
Accepted

14,000 square degrees

GrapefruitIsAwesome has already explained why the sky is significantly larger than 3300 square degrees; I'd like to explain why the sky coverage is precisely the value it is. The wording is admittedly ...
HDE 226868's user avatar
  • 36k
7 votes

Generate a uniform distribution on the sky

Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you ...
ProfRob's user avatar
  • 146k
7 votes

Open source code for the maths behind a heliostat?

I wrote the source you need some years ago: https://jumpjack.wixsite.com/progetti/sorgenti-ipsun The Arduino/Processing version was just a demo program to manually control a TENVIS camera by multiple ...
jumpjack's user avatar
  • 543
7 votes
Accepted

Is it true that the 3 body problem can't be solved using the four basic functions, radicals, and integrals?

In a sense, even solving the two body problem as a function of time is unsolvable in terms of the elementary functions. The problem is that the solution involves the solving for the inverse of Kepler'...
David Hammen's user avatar
  • 33.5k
6 votes
Accepted

Is there a rough analytical expression for the Milky Way's radial mass distribution?

For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane? The simplest ...
zephyr's user avatar
  • 15k
6 votes
Accepted

Total apparent magnitude of eclipsing binary system

Without information about stellar radii, I think it's reasonable to assume $R_A \approx R_B$. Then your equation becomes $$ m_p - m_s = -2.5 \log \frac{F_A}{F_B} $$ and you can compute $k$ and the ...
Mike G's user avatar
  • 18.2k
6 votes
Accepted

Do Astronomers really never call Lagrangian points "libration points"?

A full-text search using the astronomical-literature collection on the Astrophysics Data System (more complete and more astronomy-specific than Google Scholar) yields 10,845 hits for "Lagrange point" ...
Peter Erwin's user avatar
  • 16.4k

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