# Tag Info

13

The moon's orbit is elliptical, and it moves faster when it is closest to the Earth. This is the case for all elliptical orbits, and was discovered by Kepler, who gave a neat rule for predicting how fast an orbiting body moves. (https://kids.britannica.com/students/assembly/view/90830) For the moon, this means that when it is largest, it is also moving ...

12

It's neither the angular diameter or prallax precision that is the limiting factor, but the fact that it is difficult to get the interferometric measurements for faint stars. State-of-the-art angular diameters are measured by infrared interferometry (e.g. with the CHARA array - Gordon et al. 2019). The most precise measurements of angular diameters have an ...

9

You can calculate the angular diameter of the Earth using the equation: $$a = \arctan \frac{D}{d}$$ where $a$ is the angular diameter, $D$ is the physical diameter of the Earth, and $d$ is the distance from the Moon to the Earth. The equatorial radius of the Earth is $r_E = 6378.1 \textrm{km}$, the diameter is therefore $D= 2 \times r_E = 12756.2$. The ...

8

The plane that contains the orbit of the Earth is known as the "ecliptic". The rotation of the sun is tilted by 7.25 degrees to the ecliptic, and this value does not vary over time. The rotation of the Earth is also tilted, by 23.45 degrees to the ecliptic, it is this angle that causes seasons. An old paper Position of the Sun's axis describes how the ...

5

0.004° converts to about 14.4 arc-seconds. That's within the range typical for Mars; 5-25 arc-seconds. You'll see the sun as a point with the naked eye. Heliopause is about 14 billion miles out. Absolute magnitude of sun is 4.83. 14 billion miles = 0.0007297pc. Apparent magnitude will be -15.8543. Magnitude of the full moon, from earth, is about −12.74 So ...

5

Yes this is correct for the standard model. I believe the English language explanation would be that the angular size of a galaxy does not depend on its current distance to us, but its distance when the light we currently see from it was emitted (here using "distance" synonymous with "proper distance"). Though I also believe there are other complicating ...

5

The Earth viewed from moon will appear larger, in proportion to how much larger the Earth's diameter is versus the moons diameter. Earth diameter 7900mi Moon diameter 2100mi So the Earth-view from the moon would appear 3.75 times as large as the Moon appears in the sky on earth. I looked up the moon's typical angular diameter, it is 0.5 degree. So the ...

4

The average angular diameter of the Moon, as seen from the Earth, is about 31 arcminutes. The angular diameter depends on the distance between the two objects and the diameter of the object being viewed. Specifically, for small angles, it is the diameter divided by the distance. When the distance is the same, the angular size is proportional to the ...

3

What you need, as Aristarchus, is a frame of reference in which the Earth's shadow is stationary. Otherwise you are trying to deal with a moving thing (the Moon) relative to another moving thing (the shadow cast by the light of the Sun as it orbits the Earth). In a stationary-shadow frame of reference, the period you need is the time it takes the Moon to ...

3

While this might seem to be strictly a math problem, it's really loaded with Astronomy, let's see what we can learn! If you're far enough away that you can see essentially a full hemisphere (which you can't if you're close), the apparent angular width is twice the half-width, and that's given by $$\theta_{HW} = \arcsin\frac{r}{R} \approx \frac{r}{R}$$ for ...

3

From us, the Sun is visible around a half grad, which is $\approx$ 30 arc min = 1800 arc seconds. To get to 7 arc seconds, you need to go 1800/7 times farther away, so the result is around 250 AU.

3

The (weird for historical reasons) defintion of magnitude is that a difference of 5 magnitudes corresponds to a factor of 100 in the brightness of the source. So a difference of 3-(-27) = 30 magnitudes is a difference of $10^{12}$ in brightness. The surface brightness of the mirror and the sun is effectively equal, since the mirror reflects 100% of the ...

3

The parallax is always easier to measure than the angular size of any planet. This is also true for most stars, excluding hypergiants and some supergiants. The parallax is given by $\displaystyle \theta_{p} = \frac{d_e}{d}$ where $d_e$ is the Earth-Sun distance and is $d$ the distance of the star. Instead, the angular size of an object with radius $r$ is ...

3

To find the angular diameter of a satellite you need to find $$\arctan\left(\frac{\text{diameter of satellite}}{\text{distance to satellite}}\right)$$ As you have been using the diameter of the planet, your formulae are wrong. You should also make use of the small-angle approximations $\arctan(x)\approx x$ for small x in radians. So to get the value in ...

3

Here's a drawing that may help understand (edited to take into account the fact that the Universe expands not only along the line of sight, but in all directions — thanks to @Ed Shaya): In the distant past, at $t = t_1$, you (or your ancestors, or the Milky Way seed) was close to the other galaxy, and photons were emitted from the edges in the shown ...

3

I looked for a source without luck, so I'll just post cause I've read this answer before. The diameter is the solid surface, not the atmosphere for rocky planets. For gas giants it's the atmosphere up to 1 bar. To determine Venus' angular diameter, you'd need to add something for it's atmosphere. It's difficult to say how high it's atmosphere stops ...

2

You can see from the paper you linked that they followed the procedure you outlined in option (2). Figure 4 of that paper shows the $\theta (t)$ for a few stars. They specifically state towards the end of section 6 Figure 4 shows in the top panel the relation between the angular diameter predicted by the SB-relation versus radial variation, and in the ...

2

The number 206265 arcseconds/radian is often used in astronomy for angular conversions. It is simply derived from the product of 3600 arcseconds/degree and 57.2958 degrees/radian. Edit based on symbols as defined in comment below: With the distance to an object, $d$, and its lateral dimension, $D$, and using the small angle approximation where $D \ll d$, the ...

2

You answer your question basically yourself with the calculation - that's the right way to approach it, your math is correct. Stars and planets are big enough, that simple geometric optics work well to calculate the degree of occulatation when one body passes in front of another. This is the method used to estimate the size of exoplanets pretty accurately (...

2

You're correct, except that you should multiply, not divide, by 3600 (you're converting from degrees to arcsec, not the other way round). You get $5.7\times10^{-5}{''}$, but the correct result is 36002 times higher, i.e. $738.6''\!\!\!$. But why not just let astropy do the job: z = 0.3 x = 3.29 * u.Mpc theta = x * cosmo.arcsec_per_kpc_proper(z) ...

1

Unfortunately, I couldn't find any infos on this specific case. But as I remember, there was a similar study from NASA a while ago, observed with the Hubble telescope: https://hubblesite.org/contents/news-releases/2013/news-2013-22.html#section-id-2 There is a nice animation that show the proper motion of Proxima Centauri relative to background stars (not ...

1

Angular diameter distance is the reduced circumference of the circle, centered at our location, on which the object was located when it emitted the light (or the reduced area of the sphere if you prefer). If the universe is spatially flat then this is the same as the metric radius of the circle. In general it's related to the radius by the function $S_k$ ...

1

How does a physical distance $rR_0$ comes into play in the angular diameter distance, because from its definition ... The manner you choose to calculate it is explained on Wikipedia's "Angular diameter distance" webpage: The angular diameter distance is a distance measure used in astronomy. It is defined in terms of an object's physical size, $x$, and \$\...

1

Most lunar craters are too small to resolve with the naked eye; I would measure a published photograph. A crater near the limb appears as an ellipse whose minor axis is foreshortened but whose major axis measures about the same as if the crater were centrally located. Surface curvature has little effect on this measurement because most craters are small ...

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