# Tag Info

## Hot answers tagged astronomy

110

The premise of this question is wrong. If the moon is in between the earth and the sun (as shown on your diagram), and you can see the moon, then it is day, not night: If on the other hand, you are on the opposite side of the earth during that configuration (so that it is night), then you can't see the moon because the earth is blocking your view of it:

50

The sun doesn't just illuminate the moon directly. The moon is also illuminated by sunlight reflected from the earth. This is called earthshine. This makes the parts of the moon that face us visible even when the sun is on the other side. According to NASA, it was Leonardo da Vinci who first explained this. As an example, the brightly lit portion of ...

23

While excellent answers have already been provided (yes, it's Earthshine; yes, when the Moon is between the Sun and the Earth, you don't see the Moon at night, you see it from the daylit side of the Earth) given all the "artist's renderings" in the question and the answers, I thought it might be useful to include a diagram that demonstrates the actual scale ...

16

If it is really between the earth and the sun it is called a "solar eclipse" and and the moon's shadow falls on the earth at certain places, because it is not large enough to cover the whole sun except on a shadow path. A total solar eclipse occurs when the Moon completely covers the Sun's disk, as seen in this 1999 solar eclipse. Solar prominences can ...

12

The diagram you drew is flat, but the solar system is not. The Moon's orbit is not in the same plane as the Earth's orbit. Wikipedia has a nice diagram: Because of this, when the Moon is "in between" the Earth and the Sun, it is usually a little "above" or "below" the Sun as well. You can observe this for yourself: one or two days after the new moon, ...

5

As you are talking about a "Star map" and "current visible positions", I'll assume you are talking about a star map of the ${\sim} 5000$ stars visible to the naked eye. Most of those stars are within 1000 light years of he Earth. They have typical velocity dispersions with respect to the Earth of ${\sim} 10$ km/s, with the occasional rare star with a ...

5

Yes, there are lots of optical observations of isolated (non-pulsar) neutron stars. Such observations have been done for many, many years. An early example would be Kulkarni & van Kerkwijk (1998) who detected optical radiation from the counterpart to RX0720.4-3125. http://arxiv.org/abs/astro-ph/9803024 The authors discuss various mechanisms that ...

4

In general, yes you need to know the orbital inclination angle $i$ in order to fully solve the orbit. The radial velocity amplitude $K$ is just modified to $K \sin i$ (where $i=0$ is a face-on orbit). Combining this with the orbital period and Keplerian orbits gives you the "mass function" $$\frac{M_1^3 \sin^3 i}{\left(M_1 + M_2\right)^2} = \frac{K_{2}^3 \... 4 The apparent line-of-sight velocity (red shift / blue shift) is v\cos\theta where \theta is the angle between the plane of the stars' orbits and the line-of-sight line from the Earth. If the stars eclipse one another at a certain point in their orbit (eclipsing binaries) then we know that the Earth is in their orbital plane, so \theta=0 and the ... 3 There have been a few detected in the visible range by Hubble Space telescope and Keck observatory. The magnitude is less than 25 for the Hubble image(PSR0656+14). Neutron stars are very hot, 600,000 or more Kelvin and very small (Hubble star was 28 kilometers in diameter) so their visible luminosity is very low. The emission is redshifted due to General ... 2 Juno probe's speed is in relation to what frame of reference? The escape speed of Jupiter is ~59.5 km/s and 150,000 kph ~ 67.1 km/s, so this speed must be in reference to the sun otherwise the spacecraft would not stay in orbit. Jupiter's orbital speed about the sun is ~13.1 km/s, which subtracted from the 67.1 km/s would result in ~54 km/s, thus more ... 2 Listening to the recording that the article was derived from, it appears that the article is: Donald Lynden-Bell. Exact Optics : A unification of optical telescope design. Mon. Not. R. Astron. Soc. (2002) 2 There is actually 2 questions here. The van Cittert-Zernike theorem is essentially used to calculate the free space propagation of the electrical field/light. It can thus be used to reconstruct the original field at the astronomical object from our measurement on earth. However as mentioned correctly it only applies under certain incoherence conditions. So ... 2 I agree with Peter R. In the case of coherent astronomical sources, The radiation arrives only from one direction. All the interferometers will measure the same visibility. You get nothing more than a point. In 1989 K. R. Anantharamaiah, Tim Cornwell and Ramesh Narayan (NASA) wrote a paper on Imaging coherent and incoherent astronomical objects. Take a ... 2 The Eddington limit is a theoretical maximum accretion rate onto an accreting body under certain idealised circumstances. The limit is calculated by assuming that the gravitational acceleration felt by the accreted material is exactly balanced by radiation pressure, under the assumptions that the material is completely ionised hydrogen, that the opacity is ... 1 The Eddington luminosity is given by:$$L_{Eddington}=\frac{4\pi c GMm}{\sigma_T}. This represents the maximum luminosity of a body, when there is a balance between the two primary forces - that of the radiation, acting out of the body, and that of the gravitational force of the star, acting inwards. The terms sub-Eddington and super-Eddington refer to ...

1

The other answers seems to answer most of your questions, but I think one confusion remains: The speed of light as a maximum speed in the Universe (which is not the case). First off, redshift doesn't go to infinity for objects receding at $v = c$. We easily see galaxies recede at superluminal velocities. In fact, this is the case for all galaxies with a ...

1

The moon circles around the earth, so half the time it is between the earth and the sun and half the time the earth is between the sun and the moon. Therefore also the moon rises and sets, the same way the sun rises and sets. If it's midnight (your are on the opposite site of the earth than the sun) and the moon is betweem the sun and the earth you can't ...

1

Without seeing the particular example you have in mind it's impossible to be definitive, but without any context, for a probe in the Solar System, I'd assume heliocentric coordinates. Unless it's an orbital velocity around some other body, or an escape velocity from some other body, or of course if it's specifically stated to be in some other frame.

1

The only difference is a minute change in brightness as the earth-sun distance changes from 147000 km (perihelion) to 152000 km (aphelion). Thus, at perihelion the sunlight is slightly more intense, and hence, so is the moonlight. At aphelion it's a little less intense, and at equinox it is about midway between those two. This difference is far smaller than ...

1

Yes, this is usually what is meant by a negative longitude, it's just a convenient way to express a range that happens to straddle the zero-point. And in the context of the HiGAL survey you were looking at, this is precisely what is meant.

1

I'm not sure if this is exactly what you want, but there's a book called Practical Statistics for Astronomers by J.V. Wall and C.R. Jenkins that might fit the bill. According to the Cambridge University Press website (the book is a part of Cambridge Observing Handbooks for Research Astronomers): Astronomy needs statistical methods to interpret data, but ...

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