solar - Local Standard Time Meridian - LSTM (Units)

In a time, the word "standard" normally means the civil time, recognized by the government in the place of interest. I'm not sure about other countries, but in general conversation in the USA, "standard" also means that daylight saving time is not in effect. "Daylight" time (as in "Eastern Daylight Time") means the civil time, recognized by the government, and daylight saving time is in effect.

These can be expressed in the ISO 8601 format; the standard gives the example "15:27:46+01:00" meaning the time zone is one hour ahead of (east of) Greenwich.

In astronomy, time can also be the local time of the observer, ignoring the existence of time zones. This is local mean time, and is obtained from Universal Time by adjusting for the longitude of the observer. Usually the exact form of Universal Time that would be used would be UT1.

Local apparent solar time is based on the actual position of the sun; it is the time given by a sundial. It can be about 15 minutes ahead of or behind local mean time.

visible light - Why doesn't a lighter flame cast a shadow?

Imagine that you have a titanium screen. It's thin, so it's difficult to see. You can use a torch to heat some areas to incandescence. These areas become easily visible.

Now if we take a very bright light source and shine them at these lit areas, we won't see a shadow that gives us any information about the pattern of the light. Both the cool and hot areas interact with the bright light beam in the same way. The grid itself may be visible, but the pattern won't reflect anything about the relationship between the dark and light regions.

The flame is the same thing. It contains matter (such as fuel or soot particles), but similar matter is also present outside the flame. The imaging light may be slightly attenuated by these particles, but there is no sharp division in their location that corresponds to the shape of the flame. This lack of sharp division will prevent imaging.

amateur observing - Free source of printable star charts in the format of the Millennium Star Atlas?

I've been looking for sources of printable electronic star charts - there seem to be many different ones available, either as part of planetarium software or just plain charting packages. But there's such a richness of offerings that I'm having trouble figuring out which of those, if any, can produce printable charts in the format that I'm looking for.

I like the general format of the Millennium Star Atlas - black stars on a white sky background, and with horizontal lines representing 1 degree of declination and vertical lines representing 4 minutes of RA (see sample page at http://www.skyandtelescope.com/wp-content/uploads/GD-Sept2011-Chart.pdf ).

I'd like to be able to specify a range of RA and declination and get that chart.

Can anyone suggest any free star chart printing software that can print similar output for any desired sector of the sky?

I'm looking at Cartes du Ciel (http://sourceforge.net/projects/skychart/) which might have that capability, but if so, I haven't yet figured out how to enable it.

optics - Magnification of a telescope

I took an image of Jupiter through my 8" Dobsonian Telescope, attaching a DSLR and a 1.25" Barlow Lens where the eyepiece goes, as shown in this video: https://www.youtube.com/watch?v=reFxoF3XoaU

Through numerous online sources, I learnt that to find the angle of view of my image, the magnification of my image needed to be calculated, and that this value along with the given field view for my eyepiece.

http://www.rocketmime.com/astronomy/Telescope/Magnification.html:

The above website answers the following question, the calculations of which I attempted to mimic.

My first telescope was a Meade 6600 -- they don't make it any more -- it's a 6-inch f/5 Newtonian scope. It came with a 25mm eyepiece. So... what was the magnification I was getting with this scope?

Here are the calculations I did in hopes of getting the above result:

$$textrm{Diameter} = 8'' = 203.2 textrm{mm}$$

$$f_{textrm{ratio}} = frac{textrm{focal length of objective}}{203.2 textrm{mm}}$$

$$therefore textrm{focal length of objective} = 203.2 cdot 5.9 = 1200 textrm{mm}$$

$$textrm{Magnification} = frac{textrm{focal length of objective}}{textrm{focal length of eyepiece}} = frac{1200}{x}$$

As shown in the video (the first link above), I didn't use an eyepiece to take my picture - I used a barlow lens, a couple of adapters, and a DSLR. So at this point, I am not sure what value to use for the "focal length of the eyepiece." How can I proceed to calculate the magnification?

space telescope - How is the cost of JWST distributed on different parts of development and operations?

So most of the costs should end up as salaries to engineers.

Even in a project that is solely people-based, most of the costs do not end up as salaries to engineers. Salary is typically a bit less than a half of the cost. Those engineers get benefits. Health care in the US isn't cheap. They get holidays, vacation, and sick leave. They receive matching funds on their 401K contributions. In the US, employers and employees both contribute to the cesspool lockbox sometimes known as "Social Security". The employer share is part of the cost of having an employee (and companies charge for it). The employer also pays for unemployment insurance and other items on behalf of their employees. Those engineers sit at desks in an office or a cubicle, and have a nice computer on which they do their work. They use the employer's servers, printers, copy machines, conference rooms, break rooms, and coffee machines. All of that costs money. These overhead costs are the first of several indirect costs.

The accountants who write those engineers their paychecks aren't charging to any project. They're still paid by the company, as are the executives of the company. The executives also typically don't charge their time to a project, and those executives can be paid quite handsomely. This leads to another category that adds to quantity that companies charge the government, "general and accounting". Finally, for-profit companies need to keep their shareholders / stakeholders happy. They need to earn a profit. This leads to a third category that companies charge the government, "profit".

When you add up all of those indirect costs, they come close to and oftentimes exceed direct costs (i.e., chargeable salary). Note that this is for a project where the only costs are the people working on it.

Building a spacecraft notches things up a lot. Cleanrooms are expensive. Control rooms are expensive. The flight computers are expensive. The bolts and fasteners are expensive. The custom, one-off sensors are expensive. The spacecraft's thermal control system is expensive. The spacecraft's structure is expensive. Launch is expensive. Every piece of hardware, every fabrication facility is expensive.

Finally, cost overruns, requirements changes, and delays can represent a significant portion of the cost of a poorly managed project. JWST has experienced numerous cost overruns, requirements changes, and delays. Big contractors have learned that the only way to win big government contracts is to underbid. They would lose their shirts if the government kept to that initial contract. But it never does. The government instead changes the requirements, changes the payments, and changes the schedule. Those changes are beyond expensive.

orbit - LLR & Orbital Motion

As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a "straight line". If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.

In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.

The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?

I hope this makes sense and thank you.

the sun - How does the appearance of the analemma vary with latitude

Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have slightly different shapes. For example, see the solar analemmas taken at the same place at different times.

Solar analemma at 0700 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu

Solar analemma at 1400 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu

The best way to understand the effect of latitude on the shape of the analemma is to consider the effect of earth's axial tilt on the shape of the analemma.

If the Earth’s orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped. At the Equator, this line would be a straight line spanning from left to right or West to East.

The 23.5-degree axial tilt of the Earth affects the Sun’s apparent position in the sky – as the year progresses and the Earth continues to spin at an tilted axis and orbit around the Sun, the Sun seems to move up and down (North-South) in the sky. This has the effect of generating the two loops of the figure 8.

You're correct that the tilt of the analemma varies with the latitude. Observers in the Northern hemisphere will obtain an analemma curve with the broader loop at the bottom. This reverses for observers in the Southern hemisphere, where the broader loop is on the top of the curve. At equator, the anlaemma lies on its side. At north pole, only the top of the analemma would be visible, while it is the opposite at the south pole.

Do astronomers and astrophysicists more often use diameters or radii when discussing about planets, dwarf planets, exoplanets and stars?

Definitely radii, with one notable exception...

When observers talk about how large an object is on the sky, they usually discuss angular size, which is related to the diameter of an object, not the radius. So when discussing the angular size of, say, Alpha Centauri A (a few milliarcseconds), this is related to the diameter of the star, not the radius.

orbit - What causes objects to become tidally locked?

Tidal locking occurs because the planet deforms the satellite into an oval, with long axis pointing towards the planet. If the satellite is rotating the long axis will move away from being pointing towards the planet, and the gravity of the planet will tend to pull it back, slowing the rotation until one face is permanently facing the planet. Tidal locking isn't a result of the formation processes, but a consequence of satellites not being perfectly rigid.

In order to model the effects of tides on the orbits and rotation periods of satellites you need to know several important pieces of information.

First you obviously need to know the size of the planet and the satellite (both in terms of mass and radius) the shape of the orbit and the rotation rate of both planet and satellite. For many objects, these values are well known.

Next, and this is the tricky bit, you need to know how the satellite and planet will be deformed by the other's gravity, and how much tidal heating will occur. These are the so-called "love number" (after Augustus Love) and the dissipation function, Q.

It is hard to estimate these. For the Earth Moon system the ratio k/Q is known to be 0.0011. (but the Earth is a poor model for other planets, which don't have a substantial ocean, or a liquid core)

For other planets the value of Q varies between 10 and 10000, with larger values for the gas giants, and k can be estimated from the rigidity of the bodies.

A simple gravitation model is not able to capture the subtleties of the gravitational interaction between two mutually deforming bodies, indeed for most simulations, the planets are modelled as points, or at most as spheres, and this is good enough for all but the highest precision calculations.

Tidal locking takes a long time (by human standards) but a relatively short time compared with the age of the solar system. The time taken is very strongly dependent (order 6) on the radius of the orbit.

Direct simulation would be more or less impossible: the deformations are too small, and the time scale of locking is too large. It would be possible (though difficult) to model tidal locking in a simulation with unrealistic values for the rigidity of the satellite, and the size of the planet (think jelly world, orbiting a (Newtonian) black hole) so the deformation is greater and the locking time shorter. However modelling the elastic deformation of a body under gravity is far from trivial.

telescope - Could mirrors be replaced with CCDs?

To answer your question, we need to first show the job each mirror is doing.

First up, the Newtonian (lovingly called the "Newt", and invented by Sir Ike Newton):

https://en.m.wikipedia.org/wiki/Reflecting_telescope#/media/File%3ANewtonian_telescope2.svg

Two mirrors in this design, not surprisingly labeled as primary and secondary.

The job of the primary mirror is NOT merely to reflect light, but to concentrate the diffuse photons onto a much smaller point. This makes really dim objects brighter, and is the first step in magnification. (Further magnification is done by the eyepiece, which is similar to a small refracting telescope.)

In the case of the Newt, the secondary mirror reflects the now concentrated photons to a more convenient point for viewing. Without the secondary mirror your head would get in the way of the view. A secondary mirror is not necessary, and in fact many telescopes will place instruments, such as CCD's, at this "prime focus" point.

In the case of the Hubble Space Telescope, the secondary mirror reflects the concentrated photons to the scope's instruments, where they can work their magic.

In all reflecting telescope designs the primary mirror uses the laws of physics to give the end user, whether it's the human eye, or research gear, as many concentrated photons as possible, maximizing what we can see/detect. The bigger the primary mirror, the more concentrated the photons, and the more we have to work with.

When it comes to seeing what we call "the dim fuzzies", bigger IS better!

At the moment of total solar eclipse what are exact distances of Moon & Sun to the Earth

First off, your definition of "total solar eclipse" is different from that generally used. A total solar eclipse occurs when the moon is directly in front of the sun and is bigger so it blocks the whole of the photosphere. Also note that the sun and moon are both somewhat oblate (not perfect spheres, but flattened) and not exactly the same shape, but for the sake of estimation I'll use the diameter information from wikipedia.

The moon has an average diameter of 3,474 km, and the sun has an average diameter of 1.392684 million km. So the Sun is 400.9 times larger than the moon. When the sun is 149600000 km distant, then if the moon is $149600000/400.9 = 373200$km, the sun and the moon would appear equal in size.

When the moon is 384000km from the centre of the Earth, the distance to the Earth's surface is smallest at the sub-lunar point, about 378000km, but still greater than 373200, so when the Sun and Moon are both at their average distance, an annular eclipse occurs. Annular eclipses are slightly more common than total eclipses.

However if the distance from an observer to the moon is closer than 1/400.9 of the distance to the sun, then a total eclipse is possible, and this allows you to generate infinitely many solutions to the question "How far is the moon during a total solar eclipse"

star - Did I see another planet?

A couple days ago, I zoomed in with my 30x optical zoom camera, and after some exposure adjustments, a bright star in the night sky turned into this:

Are those other planets or other stars? Or is that a lense effect?

EDIT: The bright object in question was ~60 degrees above the horizon, and ESE of me (East-south-east). I took the picture on 3-25-16 from Madison, Wisconsin.

EDIT: Question answered, more clear picture added FYI :) Enjoy &

Probability of finding life-supporting exoplanets

The question can be rephrased into: How many exoplanets have been found that could harbour life? As of early May 2015 the number appears to be about 30, see the Habitable Exoplanet Catalog on the Planetary Habitability Laboratory's web site.

You should also note that exoplanet detection methods will bias this sample in favour of Super-Earths.