New answers tagged

0

The whole point of a pin hole projector is that it doesn't need any optics just a pin hole. If you are after magnification then that is related to the distance you project the solar image behind the pin hole, with a projection distance of 1000mm the solar disc will have a diameter of about 9mm. Increasing the distance will increase the size of the image but ...


2

A Celestron Nexstar 130 SLT is not suitable for spectroscopy and the only photography that it will be mush use for would be afocal or DSLR imaging of the Moon (or Sun with a suitable filter). For more serious work you will need a equatorial mounted scope with a better focuser. Exactly which one to choose will depend of your budget and what you want to do. ...


3

Current estimates put Apophis's diameter around 325 m and its 2029-04-13 approach about 38000 km from the center of the Earth. I figure an angular size <= 2 arcsec, almost starlike even if you manage to track it in a telescope. Ephemerides show it at apparent magnitude 3.5 or so just before closest approach - visible but not outstanding to the unaided eye ...


5

I've just rewritten this answer - @MikeG caught a glaring error by pointing out a really basic handy relationship called the Rayleigh criterion. \begin{align} {\theta}_R \approx1.22 \frac{\lambda}{D}. \end{align} It's better to read the (or any) article, but very briefly, the angular resolution is roughly the ratio of the wavelength to the diameter of a ...


0

Find a local astronomy club; they will help you get started, learn the sky, and show you different types of equipment. Also, read "Turn Left At Orion", H.A. Rey's "A New Way to See Them" and "Find The Constellations" to learn the sky. And get a pair of 10x50 binoculars. These are the best "telescope" (actually, two telescopes!) to start with: inexpensive, ...


3

Mercury's angular diameter on transit day will be 12 arcseconds. A camera obscura using a 12 mm aperture could resolve it; one lens from +0.75 diopter reading glasses, if you can get them, will project a bright 12 mm image of the Sun at a distance of 1.33 m. Note that a larger aperture or a shorter focal length will make the Sun image hotter than direct ...


6

It is possible but unlikely. Here is a really good 'Science 2.0' article about the possibility (http://www.science20.com/robert_inventor/could_you_see_moon_city_lights_or_a_greenhouse_from_earth_just_for_fun-157480). Essentially, you likely wouldn't see the light on Moon settlements because there would need to have many thousands of bright lights and ...


1

In retrospect, we now know that the peak brightness was indeed 10.5 during the first week of February, 2014: http://www.skyandtelescope.com/astronomy-news/observing-news/supernova-in-m82-passes-its-peak/


4

Stellarium identifies the mystery "moon" as a star, HIP 54057 a.k.a. HD 95848. Your telescope probably showed it on the right instead of the left. The location on Earth is insignificant since you provided the time in UTC. Jupiter passed directly in front of the same star on 2016-04-12, but the occultation was only visible from the Eastern Hemisphere.


3

Most likely would be the the main mirror needs collimating. On big things like the moon you won't notice bad collimation as much as something small like a planet or star.


0

Most likely candidate is a meteor with the brightest part cut off by the shutter. You can see that it is increasing in brightness. Forgery can be eliminated due to various sightings, and it shouldn't a satellite due to the distance covered in half a second, not to mention that it would be illuminated by the sun in that portion of the sky.


3

A seventh magnitude star. See this link: http://www.cloudynights.com/topic/533968-jovian-events-414-1516/ Further info: the star in question was probably HIP 54057 (also known as SAO 118636 or HD 95848.) On April 12th it was occulted by Jupiter (that is, Jupiter passed in front of it) for about three hours. So the star would have been in the area of ...


2

The magnitude scale is a logarithmic scale. An increase of 1 magnitude corresponds to a decrease in brightness of about 2,5 times dimmer. Vega, a bright star has a magnitude of 0, so any star that is brighter than Vega would have a magnitude that is less than 0. This is an odd system; the reason for it is historical. The ancient Greeks ordered stars by ...


2

Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us. The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6). Each ...


1

Yes you can. You just need a partly clear sky (to see the stars) and part cloud (to produce the rain). For an example of suitable conditions, look for pictures of a moonbow. (I've only seen one once, personally.)



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