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AstroImageJ
ImageJ for Astronomy

Click for other screenshots of AstroImageJ in use

The latest version has many features to enhance astronomical image processing, especially for photometry. These include a versatile image display, time-series stacks, a link to Astrometry.net for adding coordinates, built-in AstroCC to set and time and place, links to Simbad for object identification, and exoplanet model fitting. It is available for Windows, Mac OS X, and Linux by downloading an installation package . After installation, the built-in Help->Update option keeps your version current.

AstroImageJ (AIJ) ...

Runs on Linux, Windows and Mac OS

Provides an interactive interface similar to ds9

Reads and writes FITS images with standard headers

Allows FITS header viewing and editing

Plate solves and adds WCS to images seamlessly using the Astrometry.net web interface

Displays astronomical coordinates for images with WCS

Provides object identification via an embedded SIMBAD interface

Aligns image sequences using WCS headers or by using apertures to correlate stars

Image calibration including bias, dark, flat, and non-linearity correction with option to run in real-time

Interactive time-series differential photometry interface with option to run in real-time

Allows comparison star ensemble changes without re-running differential photometry

Provides an interactive multi-curve plotting tool streamlined for plotting light curves

Includes an interactive light curve fitting interface with simultaneous detrending

Allows non-destructive object annotations/labels using FITS header keywords

Provides a time and coordinate converter tool with capability to update/enhance FITS header content (AIRMASS, BJD, etc.)

Exports analyses formatted as spreadsheets

Creates color images and with native ImageJ processing power

Optionally enter reference star apparent magnitudes to calculate target star magnitudes automatically

Optionally create Minor Planet Center (MPC) format for direct submission of data to the MPC

Recently added when you update to the daily build after installation --

Nearby eclipsing binary star identification from expected depth versus light curve RMS table for TESS follow-up

Delta-magnitude versus RMS plot

Enhanced contrast controls

Annotations are retained when placing apertures

FITS header search feature

An article on AstroImageJ appears in the Astronomical Journal (2017). Additional details on the user interface are available through astro-ph in reference 1701.04817v1. The User Guide may be downloaded here.

Comments, requests for help and new features, and bug reports should be posted to the AstroImageJ Forum.

AstroImageJ is ImageJ with astronomy plugins and macros installed. It includes tools based on the Göttingen ImageJ astronomical resources with additions we find useful. Because it was necessary to modify the original ImageJ code to enable some of these features, this package should be installed in its entirety.

AstroImageJ is open source software and is distributed under the terms of the GNU General Public License. It incorporates ImageJ components which are in the public domain.

Last update: August 5, 2021
kielkopf at louisville dot edu

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AstroImageJ (AIJ) - ImageJ for Astronomy

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Click for other screenshots of AstroImageJ in use The latest version has many features to enhance astronomical image processing, especially for photometry. These include a versatile image display, time-series stacks, a link to Astrometry.net for adding coordinates, built-in AstroCC to set and time and place, links to Simbad for object identification, and exoplanet model fitting.

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Classifying stars by brightness/temperature

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Classifying Stars by Brightness and Temperature

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In this lesson, we will learn how to classify stars according to their brightness (or absolute magnitude) and their temperature (using their color).

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The weight of an object is given by W=mg, the force of gravity, which comes from the law of gravity at the surface of the Earth in the inverse square law form:

At standard sea level, the acceleration of gravity has the value g = 9.8 m/s ², but that value diminishes according to the inverse square law at greater distances from the earth. The value of g at any given height, say the height of an orbit, can be calculated from the above expression.

Please note that the above calculation gives the correct value for the acceleration of gravity only for positive values of h, i.e., for points outside the Earth. If you drilled a hole through the center of the Earth, the acceleration of gravity would decrease with the radius on the way to the center of the Earth. If the Earth were of uniform density (which it is not!), the acceleration of gravity would decrease linearly to half the surface value of g at half the radius of the Earth and approach zero as you approached the center of the Earth.
Index

Gravity concepts

Orbit concepts

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The weight of an object is given by W=mg, the force of gravity, which comes from the law of gravity at the surface of the Earth in the inverse square law form: At standard sea level, the acceleration of gravity has the value g = 9.8 m/s 2, but that value diminishes according to the inverse square law at greater distances from the earth.

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Binary stars are of great importance to astronomers because they provide virtually the only means of directly determining the masses of stars other than our Sun. As mass is such a key property of stars and to a large extent knowing a star's mass determines its life cycle and fate, being able to accurately determine stellar masses is vital in refining our models of stars.

Kepler's Laws

To find the mass of a binary system we need to apply Kepler's Laws. If we adapt them for a binary system where the masses of the component stars are similar then:

1. The stars orbit each other in elliptical orbits, with the centre of mass (or barycenter) as one common focus.

2. The line between the stars (the radius vector) sweeps out equal areas in equal periods of time (sometimes called the Law of Equal Areas).

3. The square of a star's period, T, is directly proportional to the cube of its average distance from the centre of system mass, r:
∝ . This is the Law of Periods or Harmonic Law.

Let us now see how we can apply these to determine the total system mass and the mass of the individual component stars. We shall refer to the diagram below.

Deriving Equations for Mass of Binary System

The barycenter or centre of mass of the system is where:

The forces acting on each star are balanced, that is the gravitational force equals the centripetal force so;

now (5.5) is simply an expression of Kepler's 3rd Law;
r ³/T ² = GM/4π ² (5.7)

Using equation 5.5 or 5.6 we can determine the mass of the binary system if we can measure the orbital period and the radius vector (separation between the two components) for the system. In practice most systems will not have their orbital plane perpendicular to us so we need to adjust for the observed inclination.

Whilst it is relatively straight forward to determine the total system mass, it is harder to determine the individual masses of the component stars. This requires the distance from a component star to the barycenter to also be measured. We can then use this to determine the mass of that star by using:

Once the mass of one component and the total system is known it is straightforward to calculate the mass of the other component.

Examples of Mass Calculation in Binaries

Example 1: Determining the total mass of a system.
The α Centauri system is 1.338 pc distant with a period of 79.92 years. The A and B components have a mean separation of 23.7 AU (although the orbits are highly elliptical). What is the total mass of the system?

To solve this we use equation 5.5 but before we simply substitute in we need to check units. Whilst we can use use AUs and years to give us a relative value when applying Kepler's Laws, if we want a full numerical solution we must convert all parameters to S.I. units as the constant G is normally expressed in such units (G = 6.672 × 10⁻¹¹m ³.kg⁻¹.s⁻²).

∴ M = 4.162 × 10³⁰ kg.

Example 2: Calculating the mass of one of the component stars.
As α Cen is a nearby visual binary system, careful astrometric observations reveal that the primary component, α Cen A has a mean distance of 11.2 AU from the system's barycenter. What is the mass of each of the component stars in the system?

From example 1 above we have already calculated the system mass to be 4.162 × 10 ³⁰ kg.
Now distance r A = 11.2 AU = 11.2 × 1.50 × 10 ¹¹ = 1.680 × 10 ¹² m.
We shall use this to first find the mass of α Cen A.
Using equation (5.8):
m A = M(r - r A)/r
and substituting in:
m A = 4.162 × 10 ³⁰(3.55 × 10 ¹² - 1.680 × 10 ¹²)/3.55 × 10 ¹²

The mass of α Cen A makes it about 1.1 × that of our Sun. It, too is a G2 V star with a luminosity about 55% greater than that of our Sun. α Cen B is a dimmer K0-1 V min sequence dwarf with 90% of the Sun's mass and only about half as luminous.

Of particular value to astronomers are systems that are both eclipsing and spectroscopic. The radial velocity values from the spectral data can be used to calculate absolute rather than just relative values for the stellar radii. This can then be combined with orbital inclination parameters obtained from the light curve to give the stellar masses and mean stellar densities. The relative luminosities and total luminosity of the system can be derived then used to calculate the total flux of the system. This then allows us to calculate the distance to the system. We can also infer the mass and luminosity of each star.

As already mentioned, binary stars are of vital importance as they allow us to determine stellar masses. To date (August 2004) only one single star other than our Sun has had its mass accurately determined by a means unrelated to Kepler's laws. Astronomers involved in the MACHO project at Mt Stromlo took an image in 1993 that showed microlensing of a distant background star by a closer star in the foreground. Recent parallax measurements and observations by the HST have allowed astronomers to calculate that the faint, red foreground star has a mass only one-tenth that of our Sun. For more information, read the press release.

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Binary stars are of great importance to astronomers because they provide virtually the only means of directly determining the masses of stars other than our Sun. As mass is such a key property of stars and to a large extent knowing a star's mass determines its life cycle and fate, being able to accurately determine stellar masses is vital in refining our models of stars.

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Introduction

The main goal of this lab is to give you experience putting theory into practice.

By using observations of Jupiter taken by the now retired Rigel robotic telescope formerly managed by the University of Iowa, you will determine the mass of Jupiter by applying Kepler's 3rd law.

You will also learn the power of the tool www.wolframalpha.com, which is the Google of computations. You can enter full equations with units into its search bar, and it will automatically handle all calculations!

Part 1: Kepler's Laws

Kepler's 3rd law only applies to two objects orbiting each other - like the Earth orbiting the Sun, or the moon orbiting the Earth.

It specifically links the time it takes to orbit with how far apart the objects are from each other, and how massive the objects are.

So if you can measure how far apart two objects are, and you can measure how long it takes them to orbit each other - then you can calculate their masses!

In this lab, you will measure the distance between Jupiter and one of its moons, and measure how long it takes the moon to orbit. Then using those two measurements, you will calculate Jupiter's mass.

Let's take a look at the equation for Kepler's 3rd law:

First of all - don't panic! Be patient and try to understand what's going on here - it's very simple and you CAN do it. *pom pom gets stuck in ceiling fan*

The first thing to notice is that is just a number: 4 x 3.1415 x 3.14159 = about 39.

The next thing to notice is the letter G is also just a number - called Newton's Gravitational Constant.

In other words, those numbers are constants and we can just combine them together and replace them with one number, 5.915 x 10 ¹¹:

Now, let's do an example using the Sun and the Earth to figure out how we can determine the mass of the Sun. (You will repeat this same procedure with Jupiter and one of its moons later). Start by replacing M1 and M2 with the mass of the Sun and the Mass of the Earth.

Here is where this law becomes very powerful and also very simple: because the mass of the Sun is soooooo much bigger than the mass of the Earth, adding the mass of the Sun to the mass of the earth doesn't really do anything to the equation:

The above line just says that adding the mass of the earth to the mass of the Sun gives you approximately the mass of the Sun - that squiggly equal sign means equalish.

Now the equation becomes a little bit simpler:

At this point we simply have to do two algebraic operations to solve for the mass of the Sun, exchanging a P ² for the M Sun:

This equation is easy to solve and you can maybe almost probably calculate the mass of the Sun without looking up any of the numbers, because 'a' is the distance the Earth is from the Sun - the distance between the two masses. Do you remember what that number is?

What about P? What do you put for P?

Remember that the Earth takes 1 year to orbit, so P is one year!

That's it, you just plug those two numbers in, 1 AU and 1 year into Kepler's 3rd law, and you can calculate the mass of the sun!

Instead of trying to convert all these units to get them to match, let's just head over to wolframalpha.com and put in the numbers with their units as is.

Be sure to pay attention to where you put the parenthesis, because order of operations is king.

(You will enter this formula into wolframalpha in the same way for calculating the mass of Jupiter in part 2, replacing P and a with the values for the Jupiter/moon system.)

Check it out!

Go to wolframalpha.com and enter the above to calculate the mass of the Sun.

Pay close attention to get the parentheses right: (4 AU)² is not the same as 4 AU²

What do you get for the mass of the sun when you enter that in? Report the answer on your worksheet for Part 1.

Part 2: Determining the Mass of Jupiter

For this part, you will use this image of Jupiter and its moons:

Your first job now is to figure out which of these moons is the moon Europa.

This is the moon you will use to measure the mass of Jupiter.

Download the image and open it with the Preview App. (Or MS Paint if you're on Windows).

Use Stellarium to label the moons, by entering the date, time, and location that the picture was taken from.

This picture of Jupiter was taken with the University of Iowa's robotic telescope called Rigel. It was operated at the Winer Observatory in Tuscon, Arizona. (This is because Arizona has mostly clear nights, whereas Iowa's nights are mostly cloudy. Rigel is retired, there's a new telescope called Gemini in current operation.)

The date and time it was taken was:

Date: 2012-03-09 Time: 03:39:20.848 UTC (UTC time is 6 hours ahead of MST.)

The time is in UTC time, so enter the date and time as shown into Stellarium, and then manually move the time back 6 hours.

We can find the period of Europa by watching it orbit Jupiter. In the following animation, we can see that it takes 3.55 days for Europa to go around Jupiter. This is the period of its orbit, P:

So now we know that the moon is Europa, that the period of orbit is P= 3.55 days, now all you need to do is determine 'a' for Kepler's Law. Remember 'a' is how far away Europa is from Jupiter.

To determine 'a', we need to use the small angle formula.

Recall that the small angle formula relates the distance to Jupiter (D), the distance Europa is from Jupiter (this is what we will use as 'a'), and the angle of separation between Jupiter and Europa.

In our image, if we simply count the pixels, we can measure the angular separation between Europa and Jupiter.

Here is an example of how I use the Preview app to count the pixels. I traced out 'a' with a blue line by clicking and dragging, in order to measure the angle ø:

Now look closely at the rectangle I drew:

You can see the "108 x 18" - that means the rectangle I drew has dimensions of 108 pixels over and 18 pixels down.

Now, what we actually want is the hypotenuse of that triangle, to get the distance in pixels from Europa to Jupiter. So use the pythagorean theorem to do that: √(108²+18²) = 109 pixels.
*NOTE: These numbers are not the correct numbers, I measured a random picture to show you how it works. You need to count the pixels yourself.
So now that I know Europa appears to be 109 pixels from Jupiter, I look up the manufacturer's specification for the camera used on Rigel - It tells us that each pixel has an angular size of 0.73" per pixel - that is 0.73 arc seconds per pixel. That's an angle!

So if we have 109 pixels, and each pixel has an angular size of 0.73 arc seconds, then the angular separation of Europa from Jupiter is 109 pixels * 0.73" per pixel = 80 arcseconds.

And LASTLY we need to convert 80 arcseconds into radians - there are 206,265 arc seconds in every radian:

80 / 206265 = 0.00038785 radians. (Now you see why we use arcseconds!)

This is 'ø', the angular separation between Jupiter and Europa.

Now let's use it to calculate 'a', the distance Europa actually is from Jupiter.

The Small Angle Formula says that:

ø = a / D

Rearranging to solve for a:

a = ø x D

a = 0.00038785 x D

The distance to Jupiter (D) when this image was taken can be found by querying wolframalpha.com with the following:
Distance to jupiter on March 9th, 2012
This will return a value in 'AU'.

Now plug in your values for 'a' and 'P' into Kepler's third law formula to get the mass of Jupiter!

Compare it with the true value and cry in awe at your power, human!

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The main goal of this lab is to give you experience putting theory into practice. By using observations of Jupiter taken by the now retired Rigel robotic telescope formerly managed by the University of Iowa, you will determine the mass of Jupiter by applying Kepler's 3rd law.

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When are we going to use the observatory

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When are we going to use the observatory

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Talk with mom about observing nights.

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-Images of target systems (what are target systems)
-become familiar with telescope system
-How to use binary stars to test limits of our system.

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This autumn, why not take advantage of the earlier sunsets and longer nights to get acquainted with some double stars? Double stars occur when two stars appear very close to each other in the night sky. This can simply be because they lie in the same direction, though they may actually be at vastly different distances from us. But some are in a binary star system, which means that they are gravitationally connected to each other. Binary stars (and also systems with three or more members) are in fact quite common. They can also be gorgeous to look at! Here are ten of the best to look out for:

This lies in the constellation Cygnus, can be seen with a small telescope; it marks the head of the swan. The two stars which make up the double are different in colour, one being bluey-green, the other amber. It's a beautiful sight. It is the star marked with the arrow on the map below.

This star system is sometimes called the 'double double'. It lies in Lyra, close to the bright star Vega. It is possible to make out a binary star with binoculars, but with greater magnification, it can be seen that the two stars are in fact four. you can find it marked as star 4.6 on the chart below, not far from the bright star, Vega.

Not many people realise that Polaris, one of the most instantly recognisable stars in the northern sky is actually a double star, although the companion star is much fainter, so needs a good telescope to be visible.

Is easy to locate, as it is the westernmost of the three bright stars making up Orion's belt. It is actually a complex multiple system, but can be seen with a small telescope as a binary star with one star much brighter than the other. Mintaka is the star labelled 2.2 on Orion's belt

This double star in the Capricorn constellation was known to the mediaeval Arab astronomers, who called it Algiedi, the billy goat. It is bright enough to be seen with the naked eye on a dark night.

6. Epsilon Pegasi, known as the 'pendulum star'

This double star forms the 'nose' of the Pegasus constellation. There's an optical illusion associated with this double star, noted many years ago by the famous astronomer Sir John Herschel; if the telescope is moved whilst the double star is being observed, the fainter star seems to swing relative to the brighter one. The brighter star often appears as bright orange in colour. The star is otherwise known as Enif.

In the constellation Delphinus, the dolphin. This one is bright enough to be seen with a small telescope as a pair of warm golden stars , and it forms the dolphin's snout. It is the star labelled 4.3 on the chart below

This one needs a good telescope to be seen properly, but the colours are worth looking out for - they appear as bright yellow and red. This double star is in the constellation Cassiopeia, which is one of the most distinctive features of the autumn sky. This binary is otherwise known as Achird.

Found in Aquarius, where it forms part of the 'water jar'. This one is difficult to see without a good telescope, but it has a lovely clear blue-white colour. The gap between the two component stars is slowly widening.

Located in Aries is bright enough to be seen with the naked eye, although the stars are close together and a telescope will help to resolve them. Then they appear as 'cat's eyes', two white spots of nearly equal intensity. This is another double star with a history - Robert Hooke first realised that it was two stars rather than one in 1664.

There are plenty more double stars in the autumn sky, and these are just some of the most distinctive. They are beautiful to look at and fascinating to study. Let's hope for some clear skies this autumn!

Find more on: Observing the Night Sky

Back to Learn Astronomy

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Top 10 Best Double Stars to Observe in Autumn

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This autumn, why not take advantage of the earlier sunsets and longer nights to get acquainted with some double stars? Double stars occur when two stars appear very close to each other in the night sky. This can simply be because they lie in the same direction, though they may actually be at vastly different distances from us.

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Never miss another geomagnetic storm. Sign up for Space Weather Alerts and receive a text message when auroras appear. Aurora tour guides and professional astronomers use this service. Now you can, too!

PERSEID METEOR UPDATE: Friday the 13th is a lucky day for sky watchers. The annual Perseid meteor shower is peaking as Earth passes through a stream of debris from Comet 109P/Swift-Tuttle. Last night international observers counted as many as 65 meteors per hour, while NASA all-sky cameras recorded 84 Perseid fireballs over the USA. More: photo gallery, sky map.

NOCTILUCENT CLOUDS PERSIST: Generally speaking, August is not a good month for noctilucent clouds (NLCs). The silvery clouds, made of frosted meteor smoke, begin to melt away as the mesosphere warms up in late summer. This August, however, the clouds are still being seen. Nadja Maletzki spotted them on Aug. 12th over Zürich, Switzerland:

"It was a very nice display," says Maletzki.

The significance of this picture is not that NLCs exist in August. They often do. Rather, it is the latitude of the display: +47 N. The clouds are usually retreating toward the Arctic Circle by mid-August, not showing up in the middle of Europe.

All summer long, NLCs have been descending farther south than they are "supposed to." At one point in June they were sighted near the Mediterranean coast of Spain, almost setting an all-time record. Forget about August. Will this be the year that NLCs are seen in September? Stay tuned.

Realtime Noctilucent Cloud Photo Gallery
Free: Spaceweather.com Newsletter

THIS MOONSTONE HAS TOUCHED SPACE: Are you looking for a far-out gift? Nothing says "I love you" like a moonstone from the edge of space. The students of Earth to Sky Calculus flew this one wrapped in a hand-crafted sterling silver Celtic love knot 35.1 km (115,158 feet) above Earth's surface:

You can have it for $179.95. The students are selling these pendants to support their cosmic ray ballooning program. Each one comes with a greeting card showing the item in flight and telling the story of its journey to the edge of space. All sales support the Earth to Sky Calculus cosmic ray ballooning program and hands-on STEM research.

Far Out Gifts: Earth to Sky Store
All sales support hands-on STEM education

Realtime Space Weather Photo Gallery
Free: Spaceweather.com Newsletter

Realtime Aurora Photo Gallery
Free: Spaceweather.com Newsletter

Every night, a network of NASA all-sky cameras scans the skies above the United States for meteoritic fireballs. Automated software maintained by NASA's Meteoroid Environment Office calculates their orbits, velocity, penetration depth in Earth's atmosphere and many other characteristics. Daily results are presented here on Spaceweather.com.

On Aug 12 2021, the network reported 146 fireballs.
(84 Perseids, 61 sporadics, 1 southern iota Aquarid)

In this diagram of the inner solar system, all of the fireball orbits intersect at a single point--Earth. The orbits are color-coded by velocity, from slow (red) to fast (blue). [ Larger image] [ movies]

Potentially Hazardous Asteroids ( PHAs) are space rocks larger than approximately 100m that can come closer to Earth than 0.05 AU. None of the known PHAs is on a collision course with our planet, although astronomers are finding new ones all the time.

On August 13, 2021 there were 2223 potentially hazardous asteroids.

Recent & Upcoming Earth-asteroid encounters:

Asteroid

Notes: LD means "Lunar Distance." 1 LD = 384,401 km, the distance between Earth and the Moon. 1 LD also equals 0.00256 AU. MAG is the visual magnitude of the asteroid on the date of closest approach.

Cosmic Rays in the Atmosphere

SPACE WEATHER BALLOON DATA: Almost once a week, Spaceweather.com and the students of Earth to Sky Calculus fly space weather balloons to the stratosphere over California. These balloons are equipped with sensors that detect secondary cosmic rays, a form of radiation from space that can penetrate all the way down to Earth's surface. Our monitoring program has been underway without interruption for 6 years, resulting in a unique dataset of in situ atmospheric measurements.

Latest results: Our most recent flight on June 25, 2021, confirms a trend of decreasing cosmic radiation:

Cosmic ray dose rates peaked in late 2019, and have been slowly declining ever since. This makes perfect sense. Solar Minimum was in late 2019. During Solar Minimum the sun's magnetic field weakens, allowing more cosmic rays into the solar system. We expect dose rate to be highest at that time.

Now that Solar Minimum has passed, the sun is waking up again. Solar magnetic fields are strengthening, providing a stiffer barrier to cosmic rays trying to enter the solar system. The decline of cosmic radiation above California is a sign that new Solar Cycle 25 is gaining strength.

.Who cares? Cosmic rays are a surprisingly "down to Earth" form of space weather. They can seed clouds, trigger lightning, and penetrate commercial airplanes. According to a study from the Harvard T.H. Chan school of public health, crews of aircraft have higher rates of cancer than the general population. The researchers listed cosmic rays, irregular sleep habits, and chemical contaminants as leading risk factors. Somewhat more controversial studies (#1, #2, #3, #4) llink cosmic rays with cardiac arrhythmias and sudden cardiac death.

En route to the stratosphere, our sensors also pass through aviation altitudes:

In this plot, dose rates are expessed as multiples of sea level. For instance, we see that boarding a plane that flies at 25,000 feet exposes passengers to dose rates ~10x higher than sea level. At 40,000 feet, the multiplier is closer to 50x. The higher you fly, the more radiation you will absorb.
Technical notes:
The radiation sensors onboard our helium balloons detect X-rays and gamma-rays in the energy range 10 keV to 20 MeV. These energies span the range of medical X-ray machines and airport security scanners.

Data points in the first graph ("Stratospheric Radiation") correspond to the peak of the Regener-Pfotzer maximum, which lies about 67,000 feet above central California. When cosmic rays crash into Earth's atmosphere, they produce a spray of secondary particles that is most intense at the entrance to the stratosphere. Physicists Eric Regener and Georg Pfotzer discovered the maximum using balloons in the 1930s and it is what we are measuring today.

When looking for casinos to play online when the weather is bad, you can try luotettavat nettikasinot for Finnish games. If you are not from Finland you can try the Swedish page Svenska casino online to find suitable games, check out svenskacasinoonline.net. Always check your local laws before playing with real money.

Lera Blog

Naming a Star for a loved one could not be easier with Star Registry. This novelty certification is a great addition to any wall and will last a lifetime. Buy a Star as a gift as a unique and thoughtful idea for all space lovers.

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SpaceWeather.com -- News and information about meteor showers, solar flares, auroras, and near-Earth asteroids

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Never miss another geomagnetic storm. Sign up for Space Weather Alerts and receive a text message when auroras appear. Aurora tour guides and professional astronomers use this service. Now you can, too! PERSEID METEOR UPDATE: Friday the 13th is a lucky day for sky watchers.

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Explore - The Night Sky | Hubble's Messier Catalog

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Some galaxies, nebulas and star clusters are visible with backyard telescopes, binoculars or the unaided eye. Explore a new Hubble gallery of such objects from the popular Messier catalog and learn how you can view these objects yourself!

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sky map skymap celestial astrophoto virtual telescope interactive online stars galaxy galaxies nebula constellation