Posts Tagged ‘Hubble Space Telescope’

My latest book, Astrophotography, is now available. You can order a copy by following this link. Astrophotography is a book of exquisite images of space, including some of the latest images such as New Horizons’ images of Pluto, Rosetta’s images of Comet 67P, and Hubble Space Telescope images of the most distant galaxies ever seen. Each stunning image, reproduced to the highest quality, is accompanied by text that I have written to explain the object, and any background science relating to the object.


Astrophotography is now available. You can order your copy by following this link.

One unique aspect of Astrophotography is that it emphasises the multi-wavelength approach taken to understanding astronomical objects. For millennia we could only study the Universe in visible-light (the light to which are eyes are sensitive), but for the last few decades we have used every part of the electromagnetic spectrum from radio waves to gamma rays to better understand the Universe. This multi-wavelength approach has also enabled us to discover previously unknown aspects of the Universe such as the Cosmic Microwave Background, the true appearance of Venus’ surface which lies hidden below its thick atmosphere, and huge quantities of gas between galaxies (the intracluster medium) which emit no visible-light but prodigious amounts of X-rays.

Astrophotography is split into 5 sections, namely

  1. Exploring the Solar System
  2. Exploring the Milky Way
  3. Exploring the Local Group
  4. Beyond the Local Group
  5. At the Edge of the Universe

Below are examples of some of the beautiful images found in Astrophotography, along with examples of the accompanying text. At the beginning of each page’s text I caption which telescope or space probe has taken the main image, and at which wavelength (or wavelengths).

Exploring the Solar System

Two examples from the first section of Astrophotography, the section on the Solar System, are stunning images of Mercury and of Mars. The images of Mercury were taken by NASA’s MESSENGER spacecraft. There are several pages of images of Mars, the page shown below shows an image of the Martian surface taken by the Mars Curiosity Rover, and an image of Victoria Crater taken by the Mars Reconnaissance Orbiter.


Images of Mercury taken by NASA’s MESSENGER spacecraft. The four main images are spectral scans, and show information on the chemical composition of Mercury’s surface.

The section on the Solar System also includes images of Pluto taken by New Horizons, images of Saturn and Titan taken by the Cassini space probe, images of Comet 67P taken by Rosetta, and images of Jupiter and her moons taken by the Galileo space craft.


The surface of Mars as imaged by NASA’s Mars Curiosity Rover and, at right, Victoria Crater, as imaged by NASA’s Mars Reconnaissance Orbiter.

Exploring the Milky Way

The second section of Astrophotography includes images of the Orion Nebula (Messier 42), the reflection nebula Messier 78, the Horsehead Nebula, the Pillars of Creation (part of the Eagle Nebula), and the Crab Nebula, the remnant of a supernova which exploded in 1054.

The example I show below is of the reflection nebula Messier 78, and is a visible light image taken by the Max Planck Gerzellschaft Telescope, a 2.2 metre telescope located at the European Southern Observatory’s facility in La Silla, Chile. The text describes the history of observing Messier 78, and explains what produces a reflection nebula.


The reflection nebula Messier 78 imaged in visible light by the Max Planck Gesellschaft Telescope. The text explains what reflection nebulae are, and the history of observing this particular object.

Exploring the Local Group

The third section of Astrophotography looks at the Local Group, our part of the Universe. The Local Group includes our Milky Way galaxy, the Large and Small Magellanic Clouds, and the Andromeda galaxy. Some of the images shown in this section include the Tarantula Nebula in the Large Magellanic Cloud, NGC 602 (in the Small Magellanic Cloud), the Andromeda galaxy, Supernova 1987A and the Seahorse Nebula.

The example I show here is the Seahorse Nebula, a dark cloud of gas and dust located in Large Magellanic Cloud. This Hubble Space Telescope image was taken in 2008, and the nebula is in the bottom right of the image.


The Seahorse nebula is a dark cloud of gas and dust found in the Large Magellanic Cloud, an irregular galaxy visible to the naked eye and in orbit about our Milky Way galaxy. The seahorse nebula is in the bottom right of the image.

Beyond the Local Group

The fourth section of Astrophotography looks at the rich variety of galaxies found beyond our own neighbourhood. Examples are galaxies like Messier 82, which is undergoing a huge burst of star formation in its centre, Centaurus A, which shows huge lobes of radio radiation stretching far beyond the stars we see in visible light, colliding galaxies such as The Antennae galaxies, and evidence for dark matter such as the Bullet cluster.

The example I have shown here is the spread for Messier 81, a beautiful spiral galaxy found in Ursa Major. It is one of the best known galaxies in the sky, and is visible to northern hemisphere observers throughout the  year. The main image illustrates the multi-wavelength approach astronomers take to studying many objects. The image combines visible light, infrared light and ultraviolet light to teach us far more about the galaxy than we would learn if we only looked in visible light.


Messier 81 is a beautiful spiral galaxy found in Ursa Major. Hence it is visible throughout the year to northern hemisphere observers. The main image shown here is a combination of of a visible light image (taken by the Hubble Space Telescope), an infrared image taken by the Spitzer Space Telescope, and an ultraviolet image taken by Galaxy Evolution Explorer (GALEX).

At the edge of the Universe

In the final section of Astrophotography, I show examples of some of the most distant objects known. Images include the Hubble Deep Field, the Cosmic Microwave Background, the most distant galaxy seen (GN-z11, lying about 13.4 billion light years away), gravitational lenses and the recent discovery of gravitational waves made by LIGO.

The example I show here is the spread about the gravitational lens SDP81, a galaxy lying about 12 billion light years away which is being lensed (and brightened) by an intervening cluster of galaxies which lie about 4 billion light years away. The top image was taken at millimetre wavelengths by the Atacama Large Millimetre Array (ALMA), the bottom image in visible light by the Hubble Space Telescope.


Gravitational lenses enable us to see distant galaxies which would otherwise be too faint to see, but they also provide us with a way of tracing the distribution of dark matter in clusters.

I hope these few examples from Astrophotography have whetted your appetite to find out more. I really enjoyed putting the book together, and am very pleased with the quality of the images and their aesthetic beauty.

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Earlier this week it was announced that NASA’s Hubble Space Telescope had observed evidence for water geysers shooting from the surface of Europa, one of Jupiter’s larger moons. Here is a link to NASA’s press release. I was on BBC TV talking briefly about this on Tuesday (27 September), the day after NASA’s announcement.


NASA has announced that the Hubble Space Telescope has observed water geysers emanating from the south pole of Jupiter’s moon Europa.

In fact, this announcement was additional evidence to add to a finding which had first been announced in 2013. In December 2012, astronomers used a spectroscope on Hubble to look in ultraviolet wavelengths at Europa. They found auroral activity near the moon’s south pole, and upon analysis of the spectrum of the UV emission from this auroral activity they found the spectral signatures of hydrogen and oxygen, the constituents of water.

Those 2012 observations have since been followed up using a different method. This time astronomers have observed how the Sun’s light, which is reflected from Jupiter, is affected as it passes Europa. As Europa transited in front of its parent planet, astronomers looked for signs of absorption of this light near the limb of the moon, which would be due to gases associated with Europa. Such a technique can, for example, be used to find and study the atmosphere of an extra-solar planet as it passes in front of its parent star.

Whilst not finding any evidence that Europa has an atmosphere, what the team found was that absorption features were seen near the moon’s south pole. When they calculated the amount and extent of material required to produce these absorption features they found that their results were consistent with the 2012 finding. They calculate that water jets are spewing out from the surface of Europa and erupting to a height of about 160 km from the moon’s surface.

We have had evidence since the Voyager mission in the 1980s that Europa has an ocean of water below its icy surface. This evidence was further enhanced during the Galileo mission in the 1990s. Where there is water there may be life, so it is possible that Europa’s ocean is teeming with microbial life. To find out, we need to directly study the water in this sub-surface ocean.

Unfortunately, due to the thickness of the icy crust covering its ocean, studying this water directly poses a huge challenge. We currently don’t have the capability to drill through such a large thickness of ice, although it is certainly something we would hope to do in the future. This discovery of water jets provides a much easier way to sample the water directly, and so it is quite feasible that NASA and/or ESA could send a probe to fly through the jet, take a sample of the water, and analyse it to see whether there are any signs of microbial life. This is very exciting, and is why this discovery of water geysers erupting on Europa is so important.

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This story caught my attention in the last few weeks. It is from the Universe Today website, and here is the link to the story. The two exoplanets in the story, TRAPPIST-1b and TRAPPIST-1c, are only 40 light-years away, which by cosmic standards is very close. They have been studied in near-infrared light using the Hubble Space Telescope’s Wide Field Camera 3 (WFC3), which was put onto Hubble during a last servicing mission in May 2009.


This story appeared recently on the Universe Today website.

By studying the planets in near-infrared light we can look at how sunlight is absorbed by different gases in a planet’s atmosphere. This method was pioneered by Gerard Kuiper working in the 1940s. This article mentions that the atmospheres of these two exoplanets have been shown to be “compact” like the Earth and Venus, rather than “puffy” like the gas giants Jupiter and Saturn.

How can we determine this from near-infrared studies done here? Rather than looking at reflected light from the parent star, instead these studies used the passing of the two planets in front of a background star (not the host star). By looking at the absorption lines produced by the two planets’ atmospheres, not only can the gases in them be determined, but by looking at the details of the absorption lines one can determine the temperature and pressure of the gas. This is an example of how powerful a technique spectroscopy is in determining the physical nature of gases.

To find two planets so nearby which could potentially harbour life is quite exciting. I am surprised this has not been a bigger story in the press.





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Just over a week ago, on March 3, the news broke that the Hubble Space Telescope has found the most distant galaxy seen to date, GN-z11. At a spectroscopically measured redshift of z=11.1, it breaks the previous record of z=10.7 (which is a redshift currently based on the photometric redshift method, rather than spectroscopy). The previous record for a galaxy which had been spectroscopically measured was z=8.68, so this new discovery breaks that record by some margin.

Here is an image of the galaxy. It is superimposed on an image of a survey called GOODS North to show the part of the sky where the galaxy was found.



An image of the most distant galaxy seen so far, at a redshift of z=11.1; meaning we are seeing it as it was when the Universe was only some 400 million years old.

Here is a screen capture of the summary of the Space Telescope Science Institute’s press release, you can find the actual press release here.



A screen capture of the summary of the press release announcing the discovery of a galaxy at a redshift of z=11.1, which has been measured by spectroscopy rather than based on the photometric redshift technique.

You can see in the text of this press release summary that the galaxy is referred to as “surprisingly bright”, and below I show the beginning of the preprint of the paper announcing the result (dated March 3). You can read the preprint for yourself by following this link. Again, the galaxy is referred to as “remarkably luminous” in the preprint. Today I just wanted to present this exciting story, but next week I will explain more about a “schoolboy error” (or undergraduate error 😉 ) I made in discussing the brightness of this galaxy with colleagues.



A screen capture of the preprint submitted to Astrophysical Journal announcing the discovery of a galaxy at a redshift of z=11.1

But, more on my schoolboy undergraduate error next week; as I say today I just want to present the story. At a redshift of 11.09 +0.08 and -0.12 (pretty small errors), this galaxy is being seen when the Universe was only some 400 million years old, or to put it another way we are seeing it some 13.3 billion years ago! Truly remarkable.


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One of the most iconic images to be taken by NASA’s Hubble Space Telescope in the last 25 years is the image it took in 1995 of the so-called “pillars of creation”, part of the Eagle Nebula. To celebrate the Hubble’s 25th anniversary, NASA has recently re-imaged this star-forming region, the image below shows the comparison.

A comparison of the 1995 and 2014 images of the "pillars of creation".

A comparison of the 1995 and 2014 images of the “pillars of creation”. The newer image covers a larger field-of-view and shows more detail than the 1995 image.

The differences between these two images are due to several factors. The more recent image was taken with a newer camera, WFC3, which is the 3rd generation optical camera on the Telescope (the 1995 image was taken with the 2nd generation optical camera, which was called WFPC2). The newer camera has both a better dynamic range, a larger field of view and a higher resolution than WFPC2.

Another reason for differences is more subtle – the pillars are slowly being destroyed by radiation outflows from the stars being formed in the pillars, and the newer image allows us to see subtle changes in the structure of the pillars. It is even possible to see the outflowing gas directly in some places (or rather the light emitted from the outflowing gas due to its excitation by the ultraviolet photons from the hot young stars).

NASA has also released a new near-infrared image of the pillars, and this image nicely shows how longer wavelength light is more able to penetrate the gas and dust than visible light. However, because the density of dust is so high in these giant molecular clouds, they do not appear totally transparent at these near infrared wavelengths; but rather as wispy ghosts rather than the thick pillars we see at visible wavelengths. Some background stars are visible in the near infrared, but many are still hidden, as are the actual star-forming regions which lie deep within the pillars.

This near-infrared images nicely shows how these longer wavelengths can penetrate some of the gas and dust which surround regions of star formation.

This near-infrared images nicely shows how these longer wavelengths can penetrate some of the gas and dust which surround regions of star formation.

The final image below shows the location of the pillars in the Eagle Nebula, which is otherwise known as Messier 16 (or M16). It is a nebula which is relatively easy to find with even a small telescope in the constellation Serpens.

This image shows where in the Eagle Nebula the "pillars of creation" are located.

This image shows where in the Eagle Nebula (M16) the “pillars of creation” are located.

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Last week, as part of the “tech hour” on BBC Radio 5’s “Afternoon Edition” on Wednesdays, I heard one of their reporters do an item on visiting a facility in Australia which is trying to track space debris and locate their positions precisely. After the item a listener asked about geostationary orbits, and it was clear from the reporter’s answer that he did not really understand them. Then again, if he is not a trained physicist or engineer, why should he?

In the light of the wrong answer given to the listener by the reporter, I’ve decided to blog about satellite orbits to clear up any confusion, and also to show that the mathematics of it is not that hard. This is the kind of problem that a student studying physics would be expected to do in their last year of high-school here in Wales. Mark Thompson, one of the astronomers who has appeared on BBC Stargazing Live, did clear up some of the confusion when he came on the programme later in the same hour, but he did not have time to go into all the details (nor would he want to do so on a radio show!).

Low-Earth and geostationary orbits

For those of you not familiar with the terminology, a low-Earth orbit is typically just a few hundred kilometres above the Earth’s surface, which as I will show below leads to an orbit about the Earth which takes a couple of hours at most. These are the kinds of orbits which are used by most spy satellites, and also by the International Space Station (ISS) and, when it used to fly, by the Space Shuttle. The famous Hubble Space Telescope is in a low-Earth orbit. Any satellites in low-Earth orbits will experience drag from the Earth’s atmosphere.

Space “officially” begins at 100km above the Earth’s surface, but the atmosphere doesn’t suddenly stop at any particular altitude,it just gets thinner and thiner. In fact, the “scale height” of the atmosphere is about 1 mile, or about 1.5km. For each 1.5km altitude, the amount of atmosphere roughly halves (based on pressure), so at an altitude of 3km the air is 0.5 \times 0.5 = 0.25 (25\%) as thick, and if you go up to 10.5km which is roughly the altitude at which commercial aeroplanes fly, then the thickness of the atmosphere is about 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.5^{7} = 0.0078 as thick, or 0.78 \% of the thickness at the ground. Even though the thickness of the atmosphere a few hundred kilometres up is very very thin, there is still enough of it to slow satellites down. So, satellites in low-Earth orbits cannot maintain their orbits without some level of thrust, otherwise their orbits would just decay and they would burn up in the atmosphere.

Low-Earth orbit satellites can be in any orbit as long as it is centred on the Earth’s centre. So, for example, many spy satellites orbit the Earth’s poles. If you think about it, if they are taking about 90 minutes to orbit (some take less, they will be even closer to the Earth), then each time they orbit they will see a different part of the Earth, which is often what you want for spy satellites. The ISS orbits the Earth in an orbit which is titled to the Earth’s equator, so it passes over different parts of the Earth on each orbit, but it is not a polar orbit. A satellite cannot, for example, orbit, say, along an orbit which is directly above the tropic of Cancer, and satellites do not orbit the Earth in the opposite direction to the Earth’s rotation, they all orbit the Earth in the same direction as we are rotating.


Geostationary orbits, on the other hand, are a particular orbit where the satellite takes 24 hours to orbit the Earth. So, as seen by a person on the rotating Earth, they appear fixed relative to that person’s horizons. Geostationary orbits are used for communication satellites so that, for example, your TV satellite dish can be fixed to the side of your house and point in the same direction, as the satellite from which it’s getting the TV signals stays fixed in your sky. They are also used by Earth-monitoring satellites such as weather satellites and satellites which measure sea and land temperatures. A geostationary satellite has to be at a particular altitude above the Earth’s surface, and it has to be in orbit about the Earth’s equator (so it sits above the Earth’s equator), and of course orbits in the same direction as the Earth’s rotation.

How do we calculate the altitude of these respective orbits, the low-Earth and the geostationary ones? It is not too difficult, as I will show below.

The altitude of a satellite which takes 90 minutes to orbit the Earth

To make the mathematics simpler, I will do these calculations assuming a circular orbit. In reality, some satellites have elliptical orbits where they are closer to the Earth’s surface at some times than at others, but that makes the maths a little more complicated, and it’s not necessary to understand the basic ideas.

As I showed in this blog, when an object is moving in a circle it must have a force acting on it, because its velocity (in this case its direction) is constantly changing. This force is called the centripetal force, and as I showed in the blog this is given by

F_{centripetal} = \frac{ mv^{2} }{ r }

where m is the mass of the object moving in the circle, v is its speed, and r is the radius of the circle.

For a satellite orbiting the Earth, the centripetal force is provided by gravity, and the force of gravity is given by

F_{gravity} = \frac{ G M m}{ r^{2} }

where M is the mass of the Earth (in this case), m is the mass of the satellite, r is the radius of the orbit as measured from the centre of the Earth, and G is a physical constant, known as Big G or the universal gravitational constant.

Because the centripetal force is being provided by gravity, we can set these two equal to each other and so we have

\frac{ mv^{2} }{ r } = \frac{ G M m }{ r^{2} } \rightarrow v^{2} = \frac{ GM }{ r } \text{ (Equ. 1) }

The speed of orbit, v is related to how long the satellite takes to orbit via the definition of speed. Speed is defined as distance travelled per unit time, and as we are assuming circular orbits the distance travelled in an orbit is the circumference of a circle. So, we can write

v = \frac{ 2 \pi r }{ T }

where T is the time it takes to make the orbit. Substituting this expression for v in Equation (1) we can write

\left( \frac{ 2 \pi r }{ T } \right)^{2} = \frac{ GM }{ r }

Remember we are trying to calculate r for a satellite which takes 90 minutes to orbit, so we re-arrange this equation to give

r^{3} = \frac{ GM T^{2} }{ 4 \pi^{2} } \text{ (Equ. 2)}

Notice that the only two variables in this equation are r \text{ and } T, everything else is a constant (for orbiting the Earth, obviously if the object were orbiting e.g. the Sun we’d replace the mass M by the mass of the Sun). Equation (2) is, in fact, just Kepler’s Third law (which I blogged about here), which is that

\boxed{ T^{2} \propto r^{3} }

Now, to calculate r for a satellite taking 90 minutes to orbit, we just need to plug in the values for the mass of the Earth (5.972 \times 10^{24} kg), the value of G=6.67 \times 10^{-11}, and convert 90 minutes to seconds which is T=5400 \text{ s}. Doing this, we get

r^{3} = 2.942 \times 10^{20} \rightarrow r = 6.65 \times 10^{6} \text{ m}

But, remember, this is the radius from the Earth’s centre, so we need to subtract off the Earth’s radius to get the height above the Earth’s surface. The average radius of the Earth is R = 6.37 \times 10^{6} \text{ m} so the height above the Earth’s surface of a satellite which takes 90 minutes to orbit is

6.65 \times 10^{6} - 6.37 \times 10^{6} = 281 \times 10^{3} \text{ m  } \boxed{ =  281 \text{ km} }

which is roughly 300km.

The altitude of a geostationary satellite

To calculate the altitude of a geostationary satellite we just need to replace the time T in equation (2) with 24 hours (but put into seconds), instead of using 90 minutes. 24 hours is 24 \times 3600 = 8.64 \times 10^{4} \text{ s}, and so we now have

r^{3} = 7.53 \times 10^{22} \rightarrow r = 4.22 \times 10^{7} \text{ m}

Again, subtracting off the Earth’s radius, we get the altitude (height above the Earth’s surface) to be

4.22 \times 10^{7} - 6.37 \times 10^{6} = 3.59 \times 10^{7} \text{ m  } \boxed{ = 35,900 \text{ km} }

which is roughly 36,000 km.


It surprises most people that geostationary satellites are so far above the Earth’s surface, some 36,000 km. There is no atmosphere at that altitude, so geostationary satellites really pose no threat in terms of man-made space debris. Their orbits will not decay, as there is no atmospheric drag. It is the low-Earth orbit satellites which are the source of our man-made space debris, and if their orbits are not maintained the orbits will decay and they will burn up in the Earth’s atmosphere. Of course, most satellites entering the thicker parts of the atmosphere burn up completely, but larger satellites have been known to have parts of them survive to hit the Earth’s surface. For example this is what happened to the Russian space station MIR when its orbit was decayed by switching off its thrusters.

The time it takes a satellite to orbit the Earth is entirely dependent on how far from the centre of the Earth the satellite is. Low-Earth orbit satellites orbit the Earth in a matter of hours, many less than two hours. Geostationary satellites have to be about 36,000 km from the surface of the Earth in order to take 24 hours to orbit. All satellites orbit the Earth in the same direction as the Earth is rotating, except for satellites in a polar orbit which essentially allow the Earth to rotate below them as they go around the poles. Geostationary satellites have to orbit above the Earth’s equator, but other satellites can be in an orbit which is inclined to the equator, as long as the centre of their orbit is the centre of the Earth.

Finally, coming back to the issue of space debris; one of the reasons that space debris is so dangerous to other satellites and astronauts is that objects in Earth-orbit are moving very quickly. For example, for an object taking 90 minutes to orbit the Earth, such as the orbit the ISS is in, we can work out how quickly (in metres per second) it is moving quite easily.

If T=90 \text{ minutes } = 5400 \text{ s} and the radius of the orbit (as we showed above) is r = 6.65 \times 10^{6} \text{ m } then

v = \frac{ 2 \pi \cdot 6.65 \times 10^{6} }{ 5400 } \; \; \boxed{=  7737.6 \text{ m/s } }

so, nearly 7,740 m/s or 7.7 km/s or nearly 28 thousand km/h, which by anyone’s standards is FAST!

The damage a piece of space debris can make is dependent on its kinetic energy (KE), which is given by

\text{Kinetic energy } = \frac{ 1 }{ 2 } mv^{2}

An object with a mass of 1kg moving at 7,740 m/s will thus have a KE of

KE = \frac{ 1 }{ 2 } 1 \cdot (7740)^{2} = 30.0 \times 10^{6} \text{ Joules}

(or 30 Mega Joules). This is a huge amount of energy, because the object is moving so fast. By comparison, a car, which we will assume to have a mass of 1,000 kg, moving at 100 km/h will have a kinetic energy of 0.77 \times 10^{6} \text{ Joules}, or nearly 0.8 Mega Joules, so about a factor of 40 (forty!) less than a 1kg object in low-Earth orbit. This is why such small objects can cause such damage!

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This story caught my attention last week – the discovery of water on an extrasolar planet which is as Neptune-sized. The exoplanet in question – HAT-P-11b, was discovered by the telescope Kepler in 2009 using the “transit method”. You can read more about that method here.

The star itself, HAT-P-11, is about 122 light years away in the constellation Cygnus. The planet has been determined to be about the size of Neptune, which means it is about five times the size of the Earth. Such relatively small planets can, to date, really only be found by the transit method, as the doppler shift method which discovered the first several hundred exoplanets is not currently sensitive enough to detect planets as small as Neptune.

Astronomers have discovered water in the atmosphere of a Neptune-sized exoplanet for the first time.

Astronomers have discovered water in the atmosphere of a Neptune-sized exoplanet for the first time.

A team has used the Hubble Space Telescope to observe the absorption spectrum produced as HAT-P-11b passes in front of its parent star. The light from the star, as it passes through the gases in the atmosphere of the planet, will have certain wavelengths removed, this is the principle behind the absorption spectrum that I explained here. The scientists have found that wavelengths corresponding to water vapour have been removed from the star’s spectrum, showing that water vapour exists in HAT-P-11b’s atmosphere.

This is tremendously exciting, as finding water is, with our current understanding, vital to finding life. This discovery shows that planets harbouring water are definitely out there, probably in their millions or billions. The next step will, hopefully, be finding water on a planet which is in the habitable zone, but this discovery is definitely a step in the right direction.

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A few weeks ago it was announced that a team had discovered what seems to be the most distant galaxy yet discovered. You can read the BBC story about it here, or if you like you can read the Nature science paper here to get as much detail as you could wish for. The galaxy, which has the catchy name z8_GND_5296, was discovered using the Hubble Space Telescope, with its distance being determined using the Keck 10m telescope on the summit of Mauna Kea.

In fact, what astronomers measure is not the distance of a distant galaxy, but its redshift, which astronomers denote with the letter z. Redshift is the movement of the spectral lines of a galaxy to longer wavelengths due to the expansion of the Universe, the expansion discovered by Edwin Hubble in 1929. The redshift of this newly discovered galaxy has been found by Keck to be z=7.51, beating the previous record of z=7.21. But how do astronomers translate this into a distance?


The cosmological definition of redshift

It turns out that measuring distances in astronomy is one of the most difficult things to do for several reasons. Not only are there very few direct ways to measure the distance to an object, after all we can hardly lay down a measuring tape between us and the stars and galaxies! But, to make it even worse, there also are various definitions of distance! In a future blog I will talk about the most direct ways we have to measure distance, but how we translate from these measurements to a distance also depend on the geometry of the Universe, which Einstein showed in his General Theory of Relativity is determined by the effects of gravity.

The geometry of the Universe is determined by its average density \Omega, and how this relates to something called the “critical density” \Omega_{0}, which is the dividing line between whether the Universe will carry on expanding forever, or stop expanding and start to collapse. If average density \Omega > \Omega_{0} the Universe will stop expanding and collapse. If \Omega < \Omega_{0} the Universe will carry on expanding forever, and if the average density \Omega = \Omega_{0} the Universe is on the dividing line between the two, and is said to have a flat geometry. Without going into the details here, most cosmologists believe that we live in a Universe where \Omega = \Omega_{0}, that is a flat Universe.

The preferred method for measuring large distances “directly” is to use something called a Type Ia Supernova, I will blog about this method again in a future blog. But, we can only see Type Ia supernovae out to distances corresponding to a redshift of about z=1. The galaxy in this story is much further away than this, z=7.51. So, to calculate its distance we have to use a model for the expansion of the Universe, and something called Hubble’s law.

The measured redshift of a galaxy (or any object) is just given by

z = \frac{ \lambda - \lambda_{0} }{ \lambda_{0} } \text{ (Eq. 1) }

where \lambda is the observed wavelength and \lambda_{0} would be the wavelength of a spectral line (usually for a galaxy it is a line called the Lyman-alpha line) in the laboratory.

As long as the redshift is much less than 1, we can then write that

z=\frac{ v }{ c } \text{ (Eq. 2) }

where v is the recession velocity of the galaxy and c is the speed of light. In the case of z not being less than 1, we need to modify this equation to the relativistic version, so we write

1 + z = \sqrt{ \frac{ 1+ v/c }{ 1 - v/c } } \text{ (Eq. 3) }

In our case, z=7.51, so we need to use this relativistic formula, and when we do we get that the recession velocity of the galaxy is 97\% \text{ of c }, the speed of light.

Re-arranging equation 1 we can write 1 + z = \frac{ \lambda }{ \lambda_{0} }. In principle, the distance and redshift are just related via the Hubble law

v = H_{0} d \text{ (Eq. 4) },

where v is the recession velocity of the galaxy, H_{0} is the Hubble constant, and d is the distance of the galaxy.

Things get a lot more complicated, however, when we take into account General Relativity, and its effects on the curvature of space, and even the definition of distance in an expanding Universe. I will return to this in a future blog, but here I will just quote the answer one gets if one inputs a redshift of z=7.51 into a “distance calculator” where we specify the value of Hubble’s constant to be H_{0} = 72 \text{ km/s/Mpc } and we have a flat Universe (\Omega=1) with a value of \Omega_{M}=0.25 (the relative density of the Universe in the form of matter) and \Omega_{vac} = 0.75 (the relative density of the Universe in the form of dark energy).


Putting these values in gives a co-moving radial distance to the galaxy of 9103 Mpc \text{ or } 29.7 \text{ billion light years}. (I will define what “co-moving radial distance” is in a future blog, but it is the distance quoted in this story, and is the measurement of distance which is closest to what we think of as “distance”).

The redshift also gives a time when the galaxy was formed, with z=0 being the present. We find that it was formed some 13.1 billion years ago, when the Universe was only about 700,000 years old.

A galaxy 30 billion light years away??

Going back to the “co-moving radial distance”, I said it is about 30 billion light years. A light year is, of course, the distance light travels in one year. So how can a galaxy be 30 billion light years away, implying the light has taken 30 billion years to reach us, if the Universe is only 13.7 billion years old?? This sounds like a contradiction. The solution to this apparent contradiction is that the Universe has expanded since the light left the galaxy. This is what causes the redshift. In fact, the size of the Universe now compared to the size of the Universe when the light left the galaxy is simply given by

1 + z = \frac{ a_{now} }{ a_{then} }

where a is known as the scale factor of the Universe, or its relative size. For z=7.51 we have a_{now} = (1 + 7.51)\times a_{then} = 8.51 a_{then}, so the Universe is 8.51 times bigger now than when light left the galaxy (this is what causes the redshift, it is the expansion of space, not that the galaxy is moving through space with a speed of 97% of the speed of light). It is the fact that the Universe is over 8 times bigger now than when the light left the galaxy which allows its distance measured in light years to be more than a distance of 13.7 billion light years that one would naively think was the maximum possible! So, there is no contradiction when one thinks about things correctly.

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Last week it was announced that the Hubble Space Telescope had discovered a previously unseen moon of the planet Neptune. Amazingly, this moon was not seen by Voyager 2 when it flew by in 1989.


The previously unseen moon is the 14th known moon about Neptune. It orbits once every 23 hours, and the image below shows where it is in the system of Neptunian moons. Notice the direction of motion of the moons about Neptune. The largest moon, Triton, which is also the furthest out, is orbiting Neptune in a retrograde motion, which means in the opposite direction to the direction the planet rotates. Not only does this suggest it is a captured object and did not form with Neptune, but more intriguingly it means that one day it will spiral into Neptune and get torn apart by tidal forces once it gets close enough (inside the so-called Roche limit).


This discovery just shows the power of Hubble, even as it approaches the end of its life. The camera used to find the unseen moon was the WFPC3, which had my PhD supervisor Professor Mike Disney of Cardiff University involved its design. So even though the Hubble is in Earth orbit, it was able to see something that Voyager 2 couldn’t in its fly-by.

The NASA press release, which gives more detail than the BBC story, can <http://herebe found here.


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