Beginner Astronomy

Amazing Astronomy Facts •When you look at the Andromeda galaxy (which is 2.3 million light years away), the light you are seeing took 2.3 million years to reach you. Thus you are seeing the galaxy as it was 2.3 million years ago. •Light from the sun takes 8 minutes to reach you, thus you see the sun as it was 8 minutes ago. It might have blown up 4 minutes ago and you wouldn't know about it! •The Earth is not a sphere! It actually is an oblate spheroid, it is squashed slightly at the poles and bulges out at the equator due to its rotation. •Spare a thought for the constellations that never made it into the official list... these include Machina Electrica (the electricity generator), Officina Typographica (The Printing Office), and Turdus Solitarius (the solitary thrush) •When Galileo viewed Saturn for the first time through a telescope, he described the planet as having "ears". It was not until 1655 that Christian Huygens suggested the crazy theory that they might be an enormous set of rings around the planet. •If you could put Saturn in an enormous bathtub, it would float. The planet is less dense than water. •A teaspoon-full of Neutron star would weigh about 112 million tonnes. •Jupiter is heavier than all the other planets put together. •Even on the clearest night, the human eye can only see about 3,000 stars. There are an estimated 100,000,000,000 in our galaxy alone! •The tallest mountain in the solar system is Olympus Mons, on Mars at a height of about 15 miles, three times the height of Mount Everest. It covers an area about half the size of Spain. •If the sun were the size of a dot on an ordinary-sized letter 'i', then the nearest star would be 10 miles away. •Half-a-billionth of the energy released by the sun reaches the Earth •Temperatures on Venus are hot enough to melt lead. •If you could travel at the speed of light (186,000 miles per second) it would take 100,000 years to cross our galaxy! •Only one side of the moon ever faces Earth. The moons period of rotation is exactly the same as it's period of orbit. •Betelgeuse, the bright star on Orion's top-left shoulder, is so big that if it was placed where the sun is, it would swallow up Earth, Mars and Jupiter! •If you stand on the equator, you are spinning at about 1,000 mph in as the Earth turns, as well as charging along at 67,000 mph round the sun. •On the equator you are about 3% lighter than at the poles, due to the centrifugal force of the

Earth spinning. •The atmosphere on Earth is proportionately thinner than the skin on an apple. •On Mercury a day (the time it takes for it to spin round once) is 59 Earth-days. Its year (the time it takes to orbit the sun) is 88 days- that means there are fewer than 2 days in a year! •If a piece of the sun the size of a pinhead were to be placed on Earth, you could not safely stand within 90 miles of it! •Its estimated that the number of stars in the universe is greater than the number of grains of sand on all the beaches in the world! On a clear night, we can see the equivalent of a handful of sand. •Every year the sun evaporates 100,000 cubic miles of water from Earth (that weighs 400 trillion tonnes!) •Jupiter acts as a huge vacuum cleaner, attracting and absorbing comets and meteors. Some estimates say that without Jupiters gravitational influence the number of massive projectiles hitting Earth would be 10,000 times greater. •Astronomers believe that space is not a complete vacuum- there are three atoms per cubic metre. •Saturn is not the only planet with rings- Neptune has it's own ring system.

How Do they Know That...Part 1 You've probably seen in books and magazines facts and figures they give for different things, like "Sirius is 8.7 light years away..." and "Each second the sun lets out enough energy to...". If you're anything like me you'll wonder: how on Earth do they know that? Always take claims with a pinch of salt- an astronomer's viewpoint is limited to one point in time and space in the universe, hence, we can't really be sure of anything. For instance I find it difficult to accept that we know exactly what happened 0.000000000001 seconds after the 'big bang', when we know only that the age of the universe is somewhere between 10 and 20 billion years old. However, being so limited makes deducing information about the universe a challenge, and the ingenuity of some of the methods fascinates me.

Part 1: The Solar System Earth How big is the Earth? How heavy is the Earth? How far away is the Moon? Sun How heavy is the sun? How far away is the sun?

Planets How far from the sun are the Planets? How heavy is Jupiter? Part 2: The Stars and Beyond>>

Radius of the Earth

The Greek mathematician and astronomer Eratosthenes calculated this in 200 B.C.. He knew that on a particular day in Syene, the sun would be directly overhead. He set up a stick in Alexandria 5,000 stadia away (800-900 kilometres) on this day, and measured the angle of the shadow that the sun formed at midday. He measured this to be 7.2°. He deduced that this is the same angle from the Earth's centre that the distance between the two cities represents. Since there are 360° in a circle, he worked out that 5,000 stadia is one-fiftieth of the circumference of the Earth. He therefore

suggested the value of 250,000 stadia (40,000-46,000km). Astronomers recently measured it using satellite equipment and submitted the figure of 40,008 km. From this, using the geometry of circles, he could calculate the Earth's radius:

Pretty good for a man who lived 2,200 years ago who only used only a stick!

Distance to the Moon

The distance to the moon can be determined by considering a very large imaginary triangle, and was first put forward by the Greek Hipparchus about 100 years after Eratosthenes suggested the radius of the Earth. The moon is known to be directly overhead in one place, so if the place can be found where the moon is on the horizon at this time and the distance measured, it gives us the base for a very long, thin triangle (see diagram). The angle at A is a right angle, so:

Where θ is the angle at the Earth's centre, which can be calculated by:

Where a is the measured arc length. Using this method Hipparchus estimated the distance to the moon to be 59 times the radius of the Earth (397,000km), which is not far off the modern accepted figure (382,000km).

How far away is the sun? How far are the planets from the sun? The first simple and very important question is actually quite difficult to answer, and an accurate figure eluded science until the seventeenth century. The same method as for the moon was tried, but failed due to the massive distance involved. To find the answer to the first question, we have to go the long way round and address the latter question first. Johannes Kepler studied the orbits of the planets and found that the time period of a planet's orbit squared in years is equal to the distance from the sun cubed in AU (that is, astronomical units, or the distance from the Earth to the sun):

So Earth has an orbital period of one year, so is 1 AU from the sun. Mars has an orbital period of 1.88 years (these orbital periods had been known since ancient times by observation) and so it's distance from the sun is:

This could likewise be repeated for all the planets, so that the distance from the sun for all the planets was known in terms of the distance from the Earth to the Sun (the AU): Planet Mercury Venus Earth Mars Jupiter Saturn

Distance from the sun 0.38 AU 0.723 AU 1.000 AU 1.524 AU 5.204 AU 9.582 AU

Astronomers only had to work out the actual distance to just one planet to find the distances to all the rest, including the Earth. This was achieved in 1672 when Giovanni Cassini measured the distance to Mars using Parallax. This is a similar method to the one for finding the distance to the moon, forming a tall thin triangle. The situation is more complex than the measurement of the distance to the moon however, as the two observers couldn't be on opposite sides of the world because they wouldn't both be able to see Mars at the same time and the triangle isn't necessarily a neat right-angled or isosceles triangle, but the principle is the same. (Click here if you want to see the mathematics in more detail) Cassini sent a colleague to French Guiana while he remained in Paris. At an agreed time the two measured the position of Mars against background stars (assumed to be at an infinite distance). When they met back again, they compared the two positions and measured the tiny angle of parallax. Knowing this, and the distance (through the Earth) between the two locations (which can be calculated as the radius of the Earth is known), they could form a large triangle and calculate the length of one of the sides (which gives the distance between Mars and Paris).

The figure they obtained was 93% accurate compared with the modern figure (78,000,000 km minimum between Mars and Earth).

The night they chose to do this was when Mars was at opposition to Earth, i.e. it was near enough exactly on the other side from the sun, giving the positions shown on the diagram on the left. The distance that they measured, EM on the diagram, is clearly equal to M-E. Since earlier we worked out that M=1.524 E from Kepler's laws, we can use simultaneous equations to get E, the distance from the Earth to the sun in kilometers:

This is an extremely important result as it means that we can now work out the geography of the entire solar system, as we now know that 1 AU = 150,000,000 km: Planet Mercury Venus Earth Mars Jupiter Saturn

Distance from the sun 58,000,000 km 108,000,000 km 150,000,000 km 228,000,000 km 779,000,000 km 1,433,000,000 km

(When planets were discovered later their distances were calculated likewise). More accurate estimates of the AU were made in 1771 using the same method on Venus during a transit (when it passes in front of the sun). Nowadays we can use this method but instead of measuring the distance to Venus with parallax, we can measure it directly and extremely accurately with radar. A beam of radiation is aimed at the planet and the time it takes to receive the echo is measured using an atomic clock. Because the speed of light is known exactly the distance to Venus is half the time (the signal travels there and back) divided by the speed of light.

How heavy is the Sun? A few years after Kepler died, Isaac Newton was born. In 1687 he published Philosophae Naturalis Principia Mathematica which was one of most influential books in physics. In it he published his theory of gravity. His contemporaries suspected that Kepler's laws could be a result of a centreseeking force inversely proportional to the object's distance, but no-one could prove it. Isaac Newton presented the correct mathematical proof that this was the case, and, that it was also proportional to the two masses involved (If you're mathematically-minded I've written out such a proof- that an inverse square force produces elliptical orbits amongst others- on the gravity pages). This lead to Newton's Universal Law of Gravitation:

Where G is a constant known as the Universal Gravitational Constant, which we will learn more about later. We can now apply this to work out the masses of celestial objects. If we assume that the planet moves in a circle (which is a good assumption for many orbits... the same calculation can be applied

to eliptical orbits- the mathematics just becomes a bit more complex when you allow for eccentricity) then there is a constant, centre-seeking (centripetal) force. For any object moving in a circle it is known that the centripetal force acting on it is given by:

Where m is the mass of the object, r is the radius of it's motion and T the time for 1 full revolution. In our case of objects in orbit, this centripetal force is gravity, so we can equate the two forces:

Notice the similarity between this and Kepler's law above. This was the proof of Kepler's third law. Now we have a formula relating time of orbit, distance from sun (in real measurements, not AU), and mass. Rearranging the formula to make mass the subject gives:

So we can find an object's mass if we know the time period of and distance to an orbiting object. But how do we know what the value of 'G' is? That is a very good question, to which Newton did not really know the answer. It was not until 1798 that a reliable figure was established by an experiment conducted by Henry Cavendish. He set up two masses on either end of a rod hung by a string, free to turn. He moved two very large masses near the two smaller masses and measured the twist in the string that this produced (for more detail about the experiment see this page by the Physics Classroom). His experiment resulted in a value of 6.75x10-11 for G, which is not far from it's modern value, 6.67x10-11. This is an extremely small number, which explains why we only appreciate gravitational force when it comes to very large masses. We are now free to calculate the mass of any object which has a satellite (if we know it's period and distance) This includes the Sun (by looking at any of the planets) Earth (by looking at the moon) or Jupiter (by looking at it's moons). The Earth has a period of 365.24x24x60x60 seconds (31,556,296 seconds) and orbits at a distance of about 150,000,000 km (150,000,000,000 metres) so the mass of the sun is:

How heavy is the Earth?

The Moon has a period of 2,551,443 seconds and orbits at a distance of about 384,403,000 metres (its orbit may be assumed to be roughly cicrular) so the mass of the Earth is:

However, there is a more accurate way to determine the Earth's mass that doesn't depend on the accuracy of our knowledge of the distance to the moon. An object's weight on Earth is equal to the Earths gravitational pull on it, using Newton's second law that F=ma (force = mass x acceleration):

The 'm' cancel's out and so the acceleration, (which we will call g), of any object toward Earth at a distance 'r' (the radius of the Earth) is:

The value of g can be accurately be determined by experiment to be 9.8ms-1. Rearranging then, to get M, the Mass of the Earth is:

How heavy is Jupiter?

A similar method can be applied to find the mass of Jupiter, Saturn, or any planet with an observable moon. First, though, we need to find out how to work out an object's size from observation, if we know how far away it is. Things look bigger close up than they do further away. This obvious truth is because an further object has a smaller angular diameter than a close one (see diagram). We can measure directly an objects angular diameter from where we are, so if we know how far away it is we can work out how big it is. The actual size of an object is calculated by:

Where θ is the object's angular radius, d is the distance to the object, and s is the actual radius of the object (see diagram on the right). You could think of it as parallax in reverse- the triangle is upside down. Now, we know how far away Jupiter is, so by measuring the angular size of the radius of the orbit of Jupiter's moons, we can work out it's actual radius by using the method above. Let's look at one of Jupiter's four large moons, Io. We can calculate the radius of it's orbit to be 421,600 km using this method. We observe it to take 152,841 seconds to orbit the planet. Using the same calculation for the mass of the sun, then, we can calculate Jupiter's mass:

Note that the principle that you can find the size of an object if you know how far away it is applies to anything; whether it be objects here on Earth, the Moon, the Sun, Neptune, The Orion Nebula, or even whole Galaxies. It can even be used to find the size of craters on the moon, mountains on Mars or the size of Jupiter's great red spot. Bear in mind as well that now you know the size and mass of objects, you can easily calculate their volume and average density.

Have a go yourself, try this question (you'll need a calculator, and maybe a piece of paper and pen)

When Saturn is at opposition to Earth (when the Sun, Earth, and Saturn are in a line) the radius of orbit of Titan, one of Saturn's moons, is measured. It has an angular size of 194 arcseconds (that's 194/3600ths of a degree). It's time period is observed to be 1,377,648 seconds. You know that Saturn is 1,443,500,000,000m from the sun and Earth is 149,600,000,000m from the sun (G is 6.67x10-11). What is the mass of Saturn? All the methods needed are listed above. Think- what do you need to know to work out an central objects mass?

Answer: To find Saturn's mass we need the time period and radius of an object orbiting it. To find the radius of orbit we need to know how far away it is. The Sun, Saturn and the Earth are in a straight line, so the distance from Earth to Saturn is the distance from the sun to Saturn take away the distance from Earth to the Sun:

Titan's orbit has an angular radius of 194 arcseconds, so now we know its distance we can calculate its real radius of orbit:

Now we know Titan's orbital radius and time period, we can calculate the mass of Saturn:

How Do they Know That...Part 2

Part 2: The Stars and Beyond Stars How far are the stars? How bright are the stars? How much energy do they give out? How hot are the stars? Galaxies How far away are galaxies? How far away are galaxies?

The Universe How old is the Universe? <<Part 1: The Solar System

How Far are the Stars?

This is a question that scientists have always been trying to answer, and is so difficult simply because they are so far away and appear to be at an infinite distance, because they don't appear to move relative to each other. This caused the ancients to assume that they were fixed on a sphere. When it was realised that they were all like our sun, and they all occupied specific positions in space, people tried to calculate the distances to them.

If you have read the previous article you will understand the principle of parallax- that objects appear to be in different places on a background if the position of the observer changes. Hold out your finger in front of you and cover each eye. Your finger appears to jump about- this is an example of parallax.

The angle that it appears to move through is called the angle of parallax. The greater the ratio between the distance to the object and the distance between the two observing points the smaller the angle and thus the harder it is to detect and measure. The picture on the right shows how this method can be used to calculate the distance to Mars. It can be calculated by considering the very large, tall thin triangle with the radius of the Earth as the base:

Once θ is measured and since r is known (the radius of the Earth) d, the distance to Mars, can be calculated.

How can we use this method to calculate the distance to the stars? The distances we are dealing with are so fantastically incredibly unbelievably huge the radius of the Earth is too small as a baselinethe angle of parallax is too small- we need to go much bigger. The Earth orbits at 150 million km (see previous article to find out how they knew this in the olden days) so after six months it is 300,000,000 km from where it was before. This is the biggest baseline we can use, forming an enormous tall, thin triangle all the way to a distant star (see diagram on the right), and then calculate the distance with the same formula as above.

William Herschel in the 18th century to tried to measure this parallax, but couldn't- it was just too small. As methods progressed however, these tiny angles could be measured, and the distances calculated. The largest angle of parallax belongs to Alpha Centauri (and is therefore the closest star). It is just 0.76 arcseconds- that's only about 1/5000th of a degree! Using the formula, lets work out how far away Alpha Centauri is:

Give or take a little bit- remember the smallest error in the angle can make a quite large error in the distance. A star at a distance of 3.26 light years (that's the distance light travels in a year) would have a parallax of 1 arcsecond This gives rise to a new unit- the parsec meaning "a parallax of one arcsecond". There is a satellite in orbit called HIPPARCOS, which measures stellar parallaxes all day (well actually all night). It can measure parallaxes down to 0.001 arcsecond, that's 1/3600000 of a degree! In theory that's up to 10,000 parsecs, but in reality we can only rely on this method up to 100 parsecs away. HIPPARCOS has accurately calculated the distance to 120,000 stars in our galaxy, and up to 2.5 million with a lesser degree of accuracy. So what about further out than 100 parsecs? Well, we need to rely on data obtained for the nearby stars and what they tell us. I'll come back to this question in a minute. It's convenient to answer this question first:

How Bright Are the Stars? How much energy do they give out?

Lets clear up first, by brightness I mean luminosity, that's energy per second given off. As the distance from the radiant source (the star) increases, the flux energy, that is, the energy per unit area decreases. They are related by the equation:

This because the greater distance you go from the source, the larger area the energy has been spread over. This is useful because it says

that if we know how bright it looks (F, Flux, which is measurable) and how far away it is, we can tell how luminous it actually is. An early method of measuring the luminosity of the sun was to put an oil drop on a piece of paper , so that it was translucent. If it is held up to the sun, and a light source of known Wattage is placed on the other side, and it is set at a distance from the light source so that the oil droplet and the surrounding paper are observed to have the same brightness, then the flux from the sun is roughly equal to the flux from the light source. Knowing the ratio between the distances, can tell us how much brighter the sun actually is than the light source. We were very surprised in our Physics class when the figure we calculated turned out to be exactly right, that is 4x1026 Watts. Thus, in answer to the second question, it gives out 4x1026 Joules of energy per second, that's 4,000,000,000 big Hydrogen bombs per second or enough energy to keep the USA going for a 40,000,000 years! (That's in one second remember!) There are obviously more sophisticated ways of measuring flux now, but the principle is the same, if we know how far away it is (which we do from parallax for nearby stars), and we measure how much flux we receive from it, we can calculate its luminosity or brightness.

How hot are the stars?

Something else we can work out about stars, simply by looking at them is their temperature. When something gets very hot, it glows. You can see this when a blacksmith heats metal- it glows red, then yellow, then white. If you could carry on heating it up, we would see it turn blue. We can measure the spectrum of light these objects give off, and relate it to its temperature. This sort of radiation is called black body radiation- energy radiated due to a object's heat. For most things the spectrum peaks in the Infra-red range of light; the extremely hot metal's spectrum in the blacksmith's forge peaks in the red or yellow range. Its reasonable that the same goes for stars, we can observe their colour- or more exactly measure their black body radiation spectra and relate it to a general spectrum for a particular temperature measured here on Earth. The Sun's spectra, for example, closely approximates an object at 6000K (K is Kelvin, 1K=1°C but the scale is set so that 0K=-273°C). Wien's law relates temperature to the wavelength where the spectrum peaks:

The sun emits most at a wavelength of 500nm (500x10-9m), so has a surface temperature of:

The same method can be applied to stars by analysing their spectra.

Back to "How Far are the stars"

The luminosity formula is useful in reverse- if you an objects luminosity and the flux that we receive, we can calculate how far away it is. The principle is that if we see a star to be so bright, and we know how bright it really is, we can work out how far away it is to make it appear the brightness that it does; whether its a dim nearby star, or a far-away bright star. How do we find out an objects true brightness? From our observations of nearby stars, there are some types of star which have a particular luminosity. If we can identify distant stars of the same type, we can estimate their luminosity (similar to the nearby ones) and therefore calculate the distance to them. This method is called using "standard candles", because these stars are "standard" of known and specific brightness, so if we can identify them, we can immediately work out approximately the distance to them (note the assumption in this method that these "standard candles" have the same brightness- this introduces a considerable error margin into the calculated figure, however, taking averages of lots of standard candles in a cluster can reduce this error.) Some standard candles include: Cepheid Variables. Henrietta Swan Leavitt discovered in 1904 that the time period of a type of star called a cepheid variable, (which varies in brightness over a regular period) is directly related to its luminosity. Thus if we identify a cepheid variable (don't ask me how the know its a cepheid and not some other sort of variable star) and measure it's time period (the time for one cycle of brightness), we can calculate its luminosity. Main Sequence stars. What is the main sequence I hear you cry? Well, Scientists started to notice a relationship between surface temperature and luminosity (as described above), and plotted it on a graph. Mr Hertzsprung and Mr Russell pioneered this technique so it's called, surprise, surprise, a Hertzsprung-Russel Diagram. Here are

some stars for which temperatures and luminosities are known plotted in such a fashion:

Note that most of the stars lie in very specific areas. The line down the middle is known as the main sequence, where most stars (including the sun) fit. The top right group are red giants, and the bottom left white dwarves. In reality it is a bit more complicated, with countless subgroups of different types of stars. Light from stars on the main sequence all have the same signature in their spectra. Light from the star is split into different colors and compared to light from samples of chemicals here on Earth. From this we can determine the composition of the star, and stars on the main sequence all have similar composition. So, we've looked at a distant star and calculated it's surface temperature (the x-axis on the HR diagram) from the above method (How hot are the stars?). We can also see that it belongs on the main sequence from it's spectrum. So we can look up from its temperature on the HR diagram and read off it's luminosity (with some error margin, as the main sequence isn't a perfect thin line). There are other standard candles which I won't go into, I'm sure you get the idea. Bear in mind as well that if stars are in a cluster, and are therefore all at a similar distance from here we need to find just one standard candle in the cluster and we know the distance to all the other stars in the cluster. The distances to vast numbers of individual stars in our galaxy can now be calculated (all the individual stars you

see by the way are in our galaxy). So what about beyond our galaxy...?

Some other things we can now work out:

Before I leave individual stars just a quick note that we can apply some of the a earlier methods for finding sizes of distant objects to star clusters. Now we know the distance to them, by measuring their angular size (sizes of stars themselves cannot be calculated this waythey are too far away)we can work out their actual size. This is how they can tell some star clouds and clusters are light-years wide! Also, there are lots of other relationships on the HR diagram that tell us about stars: their mass and size in particular- there are relationships between mass, temperature and luminosity, and size, temperature and luminosity which show up on the HR diagram. I'm not going to go into this now, if you want information in more detail see this useful page on the HR diagram. So, now we have found (albeit with some uncertainty due to the inaccuracies, limitations and assumptions of the methods) the mass, size, luminosity, surface temperature and distances of millions of stars in our galaxy! I think that's pretty good seeing as all we started out from is a bit of basic geometry!

How far away are Galaxies?

The same "standard candle" method can be used on galaxies. Cepheid variables are quite bright so can be picked up even in other galaxies. The above method can be used to calculate the distance to the whole galaxy. Another standard candle useful in galaxies are type 1a supernovae. This is when a star of a particular type dies spectacularly, letting out so much energy that it can briefly outshine it's whole galaxy. These are rare events, but as there are so many stars in galaxies, they happen regularly somewhere in some galaxy out there. Scientists reckon that Type 1a Supernovae have a fixed maximum luminosity, so once one is identified (by its spectrum) we can use this known luminosity to calculate an estimated distance to the galaxy. This method works until the galaxies get so far away we can't pick up these stars. So how do we work out distance to far-away galaxies? Edwin Hubble (yes folks, he had a telescope named after him) was

analysing some galaxy spectra, and noticed that the peaks and troughs matching elements on Earth didn't occur at exactly the same wavelengths- they had been shifted toward the red end of the spectrum (you've probably heard of red-shift- more about it later). He theorised that this redshift is caused by the source moving away from us, an effect known as the doppler effect. You have probably heard when a car is approaching you quickly it's engine sounds high pitched, then when it goes past and is therefore moving away, it sounds low pitched. This is because when an object's coming toward you, waves emitted from it (whether sound or light) are 'squashed up', reducing their frequency. With light, this is shifting it toward the blue end of the spectrum, so it's called blue-shift. When an object is moving away from you, the opposite occurred, the waves are 'stretched out', with light this would be a red-shift:

A simplified form of the doppler equation is:

Where v is the speed of the wave and u is the speed of the source relative to the observer, and z is doppler shift. Going back to our situation, z is the redshift of the galaxy, u is the speed of the galaxy (which we'll call v), and v is the speed of light, c. So the speed of a galaxy moving away from us is:

Hubble calculated the distance to lots of nearby galaxies (using cepheid variables) and also measured their redshift, thereby calculating their velocity. He plotted a graph of velocity against galaxy distance for nearby galaxies, and found that speed of recession is directly proportional to distance:

v = H0d Where v is speed, d is distance and H0 is a constant known as the Hubble constant- a hugely important number, which we'll come on to later. Combining the previous two equations eliminating v:

So now distance to faraway galaxies can calculated by measuring their redshift. In theory this gives us the distance to any galaxy in the universe. However, its not really that simple- we've made a large assumption that this linear relationship applies throughout the whole universe. Also, when plotted on a graph, the points for nearby galaxies aren't in an exact straight line, so there is some doubt about the value of H0- values range from 50 km s-1 Mpc-1 to 100 km s-1 Mpc-1- a factor of two which is quite a large uncertainty (although Hubble- I'm talking about the telescope now, not Edwin Hubble- recently measured it to be about 70 km s-1 Mpc-1 to an accuracy of 10 percent which is a great improvement).

How heavy are Galaxies?

Remember how we found the mass of the Sun and Jupiter on the first page- using the equation for gravity and circular motion for it's orbit. The Mass of a central object orbited by some satellite is given by:

Galaxies rotate- we can easily measure the speed of this rotation of a side-on galaxy (allowing for the galaxy's overall red-shift and taking it as fixed) as the stars on one side are moving toward us, and the stars on the other side are moving away from us. We can calculate the speed of either side, and thus calculate it's rotation speed.

We can also easily work out the size of the galaxy by measuring its angular size, and, knowing the distance, we can calculate it's size. Taking an individual star near the edge, we can work out it's speed and distance from the center, and put it into the formula (as the satellite), and calculate the mass of all the material inside it's orbit, which is most of the galaxy (the central object). This leads on to the matter of dark matter. When astronomers do these sums they find that the galaxy is far too heavy for the amount of matter we can see- the stuff that's luminous, radiating light- by some estimates 90% of the mass in the universe is missing!. Thus they infer that there must be some "dark matter" matter that doesn't radiate at all. There are several suggestions for the nature of dark matter; black holes, brown dwarves, neutron stars, WIMPs (weakly interacting massive particles), MACHOs (Massive Compact Halo Objects- that is matter in the 'halo'- the mass surrounding the central part of a galaxy) and kinds of other crazy stuff, but I'm not going to really going to go into that, as I'm digressing (if you want more information about dark matter here's a good page)

How old is the universe?

OK, the big one- the age of the universe. When Hubble showed that all the galaxies in the universe are receding from us, he realised that this indicates that the universe is expanding. If its expanding, the further back in time we go the smaller the universe is, and since this rate is constant it must have had zero size at some point- the universe must have had a beginning, a start point, an instant of creation. How long ago was this? Well, since: v = H0d

And if a galaxy's speed is constant, the distance it has traveled after time t at speed v is: d=vt Putting these two together to eliminate d, v also cancels out leaving: H0t=1 Therefore the time that the galaxy has been moving for, t, i.e. the age of the universe, is:

Isn't that neat? I told you H0 was important. However, this has its limitations- it depends on the value of H0, the value of which as I mentioned earlier is in doubt. Here's the calculation using the upper and lower bounds for H0 (in s-1):

This gives us an order-of-magnitude estimate, but clearly, we can't know for certain. Remember also that it is also based on assumptions about the motion of galaxies and the expansion of the universe, about which it is difficult to know for certain whether these are correct. Astronomers are currently working on more reliable ways to find a figure for the age of the universe, usually involving observing very distant galaxies at the edge of the universe- you may have seen the famous deep-field Hubble picture:

These galaxies are so far away that light has taken billions of years to get here, so we see them as they were billions of years ago, near the beginning of the universe. Astronomers get more idea of what the early universe was like by studying such photographs. More accurate estimates of H0 are being sought as well to give more reliable estimate the age of the universe (as I said Hubble has recently obtained a figure to within 10% accuracy). Current estimates of the age of the universe are round about 15 billion years, but this is constantly being revised. There you go- hopefully now you understand where astronomers get some of their figures from. Remember, though the limitations of the experiment- very little, if anything, is proven in astronomyall we can do is look up from this one point is space and gather what evidence we can, speculate on, theorise, and infer information from this evidence. We certainly don't know it all at the moment, there's still (and will always be) a lot more out there to learn!

Gravity Introduction You probably already know that the force of gravity is in inverse square to the distance between the two objects involved:

Analysing a circular orbit, where this force is constant, is easy. In the general case it is not however, and the orbit is often elliptical. This was one of the great questions of science in the 17th century. Kepler had shown that planets orbits are ellipses with the sun at one focus, but did not know why this should be the case. He was of the opinion (as was everyone else of his time) that for an object to move, it must have a force pushing it from behind. It was suspected later that an inverse square force toward the focus (the sun) could be causing the elliptical orbits, but no-one could prove it. Isaac Newton proved it in 1687 in his book Philosophae Naturalis Principia Mathematica, and named the force gravity. This article sets out the proof (probably not the same as Newton's!) that if an object moves under the influence of a central inverse square force, its path is a conic section (don't worry, conic sections will be explained later), also how in real life this relates to gravity (and hence how my orbit simulator works!) First of all you'll need to understand motion described in polar co-ordinates and conics. I know this site is called astronomy for beginners but you don't mind if I put one complicated bit in do you? Good. This one's for you mathematicians and/or physicists out there, knowledge of calculus and a grounding in mechanics is needed to understand the following page (in particular you'll need to know about conics, the polar equation of a conic and using radial unit vectors) that is, proof that an inverse square attractive force produces conic motion. If you don't feel up to that you might be able to skip the proof and just have a look at the last page, the actual maths behind orbits.

Gravity as an Inverse Square Force We showed on the last page that an object moving under an inverse square force...

...moves in the polar equation:

r is the distance from the centre of the force, θ is the angle between the object, the centre and the (arbitrary) axis, A is an arbitrary constant and h is a constant such that:

That is that at any point in its motion, it's angular speed times the distance from the centre is constant. This formula is remarkably similar to that of the general polar equation of a conic:

Thus the objects path is a conic with the central object (e.g the sun) as its focus, where:

(Interesting fact: Focus is the latin word for 'fireplace'. In his study of conics and orbits, Kepler realised that the sun was at one focus of the conic, hence, he called it fireplace.)

Conics Conics is the name given to a set of functions, because all of them can be acheived from cutting up a cone. They are a circle (cutting horizontally) an ellipse (cutting at an angle, but an angle less that the slope of the cone) a parabola (cutting parallel to the slope) and a hyperbola (cutting vertically). Remarkably, all of these shapes have the same polar equation. Defining a point as the focus, the distance r from the focus of a point on the line at an angle Θ with the axis is:

l is just a scaler, determining the size of the shape, the type of shape is determined by e, the eccentricty of the shape. Here are the four conics and the ranges of values of e which produce them:

Circle e=0

Ellipse 0<e<1

Parabola e=1

Hyperbola e>1

Gravity Now to apply these last two sections to gravity. We know that the force of gravity is given by:

Comparing this to our general formula, we can see that G M = k We found that the centre of the force was the focus of the conic- if we are equating this to gravity, the sun is at the focus of a planet's orbit. This proves Kepler's first law.

What is h in the specific case of gravity? Remember that h is the constant equal to the angular speed of the object times its distance from the centre. Taking the point nearest the object, where its velocity is wholly transverse:

Where r0 is the closest the object comes to the centre (θ = 0) and v0 is the speed at that point, it's initial speed if you like. Its also worth noting at this point that we have just proved another one of keplers laws, that planets 'sweep out' equal areas in equal times. It can be shown by considering a small time Δt and calculating the area of the resulting 'triangle' that since r Θ (dot) is constant, the area it covers in a given time is constant, since it must travel faster when r is small and slower when r is large. So how do we find e? Substituting A h2 / k for e in the inverse square path, and setting Θ = 0 and hence r0 = 0:

We said earlier that k is G M in the case of gravity (that is the mass of the cetral object times the universal gravitational constant). Remember that e determines the type of conic produced- we now have e in terms of real-life varaibles such as "initial" distance (closest approach) r0, "initial" speed (speed at closest approach) v0, the universal gravitational constant G and the mass of the central object, M. We can now tell whether an object will orbit, and what shape orbit it will have.

=0

1>

>0

Object orbits in a circle: such as the Earth (nearly!).

=1

Object does not orbit: It swings round the central object and it is sent nearly in the opposite direction.

Object orbits in an ellipse: such as planets (nearly zero) or comets (nearly 1).

>1

Object does not orbit: It is deflected slightly off its course by the central objects gravity.

Amazing isn't it- the elegance of it that something so complicated should reduce to neat little constant which determines whether and how orbits, and that such theoretical mathematics applies s perfectly in the real world. Funny as well, that the secrets to something so vast as the solar system and even the universe lay hidden in the humble cone. Of course you can substitute in the other variables to get the equation fully in terms of real-life things, here it is if you want.

Its also possible from this to calculate the time period of the orbit. This ties in perfectly with Kepler's laws, as it produces something of the form T2 = w a3, (where a is the distance from r0 to the centre of the ellipse, not the focus). Newton showed that the actual value of w, the constant is:

There you go- all three of Keplers laws are proved by this method:

• • •

A planet's orbit is an elipse with the sun at its focus. A planet's orbit sweeps out equal areas in equal times. The square of semi-major axis (a) of a planets orbit is proportional to the cube of it's orbital time period.

We've shown that these are true if the object in orbit is under a centripedal, inverse-square, attractive force, which we call gravity. This is how my orbit simulator works- you can type in the variables (r0 and v0), and see if you can get the satellite orbiting the Earth in a circle. In thoery it should be an exactly scaled down model, with seveal million miles per pixel and so many months per second., using the equations above and the correct values of G and M (the mass of the Earth). Sorry but it only seems to work in Internet Explorer and Opera, but you can try other browsers (doen't work in Mozilla Firefox, but let me know if it works in any others)

Planet Data Reference

Mercury

Venus

Earth

Mars

0.3302

4.8685

5.974

0.6419

1,898.6

468.46

86.83

102.43

0.0125

Volume 1010km3

6.083

92.843

108.321

16.318

143,128

82,713

6,833

6,254

0.0125

Mean Radius km

2,440

6,052

6,371

3,370

69,911

58,232

25,362

24,624

0.715

Surface gravity (g) ms-2

3.7

8.9

9.8

3.7

24.8

10.4

8.9

11.2

0.58

Number of Moons

0

0

1

2

63

33

26

13

1

Distance from Sun (semi-major axis) 106km

57.8

108.2

149.6

227.9

778.6

1,443.5

2,872.5

4,495.1

5,869.7

Min distance from sun (perihelion) 106km

46.0

107.5

147.1

206.6

740.5

1,352.6

2,741.3

4,444.5

4,435.0

Max distance from sun (aphelion) 106km

69.8

108.9

152.1

249.2

816.6

1,514.5

3003.6

4,545.7

7,304.3

Eccentricity

0.21

0.0067

0.017

0.094

0.049

0.057

0.046

0.011

0.24

Inclination of rotation to orbit

0.01°

177.36°

23.45°

25.19°

3.13°

26.73°

97.77°

28.32°

122.55°

Length of "Day" (Rotation period) hrs

4,223

2,802

24.0

24.7

9.9

10.7

17.2

16.1

153.3

87.969

224.701

365.256

686.980 4,332.589

10,759.22

30,685.4

60,189

90,465

Erm...

Unknown

Unknown

Unknown

William Herschel (1781)

Johann Galle (1846)

Clyde Tombaugh (1930)

1

0.006

>>1000

>>1000

>>1000

>>1000

10-6

CO2, N2<>

H2, He

H2, He

H2, He

H2, He

CH4, N2

-63

-108

-139

-197

-201

-223

Mass 1024kg

Length of "Year" (Orbital period) days Discoverer Atmosphere Pressure bar Atmosphere Composition Average

Unknown Unknown 10-15

92

O2, Na, CO2, H2 N2 167

464

N22

15

Jupiter

Saturn

Uranus Neptune

Pluto

Temperature °C

Source: National Space Science Data Center/NASA

What To See? The Moon The obvious place to start is the moon- big, bright and very easy to find. It's probably the first thing you looked at when you bought your first telescope. Through a small telescope, very close detail can be seen. The lunar landscape is pitted with craters, and these appear different every day in the moon's cycle as the light catches them at different angles. A full moon, although spectacular,

through the telescope doesn't really look so impressive (plus the brightness hurts your eyes), but a in a half moon, the sun casts wonderful shadows which look great in the eyepiece. The darker areas are called mare (plural of maria), which means 'sea'- these are great flat plains on the moon's surface. One of these, Mare Tranquillitatis (sea of tranquility) is where the Apollo mission landed. The very large crater towards the bottom is called Tycho. When the moon is full you can see the impact lines extending out over what must be hundreds of miles. Look at it when the moon is half-full as well- it looks magnificent as the light casts shadows among the craters. If you want to go into more detail studying the lunar landscape, an excellent photographic map can be found at moon-phases.com

Crater Tycho at Full Moon Crater Tycho at Half Moon

A Lunar eclipse occurs when the Moon passes through the Earth's shadow. Light shining through the Earth's atmosphere is the only light to get through onto the moon, and when light has to travel through a lot of air, it appears redder (that's why sunsets are red). This makes the moon turn bright red, a spectacle that occurs every six months or so. Click here to find out if there are any coming up.

Planets The main planets that are clearly visible to a small telescope are Venus, Mars, Jupiter and Saturn. These appear in the sky to be very bright stars, but they move relative to the fixed background stars. How do you know you're looking at a planet? Stars appear as pinpoints of light however much magnification you use, but the major planets have a definite size, appearing as small spheres through the telescope. Planets are best viewed with high-power eyepieces.

Venus

Venus is the brightest object in the sky (apart from the moon and sun of course). It can be seen in the early evening or the early morning earning it the name morning/evening star. You can only see it at these times as, because it's orbit is inside Earths, it can never be too far from the sun (it can never pass over the night-time side of Earth), only revealing itself after the sun has set or before it has risen. It will rise up, reach it's highest point in the sky, then dip down and pass across the sun then become the "morning star" then the same in the morning. Through the telescope it is dazzlingly bright, and not much actual detail can be seen on it as the planet is completely covered be cloud, but with a high-power eyepiece you can observe it's changing phases, in the same manner as the moon.

Mars

Mars is the next planet in our solar system out from Earth. Mars can be seen as having a very deeporange colour in the sky, like a bright star but not brilliant like Venus. It passed very close recently, and I was able to see it's polar regions, a white icecap at the top with my Meade ETX-70. Mars has seasons, much like Earth, meaning sometimes it's South pole is facing us and sometimes the North pole. With a slightly larger telescope further details may be seen, such as dark patches which come and go on the planet's surface.

Jupiter

Jupiter is the biggest planet in the solar system, so appears quite large through the telescope. It's very bright, probably second only to Venus, so is quite easy to find. Through any telescope, four of it's largest moons are visible, Io, Europa, Ganymede and Calisto. Galileo discovered these as soon as he turned his newly-invented telescope to the planet. On higher powers on a very clear night through a small telescope, the cloud bands on the planet itself can be seen as well. The moons are in a different position every night, making it an interesting object to observe. On larger telescopes you can see the great red spot and also the shadow of the moons from time to time as they pass in front of the planet- this is called a transit, and they happen very frequently due to the fast orbital speed of the moons.

Saturn

Saturn appears as a yellowish, bright star, though not brilliant like Venus or Jupiter. And although Saturn is much further away than other planets, it's good to observe because it's very big, and because of it's magnificent rings, and it's these rings which make it unmissable for anyone with a telescope. The rings are just visible at 40 x magnification, so should be within the range of most telescopes, through a slightly larger telescope on a very good night you may be able to see the division between the two rings, called the Cassini division. The rings are actually hundreds of tiny rings made up of lumps of ice and rock. Galileo, on turning his telescope to the planet, concluded that it must have ears! It was not until much later that someone suggested the implausible notion that they were giant sets of rings around the planet. The angle that Saturn faces us at changes throughout it's orbit, so when you see it once you may be looking straight down on the rings, and on another occasion looking at them on their side.

Deep Sky Objects These are objects outside the solar system, which remain fixed in their positions on the 'celestial sphere'. These include galaxies, nebulae, and star clusters. When you first look through your telescope on a clear night your mind will be boggled by the sheer number of stars even in barrenlooking areas of sky. Generally, they are best viewed through wide angle, low-power eyepieces.

The Pleiades (M45)

Picture by Joe Roberts

The Pleiades are an easily recognisable star cluster, resembling a tiny version of the 'big dipper'. Also called the seven sisters, usually about 6-8 stars are visible with the naked eye. However, through the telescope, 50-100 stars are visible, making it an incredible sight. You'll need an eyepiece with a wide field of view to get the whole cluster in view. It is a group of very young, hot, bright stars, very close together. The stars are surrounded by a slight nebulosity which is illuminated by the star's light, and is thought to be the remnants of the larger nebula from which the stars were formed. The Pleiades are best seen about October-February.

Andromeda Galaxy (M31)

Picture by Nasa/Hubble

The Andromeda Galaxy is the nearest galaxy to our Milky Way, and is one of very few that can be seen with the unaided eye. It is thought to contain about 400 billion stars and is about 2.2 million light years away. Don't expect to see it like the photograph unless you have a large telescope. To most of us through a small telescope it will appear as quite a large fuzzy blob. When it is at it's highest and on a good dark night, you realise what you were seeing before was just the bright

nucleus in the picture- you begin to see the edges of it very faintly and it is much larger than you thought. It is best seen about September-January.

Great Orion Nebula (M42)

Picture by Joe Roberts

The Orion Nebula is one of those 'must see' objects for anyone with an interest in astronomy. It is bright enough to be seen with the naked eye as the middle 'star' in Orion's sword. With a small telescope you can see a dark lane running through a fuzzy patch. Try diverting your eyes away from it, and you'll be able to see more structure in it (for some reason our peripheral vision seems to be better at picking up faint light than looking straight at a dim object.) You don't see it in red like it comes out in pictures, but on a very clear night you can recognise its structure. It is best seen in the winter months, from about October to January.

The Double Cluster

Picture by Joe Roberts

The double cluster can be found in Perseus and appears to the naked eye as a large, indistinct patch of dim stars. Through the telescope, however, it looks wonderful. You can see two nuclei about which all the stars are gathered in the cluster.

Beehive Cluster (M44)

Picture by Joe Roberts

An open cluster, similar to the double cluster, a dense patch of dim stars barely visible to the naked eye, but again magnificent through the telescope.

Great Hercules Cluster (M13)

Picture by Joe Roberts

The great Hercules cluster is a globular cluster similar to cluster M5 shown in the picture, but different to the beehive 'open cluster' in that it's very densely packed stars in a smaller sphere, rather than a loose grouping of several stars over quite a large area. It appears as a patch of fuzz through the telescope, and on a slightly higher power I could almost pick out individual bright stars in the centre

The Celestial Sphere

The Messier Catalogue is a list of interesting objects in the sky, distinguished from ordinary stars. In 1758 French astronomer Charles Messier began compiling a list of nebulae- diffuse objects so that

they wouldn't be mistaken for comets. Each object was designated an 'M' number- and astronomers still use this today to identify objects. For instance, the Pleiades are known as M45. There are 110 Messier objects. The Messier catalogue is of interest to amateur astronomers because, as these objects were discovered in the 18th century, they are usually bright enough to be enjoyed with a modest telescope. Some people claim to have spotted every single one with such equipment.

An excellent site about the Messier Catalogue can be found at SEDS.org, with detailed information and pictures of each object.

Some Highlights of the Messier Catalogue • • • • •

M13 Great Hercules Cluster M31 Andromeda Galaxy M42 Great Orion Nebula M44 Beehive Cluster M45 Pleiades

Star Catalogues Since ancient times certain stars have been given names. However not all stars can be given a proper name, and such a system would only prove confusing, giving us no information about the whereabouts of the star. Here's some of the ways that you may hear a star referred to, and their meanings.

•

•

•

•

Greek-letter system One of the earliest attempts to catalogue the stars was the greek-letter system, devised in the early 17th century. According to this system stars are named according to their constellations, and assigned a greek letter, usually according to it's brightness. For example, the brightest star in Lyra (Vega) is given the name Alpha Lyrae. The second brightest is called Beta Lyrae then Gamma, Delta Lyrae and so on. This is useful to amateur astronomers as it mainly covers the stars we can see, and gives us some idea of it's whereabouts and it's brightness. Flamsteed numbers Later on, in the 18th Century astronomer John Flamsteed introduced a numbering system for stars in each constellation, where visible stars are numbered Eastwards, so the Easternmost star in Centaurus is 1 Centauri, the second 2 Centauri etc. BD system BD, the Bonner Durchmunsterung (Bonn Survey) system came next as telescopes were revealing hundreds of thousand of stars. The sky was divided up into 1 degree bands of Declination, and stars numbered according to where they came around the circle. For instance the star BD + 52°1245 is the 1,245th star (counting from Right Ascension 0°) in the area between 51° and 52° declination. SAO catalogue SAO, the Smithsonian Astrophysical Observatory star catalogue, is one of the most widely-used catalogues today. It covers most stars down to about the 9th Magnitude, in which stars are numbered according to their Right Ascension co-ordinates.

As people first started to map out the night sky, it became necessary to have a standard, universal way of plotting positions of objects in the sky. As the stars appear to occupy fixed positions in the sky relative to each other, a convenient way of thinking about the situation would be to imagine the Earth placed at the centre of a larger sphere. The stars occupy fixed positions on the surface of this 'celestial sphere', and the Earth rotates within it. This is obviously not the case in real life but it is a good model because this is pretty much how things look from our point of view.

The Earth rotates within this sphere, but of course, to us it appears as though the celestial sphere is rotating, on the same axis as the Earth. If the line of the Earth's axis were extended to infinity, the point where it meets the celestial sphere is called the celestial pole. If we take the Earth as fixed, all the stars appear to rotate around this point. In the Northern Hemisphere, we are fortunate to have a star more or less exactly on the celestial pole, Polaris. If you go out on a clear night and look at a certain constellation several times over a period of hours you will notice it appears to rotate around the celestial pole.

Declination Now imagine the Earth's Equator extended out to infinity, the corresponding point on the celestial sphere is the celestial equator. Just as the angle between the pole and the equator on Earth is a right angle, the same is true of the celestial sphere. For this reason points on the celestial equator are said to have a declination of 0° and the celestial pole is said to have a declination of 90°. Note that declination is analogous to latitude on Earth, where the pole has a latitude of 90° and the equator has a latitude of 0°. Of course, we can also imagine lines in between of declination 10°, 20°, 30° etc., corresponding to latitude on Earth, and note also that lines South of the equator have negative declinations (e.g. -76° is 76° South). Parts of a degree are measured in arcminutes (60 arcminutes per degree) and seconds (60 arcseconds per arcminute). So you may see a particular declination given as -7° 27' 03", which means 7 degrees 27 minutes and 3 seconds South, which is just under 7 and a half degrees South of the celestial equator. In reality arcseconds are much too tiny for us to worry about, rounding to the nearest half-degree is usually more than accurate enough.

The relationship between declination and latitude does have a real significance, as it enabled sailors for thousands of years to find out their latitude, but I'll come on to that later.

Right Ascension

Obviously we can't set points with only one coordinate, another set is needed, just as longitude was needed on Earth to be able to define a point completely. Right Ascension (RA) is the celestial equivalent of longitude. Just as longitude doesn't have so much of a natural significance, so was arbitrarily centred on Greenwich Observatory, London, RA was defined to be zero at the first point in Aries.

Because the Earth is turning the stars appear to turn- but since there there is no movement NorthSouth lines of declination stay fixed- this is why they are more significant. As the stars appear to rotate East-West, lines of RA move- they complete one full turn per day (well, actually a little bit more due to the fact the Earth has moved about 1° round the sun in a day, but this is small enough to be ignored day-to-day). This means that a particular star moves through 15° per hour. So, it is convenient for us to measure Right Ascension in hours, where 15 degrees is an hour and 360 degrees is 24 hours. This is further divided up into minutes and seconds (not to be confused with arcminutes and arc seconds, which are not the same thing) in the same way that time is divided, i.e. an hour is 60 minutes, a minute is 60 seconds etc. If you were looking at a star through your telescope say at 50° declination and 0hrs RA, in 1 hours' time you would be looking at 1hr RA, then 2 hours later at 2hrs RA and so forth. In fact, after say, 3 hours, 56 minutes and 12 seconds, you would be looking at 3hrs 56mins 12sec Right Ascension! So now we have two co-ordinates, we can plot any point on the surface of the celestial sphere. For instance the Orion Nebula has co-ordinates -5° 27' Dec, 5h 35m 24s. RA The Andromeda Galaxy has coordinates 41° 16' Dec, 0h 42m 42s RA. Important to grasp is that although these objects appear to move, their co-ordinates are fixed because they stay the same relative to each other.

The View From Here... How does this look from where we are? Well, this depends on your position on the Earth.

Observer A is at the North Pole (90°), Observer B is at 45° latitude and Observer C is on the Equator (0°). Here is what they see:

Observer A has the Celestial pole directly above him, being at the Earth's pole. The stars simply rotate in a circle around our observer. The celestial equator is on the horizon- he can only see stars in the celestial Northern Hemisphere. Observer C has the celestial equator directly above him, being at the Earth's Equator. The celestial poles are at opposite ends of the horizon and the stars arc over head, directly East-West. From here is the only latitude you get to see all the stars because equal amounts of the Northern and Southern Hemisphere are visible, and as Earth turns all positions on the celestial sphere are visible at one point. I came to Observer B last, because most of us are in his position. Our observer is at 45° latitude but this applies to anywhere between the two extremes of pole and equator. The observer at 90° latitude had 90° declination directly above him (at his zenith), and the observer at 0° latitude had 0° declination at his zenith. The same applies to all latitudes, if our observer at 45° latitude looked straight up, he would be looking at a point on the 45° line of declination. I live on 51° latitude, so the line of 51° declination passes directly overhead from my position. Generalising this, the angle between the celestial equator and the zenith, (that is, the declination) anywhere on Earth is equal to the latitude at that point. This leads us to some useful results:

From geometry, it is easy to see that angle p, the elevation of the celestial pole from the horizon, is equal to angle l, the latitude. This is very important in navigation, as it allowed ancient sailors to tell their latitude by measuring the elevation of the celestial pole (in the Northern Hemisphere, this is easy because Polaris, a bright star lies near enough on the pole.) So if you're ever lost in the snaky swamps of some strange country, or get shipwrecked and you're marooned on some desert island, you know how to find your latitude- the elevation of the celestial pole. Note also that the celestial equator always intersects the horizon exactly at East and West, wherever you are. To find the celestial equator, find the pole first. Look 90° one way, then the other, these are East and West. Then turn round through 180° so you're facing the opposite way. At this point (opposite the pole) the equator is at an elevation from the horizon of 90° - your latitude, and it runs from East, through this point, then down to meet the horizon again at West. It will help you if you learn where the celestial pole and equator are where you live. Now you can see how the stars appear to move: following the line of the celestial equator, rotating around the pole. If you want to see a demonstration of this look at these time-lapse photographs taken from all over the world, showing the apparent motion of the stars. Note how the position of the celestial pole varies with latitude, e.g. on Mt Kilimanjaro 3° south the (south) pole is 3° above the horizon.

Yearly Motion As I mentioned earlier, the stars daily rotation is just over a full circle, about 1° over- because the Earth has moved about 1° round the sun. This means that over the course of a year, the stars appear to complete one whole turn. If you went out and looked at a particular constellation over a period of months at the same time, it would appear to slowly move around the pole in the same direction as the daily motion- about 30° (or 2 hours of Right Ascension) a month. This also means that the constellations obscured by daylight slowly come round to be visible at night as the year progresses. So do I get to see all the stars on the celestial sphere at some point? Well, not usually, unless you live exactly on the equator. The pole is inclined at an angle equal to your latitude, so the equator is below the horizon at an angle of 90° - your latitude. Stars above the declination of 90° - your latitude never set. For instance, I live at 51° latitude, so the pole is at an elevation of 51°. Stars at 39° declination at their lowest point just touch the horizon, and all the stars at a declination above 39° never set:

(Note the diagram is drawn illustrating the Northern Hemisphere, but it applies equally in the Southern Hemisphere, if you reverse the directions.) Angle l is your latitude. The stars inside the white line never set, either daily or yearly. Outside this line the stars rise and set at different times of the day and year. Conversely, looking opposite the pole:

The stars below the white line, those with a declination greater than 90° - your latitude below the equator at their highest point still are below the horizon, so they never rise, either daily or yearly. This means you only ever see stars in your Hemisphere (declination greater than 0° in the Northern hemisphere and declination less than 0° in the Southern hemisphere), and the stars in the other hemisphere with a declination of 90° minus your latitude.

Equatorial Mounts Now you might understand the equatorial telescope mount better. It aligns the telescope with the Earth's axis, so that instead of moving up and down and turning around (alt-az mount), it moves in the Declination direction and the Right Ascension direction. This means you only have to turn the RA axis to keep an object centred. It also means you can look up the celestial co-ordinates of an object, then set the telescope's dials to the right values and you should be looking right at it. To align an equatorial mounted telescope, first of all you need to make sure that it is set at an angle equal to 90° - your latitude facing the pole. You can use a compass to get the right direction in the Southern Hemisphere, in the Northern Hemisphere, just point it at the North star. It obviously

involves a little more care than that, and not owning an equatorial telescope, I'm not really qualified to give an expert tutorial on it, however here are some links to pages to help you.