Moon's Orbital Radius Increasing

In an unexpected and unprecented move, the European Space Agency announced this morning that their data shows that the Moon has started to increase its mean orbital radius at a rate considerably higher than previously believed.  It's been known for some time that the distance from the Earth to the Moon is increasing, but the rate of increase is worrying.  In other words, the Moon is moving further away from the Earth faster than we thought, and in a few year's time will leave the Earth's orbit completely.  

This finding comes after a series of measurements of the Earth-Moon distance (carried out using a more accurate process than this one), using a laser beam pointed at a network of mirrors which Neil Armstrong placed on the Moon's surface over 40 years ago, during the Apollo 11 mission.  By measuring how long it takes for a beam of light to travel to the Moon and back, ESA scientists have been able to determine that the distance from the Earth to the moon has, over the last three years, increased by an average of 140 metres per year.  

However, more alarming is the fact that the rate of increase is also going up - the Moon is, on average, moving further away at a faster rate with time.  The increase over the last six months is about 1.4% more than the average increase over the previous six months.
 

Exact forecasts vary, but ESA scientists are all agreed that within 14 years the Moon's orbit will have extended so far that it will leave the Earth's gravitational field completely, and head off into space.  The team of scientists have proposed various reasons for the Moon's recent moves, and the most common suggestions are related to the recent tectonic activity on Earth - the tsunami of 2004, the volcano in Iceland during 2009-10, and possibly the recent earthquakes near New Zealand, an in particular Japan, which has left to a shortening of the length of the Earth's day.  The recent earthquakes have coincided with the moon coming towards a particularly close approach (perigee) and the theory proposes that this has caused the moon to increase its speed while making this close approach, which will lead to it reaching a larger distance at its furthest point 14 days later.

Other scientists have yet to confirm the team's findings, which have sparked considerable controversy in astronomical circles.  Teams in the southern hemisphere have carried out measurements into the exact time for the moon's orbit and have not noticed a significant change in this - either an increase or a decrease, and therefore have concluded that the moon's orbital radius has not changed.  Other teams are preparing to carry out their own measurements using Armstrong's mirrors, and will be sharing their results later next week.

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity

What are Constellations?

Astronomy 2:  Constellations

Constellations are man-made dot-to-dot pictures in the night sky, connecting the stars in the sky into pictures of people, animals and other objects.  The stars we see were grouped into constellations by the Greeks, who made up stories about their gods and then used the characters from these stories as the basis for grouping stars together.  For example, a group of stars might look like two people standing side by side, and so they'd be identified as twins.

The stars that we group into constellations are not always close together in space.  Although two stars might look close together, one could be considerably further away than the other, but might seem to be next to each other because we have no sense of perspective in space.  We can't tell if one star is closer to us than another - and brightness is no help either.  A star that looks bright might be close to us, but a star brighter might be an extremely bright star that's actually further away.

Anyway, treating the stars as points on a flat canvas, the ancient Greeks started to group stars together into pictures, characters, animals and so on.  They didn't have to contend with light pollution, and tended to have clearer skies than we do in Britain (I've missed a number of eclipses due to clouds) so they were able to see more stars at night.  This makes it easier to draw their imaginary dot-to-dot pictures in the sky.

The Greeks got to name the constellations that we talk about today, because they were the first to classify them.  However, there's nothing wrong with devising your own constellations, using the stars that you can see at night.  For example, here's a constellation called Ursa Major (Greek for the "Great Bear").  

This part of the constellation is also known as the Plough.  But they could just as easily be called the Saucepan or the Ladle.

The saucepan...

Or the ladle...

A few things to consider when looking for constellations:  they're not always the same way up.  The Earth is rotating all the time, and this means that the stars (and the constellations) rise in the east and set in the west, in the same way as the Sun (and the moon).  The Earth's axis is tilted - what this means is that the Earth doesn't spin with a vertical axis (like spinning a basketball on your finger), but it's tilted so that it spins with a tilt


The effect of this is that the constellations appear to rotate around a point in the sky - in fact, there's a star in the sky which doesn't rotate.  The earth's axis points directly at it, as it's above the North Pole, and the star is called Polaris.  The photo below was taken near the equator, and shows the stars rotating around the pole star (the dot near the centre of the horizon).

So, although pictures in a book or on a website might show an 'upright' version of a constellation, bear in mind that it might not always look like that in the sky.  It might be at a slightly different angle, and parts of it might be obscured by clouds, and may have fainter stars hidden by light pollution.  One very important consideration is the time of the year; some constellations are only visible at certain times of the year.  I'll explain this some time soon, but as an example, Orion is only visible during the autumn and winter months.

And in all honesty, constellation spotting is sometimes an exercise in imagination.  Some constellations look nothing like the objects or characters that they're meant to represent, and require a serious leap of faith to identify.

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity

Physics discussion: Escape Velocity

The story goes that Isaac Newton was sitting under an apple tree, when an apple fell on his head, and prompted him to wonder why it fell downwards, and not upwards or even sideways.  However, what history doesn’t tell us is that he probably got quite upset at having his afternoon nap interrupted by an apple, and, in his annoyance, threw the apple away as far as he could, declaring, “Stupid apples!”  He then wondered why the apple fell back to the earth, despite him throwing it away as hard as he could.

The same applies today (gravity hasn't changed much since then).  Consider throwing a tennis ball:  the harder you throw it, the higher it goes.  How about throwing it upwards, or even aiming for the moon (it’s not a million miles away, you know)? How fast does it have to be travelling, or how quickly do I have to throw it, so that it doesn’t come back down again?  We call this initial speed (how fast you have to throw it) the escape velocity.

Thinking in scientific terms, we can say that the apple (or the tennis ball) has escaped from the Earth’s gravitational pull, and will not fall back down to the earth.  It has maximum gravitational potential energy, and no kinetic energy (i.e. it stopped moving).  This happens at the edge of the gravitational field.

Since the kinetic energy at the start (i.e. from the throw) has all been converted into potential energy, we can say that the two are equal.

The potential energy is:

And the kinetic energy is:

where m₂ is the mass of the object being thrown, and m₁ is the mass of the Earth. 

 I’ll explain a bit more here about how this works, because at school I was taught that gravitational potential energy = mgh¸ where m is mass, g is acceleration due to gravity, and h is height – so that potential energy continues to increase with height.  So, when does PE = mgh stop being correct?  PE = mgh is not true when h becomes large, and g becomes very small.  The value of g changes with height; close to surface of the earth, mgh is an acceptable approximation, however at high altitude, g becomes

very much smaller.  It’s different at the top of a mountain than it is at sea level for instance.

So, the definition of potential energy is something else, it’s not mgh, it’s taken as something else.PE for all locations is equal to the formula given above.

Since we can equate these two energies, we have that:

Solving for this revised equation gives an expression for escape velocity, v, as:

Where m₂ is the mass of the Earth (in this case) and r is the Earth’s radius from centre to surface (i.e. from the centre of gravity to the point we’re launching from), since we have a bit of a head-start on gravity (we don't have to launch from the centre of the Earth).

Solving for all the numbers gives us an escape velocity of 11,181 metres per second, which is 34 times the speed of sound (Mach 34).  If you tried to throw an object at this speed, you'd probably either break your arm, or suffer friction burns from the air resistance as the air particles tried to move out of the way of your arm (and failed).  

It's also worth mentioning that I've not looked at air resistance, which at Mach 34 is considerable.  The sonic boom caused by the apple (or the tennis ball) would be extremely loud... in fact, I imagine the apple would turn into apple sauce, and the tennis ball would melt into a sticky, furry goo before it got anywhere near earth orbit.

I should explain at this point that escape velocity isn’t the speed that space rockets travel at when they take off.  This is really important.

An important point about escape velocity

Remember at the start that we were talking about throwing objects – where all the energy, and force is transferred to the object at the start of its flight.  With space rockets, the engines keep pushing the space rocket while it’s in flight, so they don’t have to travel as quickly, they just have to push upwards with a force that constantly exceeds their weight until they achieve an earth orbit.  This means that space shuttles, and space rockets, don't have to reach escape velocity.  Instead, they just have to keep pushing upwards with a force that is greater than their weight, until they reach an orbital height.

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?
Calculating (and explaining) escape velocity

Astronomy 1: Stars, Planets and Moons

Following last night's visit to Keele Observatory, I thought it might be helpful to cover some of the basics of astronomy, and then move onto some more detailed topics.  Everybody's got to start somewhere, so I figure it's best to start with home, and move on from there.

The Earth spins on its own axis, taking one day to complete one revolution (one full turn).  This gives us day and night.

The Earth orbits (goes around) the Sun, going around the Sun in one year.  One year is 365.25 days.

Stars:  Stars are huge (very, very big) balls of gas that are carrying out nuclear reactions.  It might be easier to think of a star as an enormous nuclear reactor, constantly going out of control.  The Sun is a star.

Planets:  Planets are smaller balls of rock or gas that go around stars.  There are nine planets that go around our star, the Sun.  The nine planets are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto (going in order from nearest to the Sun to furthest away).  

Moons:  Moons are smaller than planets, and go around planets in their own orbits.  Our Moon goes around the Earth in just under 28 days; some planets (such as Mercury and Venus) have no moons, while other planets (such as Jupiter and Saturn) have over 10 moons each.

One of the basic principles of astronomy is that smaller (lighter) objects go around larger (heavier) objects, and that's all due to gravity.  Galileo, who was one of the first people to make serious use of a telescope, saw Jupiter and four of its moons going around it, and started to wonder if the Earth goes around the Sun.  It wasn't a popular theory at the time, but a serious step forwards in our understanding of astronomy.

Our star, and its nine planets, are all part of a bigger group of stars (about 10 billion stars, roughly) that are all held together by gravity, in a group called a galaxy.  Our galaxy is called the Milky Way.  It's called the Milky Way because, if and when you can see the faint stars in our galaxy in the sky, they look like a milky cloud stretching across the sky.  Almost all of the stars that we can see in the night sky are in our galaxy.  Our nearest neighbouring galaxy is called Andromeda, and in the right conditions, it can be seen without a telescope or binoculars.

Why don't the planets crash into each other?
Because they're all going around the Sun at different distances.  Mercury is closest to the Sun, and completes one orbit in 88 Earth days, while Pluto, which is furthest away from the Sun, takes 220 times longer than the Earth to go around the Sun.

What is a light year?
A light year is a measurement of distance, and it's equal to the total distance that a ray of light would travel in a year.  The speed of light is 300,000,000 metres per second, or 186,000 miles per second, and there are 31 million seconds in a year (60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, 365.25 days in a year).  This means that in 31 million seconds, light would travel 9,467,280,000,000,000 metres, or 9,467,280,000,000 kilometres, and this distance is called a light year.  The distances in space are so far, that we need a meaningful measurement that we can use to compare distances between objects.  

The Sun's nearest neighbour is called Proxima Centauri ("proxima" meaning "close") and that's 4.22 light years away.  This means that light shining from Proxima Centauri takes just over four years to reach us, and that means that we're seeing what it looks like four years ago.  This, it is true, is a very strange situation, and that's because we're used to looking at objects that are much closer, where we can assume that we're seeing things as they are now (because the speed of light is very, very fast, and it takes fractions of a second for the light to travel from the object to our eyes).

I should make it clear that a light year (despite its name) is not a measurement of time, it's a measurement of distance!

In my next post, I'll try and move onto some more specific details, and answer a few questions that I've heard or been asked about astronomy.
Other astronomy related articles you may find interesting:
Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?
Calculating (and explaining) escape velocity

Travelling on the surface of a star

As a follow-up to my last post calculating the distance to the Moon I was asked to calculate the following:

"Here is a question for you Mr Science. How long would it take a plane flyng at approximately 900km per hour across the surface of the biggest known star in our galaxy to travel full circle?"

Good question.  Let's assume that the star is perfectly spherical, which seems reasonable enough.  Now, to find the largest known star in our galaxy.  According to Wikipedia, the largest star in our galaxy is VY Canis Majoris found in the constellation Canis Major (meaning 'large dog').  It's a particularly bright star, which appears faint in the night sky because it's so far away.

VY Canis Majoris has a radius of 1800~2100 solar radii (it's 1800-2100 times wider than the Sun) - the figure varies as the star is surrounded by a nebula, which also makes it difficult to get an exact figure.  Taking 1800 solar radii as a minimum figure, to give us an approximate idea, this means that the star has a radius of:

1 solar radius = 695,500 km
1800 solar radii = 1.251 billion kilometres

Now, the circumference (i.e. the distance around the edge of the star) is 2 π r which gives a circumference of 7.866 billion kilometres.

And that's just the distance around the star's equator...  it's enormous.

Travelling at 900 km per hour, this would take 7.866 billion / 900 = 8.74 million hours.
8.74 million hours = 364,163 Earth days = 997 Earth years (assuming 365.25 days per year).

The exact figure depends on the value of solar radius, the rest is maths, but a round figure would be 1000 years.  Having said that, 900 km/h is not that fast - the speed of sound (Mach 1) is 1193 km/h.  The land speed record is held by Thrust SSC which achieved 1240 km/h in 1997, while Concorde used to reach 2170 km/h.

Still, doubling the speed from 900 km/h to 2170 km/h is only going to reduce the journey time to 500 years... so perhaps the question of time should be put aside.   The real question should be, if you're going to fly or travel on the surface of a star with a temperature of 3000 K, how are you going to keep the pilot flying, and stop him from frying?

Calculating the Earth-Moon distance

This post follows up my previous post on geostationary satellites.  Long before we were launching satellites (even non-geostationary ones), our natural satellite, the Moon, was orbiting the Earth.  As the moon goes around the earth, its phase (shape) changes, and in fact, the word "month" derives from "moonth", the time taken for the moon to go from new to full to new again.  This time is the time taken for one complete orbit around the Earth - the different phases of the moon are a result of us seeing a different amount of the lit half of the moon (I once based a very neat science lesson on this principle - in fact I used it in my interview lesson  and subsequently got the job).

One of my photographs of the moon, taken through a telescope.
The darkening at the bottom of the image is the edge of the
telescope's field of view

We can use physics, and our knowledge of the mass of the Earth, the value of pi and the time the Moon takes to complete one orbit, to work out how far it is from the Moon to the Earth.

Back to the two key equations that we'll need, which are the force on a body moving in a circular path:

where

And Newton's Law of Gravity

Equating the two, and rearranging to find r, gives us

This is the same equation used for geostationary satellites, and describes the basic relationship between the distance between two bodies (a planet and a moon, for example, or a star and a planet).  This gives it great power as it can be used in many different situations.

Turning to the current situation, then:

Calculation of the Earth-Moon distance:

G is the universal gravitational constant, 6.67300 × 10^-11 m³ kg^-1 s^-2

M is the mass of  the Earth, 5.9742 × 10²⁴ kg

T is the time to complete one orbit, which for the Moon is 27.32166 days, which is 2,360,591 seconds.

Plugging the numbers into the formula above gives the distance as 383,201 km

However, this is not the distance to the Moon from the Earth's surface.  Newton's law of gravity gives the distance between the centres of gravity of the two bodies.  I'm ignoring the radius of the Moon (which is perhaps an oversight on my part, you decide) but we must subtract the radius of the Earth from this value, to give the orbital height.  Radius of Earth = 6378.1 km, so the distance to the Moon is calculated as  = 376,823 km, or, if you prefer, (at 1.61 km to the mile), 234,147 miles.

Previously, I've learned that the Moon is about a quarter of a million miles away, so I'm glad the method I've used shows a figure which is 'about right' without any checking.  Looking at other sources, it looks like my figures are close enough, considering the assumptions I've made.  One key assumption I've made is to suggest that the moon travels in a circular orbit, and it doesn't.  It has an elliptical orbit, which means the distance from Earth to Moon changes during the orbit - so I've calculated an average distance.  Still, my figure is pretty close, and not a million miles away (and next time somebody reliably informs you that their opinion is not a million miles away, you can tell them that not even the Moon is that far away).

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity
Measuring g using a pendulum

Maths Problems 5: Geostationary Satellites

Now that I've shown how to estimate (or calculate) pi with a high degree of accuracy and precision, it's time to start using it!

Modern communications around the world rely on being able to bounce a signal of a satellite in orbit around the earth, so that it can be relayed beyond the horizon of the person sending it (or better still, over a specific area of the earth). The ability to bounce the signal off the satellite is very important, and relies on one important factor - knowing where the satellite is, and better still, having it stay directly above the same point on the earth, so that its apparent position in the sky is fixed.

In order to do this, the satellite needs to occupy a geostationary orbit, meaning that it goes around the earth at the same rate as the earth rotates on its axis - once every 24 hours. A geostationary satellite appears to stay in the same point in the sky because it's rotating at the same rate as the earth.

The question is - how can this be done? The answer is to put the satellite in orbit at a specific height above the earth, and above the earth's equator. The rest is physics. Well, it's not rocket science, is it?

Firstly, there are two key formulae to use - the first is Newton's universal law of gravity, because gravity is what holds a satellite in orbit, and the second is the laws of motion for an object moving in a circle (we'll be assuming that the satellite follows a circular orbit).

The first formula is Newton's universal law of gravitation, describing the force of attraction between two bodies.  It is:

where:
G is the universal gravitational constant, 6.67300 × 10^-11 m³ kg^-1 s^-2
M is the mass of the heavier body (in this case, the Earth)
m is the mass of the lighter body (in this case, the satellite)
and r is the distance between the centres of gravity of the two bodies.

The second formula is the equation which describes the force required to keep an object moving in a circular path.

In this formula, F is the force which points towards the centre of the circle, m is the mass of the object (in this case, the satellite), v is the angular velocity (how quickly it's moving in a circle), and r is the radius of the circular path being described by the body.

The first thing I'm going to do is unpack v in the second equation.  v, the angular velocity, is defined in the same way as all speeds as the distance travelled divided by the time taken.  For a circular path, the distance is the circumference of the circle, and the time is the time taken to complete one circle (or orbit), T.

We can plug v² into the formula for circular motion, and we can also equate the two formulae, since the force that will hold the satellite in orbit is the force of gravity.

By cancelling and rearranging, we have one expression for the radius of a geostationary satellite held in orbit by the Earth's gravity.

One thing to notice here is that the mass of the satellite, m, does not appear in our final expression.  The period of rotation of a satellite depends only on the height of its orbit.  The time taken to complete an orbit is only affected by the height of the orbit - the mass of the satellite is not important.

Anyway, we know the time taken for one orbit has to be 24 hours, and all the other components of the formula are constants, so we can plug them in and calculate r.

G is the universal gravitational constant, 6.67300 × 10^-11 m³ kg^-1 s^-2

M is the mass of the Earth, 5.9742 × 10²⁴ kg

T is the time to complete one orbit, 24 hrs = 86 400 s

Therefore, r = 42,243 km.

However, this is not the altitude of the satellite from the Earth's surface.  Remember that when I first gave Newton's law of gravity, the r was the distance between the centres of gravity of the two bodies.  The size of the satellite is small enough to be ignored, but we must subtract the radius of the Earth from this value, to give the orbital height.  Radius of Earth = 6378.1 km, so orbital radius = 35,865 km.

Later edits of this post will include some nice diagrams, but for now I'm happy to have posted the result!

NASA gives the height as approx 35,790 km using a more exact value of the length of one day, while Wikipedia has a similar figure and more information on geostationary orbits.

In my next post... I'm not sure.  Possibly more on orbits generally, including calculating a Moon - Earth distance, or a Sun - Earth distance, or a practical experiment to determine g (acceleration due to gravity on Earth).

Other astronomy related articles you may find interesting:

Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity