Feeds:
Posts

## Water found on an extra-solar planet

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

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

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

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

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

## Derivation of Newton’s form of Kepler’s 3rd law – part 2

In part 1 of this blog, I showed how Kepler’s third law, $T^{2} \propto a^{3}$ could be written as

$T^{2} = \frac{ 4\pi^{2} }{ GM } a^{3}$

Today, in part 2, I will show how this can also be written as

$(M + m)T^{2} = a^{3}$

The key to doing this is to use something called the reduced mass, but to understand the reduced mass we first need to know about the centre of mass of a two-body system.

## The centre of mass

If you have two objects of masses $m_{1} \text{ and } m_{2}$, the centre of mass is the point between them where, if the two masses were on a balance beam, the beam would be balanced. When the two masses are equal, this is just the mid-point between them.

The centre of mass for two objects with the same mass is the mid-point between them.

If the masses are not equal common sense tells us the centre of mass will be closer to the more massive object.

If we have a baby and a cow, the centre of mass will be much closer to the cow, at is much more massive.

The exact position of the centre of mass can be found from the fact that

$m_{1}r_{1} = m_{2}r_{2}$

where $r_{1} \text{ and } r_{2}$ are the distances of objects 1 and 2 respectively from the centre of mass. Re-arranging this gives that

$\frac{ r_{1} }{ r_{2} } = \frac{ m_{2} }{ m_{1} }$

Clearly, when the masses are equal $\frac{ r_{1} }{ r_{2} } =1$ and the centre of mass is midway between the two masses. In the case of the baby and the cow, let us suppose the baby has mass of 5kg and the cow a mass of 500kg, then

$\frac{ r_{1} }{ r_{2} } = \frac{ 500 }{ 5 } = 100$

Suppose the baby and the cow were separated by 3 metres, this is the vlaue of $r = r_{1} + r_{2}$. We can write

$\frac{ r_{1} }{ r_{2} } = 100 \text{ and } (r_{1} + r_{2}) = 3$

and so

$r_{1} = 100 r_{2} \rightarrow (100+1)r_{2} = 3. \text{ So } r_{2} = (3/101) = 0.03 \text{m, and } r_{1} = (300/101) = 2.97 \text{m}$

The centre of mass is only 3cm from the centre of mass of the cow, and 297cm from the centre of mass of the baby!

## The reduced mass

When two objects are acted upon by the same central force, such as in the case of gravity, it is useful to use a concept called the reduced mass.

If we have two objects acted upon by a force towards the centre of mass, we can use the concept of the reduced mass.

To determine the expression for the reduced mass, we note that the two forces, which I will call $\vec{F_{1}} \text{ and } \vec{F_{2}}$ which act on objects 1 and 2 are equal and opposite (from Newton’s 3rd law). We can therefore write that $\vec{F_{1}} = -\vec{F_{2}}$.

But, we can also write

$\vec{F_{1}} = m_{1} \vec{a_{1}} \text{ and } \vec{F_{2}} = m_{2} \vec{a_{2}}$

Although the two forces are equal, the accelerations in general are not, they will only be equal if the two masses are equal. We can write that the relative acceleration between the two objects is just the difference in their accelerations, that is $\vec{a} = \vec{a_{1}} - \vec{a_{2}}$. But, we can also write this as

$\vec{a} = \frac{ \vec{F_{1}} }{ m_{1} } - \frac{ \vec{F_{2}} }{ m_{2} } = \vec{F_{1}} \left( \frac{ 1 }{ m_{1} } + \frac{ 1 }{ m_{2} } \right) = \vec{F} \left( \frac{ m_{1} + m_{2} }{ m_{1}m_{2} } \right)$

(as $\vec{F_{2}} = -\vec{F_{1}}$). This then leads to

$\vec{F_{1}} = \vec{F} = \vec{a} \left( \frac{ m_{1}m_{2} }{ m_{1} + m_{2} } \right) = \mu \vec{a}$
where $mu$ is the so-called reduced mass. The reduced mass is the effective inertial mass appearing in two-body problems, such as when a planet orbits a star (if we ignore the gravitational influence of any other planets).

## Newton’s form of Kepler’s third law

We are now ready to derive what is known as Newton’s form of Kepler’s third law. Remember, we are talking about a planet orbiting the Sun. Our starting point is to realise that, strictly speaking, the Earth does not orbit the Sun. More correctly, they each orbit their common centre of mass. Because the mass of the Sun is about 1 million times the mass of the Earth, the centre of mass of the Earth-Sun system is pretty close to the centre of the Sun, in fact it lies within the body of the Sun.

But, the centre of the Sun does orbit this centre of mass point. Or, it would if the Earth were the only planet in the Solar System. In fact, the Sun orbits a point which is dominated by the centre of mass in the Jupiter-Sun system, but you get the idea.

The figure below shows a diagram of our two objects, 1 and 2 (the Sun and one of the planets), orbiting their common centre of mass. They orbit with speeds $v_{1} \text{ and } v_{2}$.

Any two objects in orbit actually orbit their common centre of mass. To make things easier we will assume the orbits are circular about this centre of mass, but the same equations can be derived for elliptical orbits too.

As can be seen from the diagram, the distance between the two objects, which we will call $r$, is given by $r = r_{1} + r_{2}$. We can write an equation separately for each object equating the gravitational force felt by each object to the centripetal force.

$\frac{ G m_{1} m_{2} }{ (r_{1}+r_{2})^{2} } = \frac{ m_{1} v_{1}^{2} }{ r_{1} } = \frac{ m_{2} v_{2}^{2} }{ r_{2} }$

But, we can also write this as

$\frac{ G m_{1} m_{2} }{ r^{2} } = \frac{ \mu v^{2} }{ r }$

where $\mu$ is the reduced mass. Re-writing $\mu$ and remembering that $v = (2 \pi r)/T$ we have

$\frac{ G m_{1} m_{2} }{ r^{2} } = \left( \frac{ m_{1}m_{2} }{ m_{1} + m_{2} } \right) \left( \frac{ 4 \pi^{2} r }{ T^{2} } \right)$

$T^{2} (m_{1} + m_{2}) = \left( \frac{ 4 \pi^{2} }{ G } \right) a^{3}$

where we have replaced $r \text{ with } a$. If we express $T$ in years and $a$ in Astronomical Units, this reduces to

$\boxed{ T^{2}(m_{1} + m_{2}) = a^{3} }$

which is Newton’s form of Kepler’s third law.

## Derivation of Newton’s form of Kepler’s 3rd law – part 1

In 1619, Johannes Kepler published a relationship between how long a planet takes to orbit the Sun and the size of that orbit, something we now call his 3rd law of planetary motion, or just “Kepler’s 3rd law”. It states that

$T^{2} \propto a^{3}$

where $T$ is the period of the orbit and $a$ is the size of the orbit. Kepler also found that the planets orbit the Sun in elliptical orbits (his 1st law), and so the size of the orbit $a$ that we refer to is actually something called the “semi-major axis”, half the length of the long axis of an ellipse.

Any proportionality can be written as an equality if we introduce a constant, so we can write

$T^{2} = k a^{3} \text{ (equation 1)}$

where $k$ is our constant of proportionality.

Kepler found that the planets orbit the Sun in ellipses, with the Sun at one of the foci. The long axis of an ellipse is called its major axis. The $a$ in Kepler’s 3rd law refers to the length of the semi-major axis of a planet’s ellipse.

Newton was able to show in his Principia, published in 1687, that this law comes about as a natural consequence of his laws of motion and his law of gravity. How can this be shown?

## Why is Kepler’s law true?

To show how Kepler’s law comes from Newton’s laws of motion and his law of gravitation, we will first of all make two simplifying assumptions, to make the mathematics easier. First we will assume that the orbits are circular, rather than elliptical. Secondly, we will assume that the Sun is at the centre of a planet’s circular orbit. Neither of these assumptions is strictly true, but they will make the derivation much simpler.

Newton’s law of gravity states that the gravitational force between two bodies of masses $M \text{ and } m$ is given by

$F = \frac { G M m }{ r^{2} } \text{ (equation 2)}$

where $r$ is the distance between the two bodies and $G$ is a constant, known as Newton’s universal gravitational constant, usually called “big G”. In the case we are considering here, $r$ is of course the radius of a planet’s circular orbit about the Sun.

When an object moves in a circle, even at a constant speed, it experiences an acceleration. This is because the velocity is always changing, as the direction of the velocity vector is always changing, even if its size is constant. From Newton’s 2nd law, $F=ma$, which means if there is an acceleration there must be a force causing it, and for circular motion this force is known as the centripetal force. It is given by

$F = \frac{ m v^{2} }{ r } \text{ (equation 3)}$

where $m$ is the mass of the moving body, $v$ is its speed, and $r$ is the radius of the circular orbit. This centripetal force in this case is provided by gravity, so we can say that

$\frac{ G M m }{ r^{2} } = \frac{ m v^{2} }{ r }$

With a little bit of cancelling out we get

$\frac{ G M }{ r } = v^{2} \text{ (equation 4)}$

But the speed $v$ is given by the distance the body moves divided by the time it takes. For one full circle this is just

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

where $2 \pi r$ is the circumference of a circle and $T$ is the time it takes to complete one full orbit, its period. Substituting this into equation (4) gives

$\frac{ G M }{ r } = \frac{ 4 \pi^{2} r^{2} }{ T^{2} }$

Doing some re-arranging this gives

$\boxed{ T^{2} = \frac{4 \pi^{2} }{ GM } a^{3} } \text{ (equation 5)}$

where we have substituted $a$ for $r$. This, as you can see, is just Kepler’s 3rd law, with the constant of proportionality $k$ found to be $(4 \pi^{2})/(GM)$. So, Kepler’s 3rd law can be derived from Newton’s laws of motion and his law of gravity. The value of $k$ above is true if we express $a$ in metres and $T$ in seconds. But, if we express $a$ in Astronomical Units and $T$ in Earth years, then $k$ actually comes out to be 1!

## Newton’s form of Kepler’s 3rd law

A web search for Newton’s form of Kepler’s 3rd law will turn up the following equation

$(M + m) T^{2} = a^{3} \text{ (equation 6)}$

How can we derive this? I will show how it is done in part 2 of this blog, as we will need to learn about something called “reduced mass”, and also the “centre of mass”.

I mentioned in this blog here that I would be on TV talking about the calculation that the Milky Way galaxy contains some 17 billion Earth-like planets.

Here is a youtube video capture of my appearance on the TV show. My apologies that the subtitles lag behind what is being said, and for the subtitles only being a summary of what is said. But at least if will give you a vague idea of what I’m saying if you cannot understand Welsh.

## 17 billion Earth-like planets?

Could there be as many as 17 billion Earth-like planets in our Milky Way galaxy?
This has been suggested in a paper presented in this last few weeks at the semi-annual meeting of the American Astronomical Society. The lead author of the papaer is Dr. Francois Fressin, who was part of the team that discovered the first Earth-sized planets in late 2011.

An artist’s representation of extra-solar planets

Dr. Fressin has analysed data from the Kepler mission to come up with his startling figure. But, it should be pointed out that a lot of assumptions come into this number, so I thought I would explain what some of these assumptions are, as well as explaining a little about the Kepler mission.

## Wobbling stars

The history of detecting extra-solar planets (exoplanets) goes back to the mid 1990s. The detection technique used for the vast majority of the early discoveries was to detect the wobble that an orbiting planet produces in the position of its host star. When a planet orbits a star they in fact both orbit the system’s centre of mass. This may be a point actually within the body of the star. The larger the ratio between the mass of the star and the mass of the orbiting planet, the closer the centre of mass will be to the centre of the host star.

The dopper shift in the spectrum of a star produced as an unseen planet orbits it

Our Earth produces a wobble in the Sun, but it is too small a wobble for us to be able to detect. However, Jupiter, the most massive planet in our Solar System, produces a wobble in the Sun’s position that is detectable.

If we are looking at a star with a Jupiter-mass planet going around it then, as long as this planet is not too far from its host star, we should be able to detect the wobble in the position of the host star. But, only if we are looking at it with the right orientation. This is because we do not directly see the wobble in the host star, what we observe is a rhythmic Doppler shift in the spectral lines of the star, which shows that it is moving towards and away from us in a regular manner.

If we were to look at such a system face-on (at right angles to the plane of orbit of the planet) we would not detect any wobble, as the wobble would be side-to-side. The effect is maximum when we view the system edge-on, and will be less for other angles. More precisely, if $\theta$ is the angle between the plane of orbit of the planet and our line of sight, then the observed wobble to our line of sight will vary as $\cos\theta$, maximum when $\theta=0^{\circ}$ (edge-on) and zero when $\theta=90^{\circ}$ (face-on).

Because of the accuracy with which we can measure Doppler shifts, this technique for finding exoplanets tends to predominantly find planets with masses as large or larger than Jupiter orbiting often much closer than our Earth orbits the Sun.

## The Kepler mission

The Kepler mission was launched in 2009 and uses an entirely different technique, known as the transit technique. If a planet passes in front of its host star, the light coming from the star will be reduced a small amount as the planet passes across the disk of its host star. Often the amount can be less than 1%, but with our modern-day high accuracy cameras we can detect such tiny dips in brightness.

The dip in light from a star when a planet passes in front of it.

Of course, we will only see such a dip if we are viewing the system edge-on or close to edge-one. Depending on how close the planet is to its host star, once the viewing angle is more than a few degrees away from being edge-on, the planet will no longer be seen to pass across the disk of its host star and so no dip in light will be observed.

Although this is a severe limitation, Kepler gets around this by viewing many stars simultaneously. Kepler constantly stares at the same small patch of the sky (some 1/400th of the sky), but in its field of view there are over 150,000 stars. To date, Kepler has found some 2,740 candiate exoplanets, a much larger figure than the number of exoplanets found using the wobble technique.

The Kepler mission telescope, which was launched in 2009.

I should also point out that not only orbiting planets can cause a dip in a star’s light. Some stars are intrinsically variable, but we know which kinds of stars these are so can ignore those. Also, something else could come between us and the star, such as a clould of gas and dust, or another passing star. So, the dip in light of a particular star needs to be observed to be repeating for us to know that it is due to a planet in orbit about it.

In addition to its greater number of detections, the transit technique is able to observe planets as small as the Earth orbiting their host star, because we are capable of detecting even such tiny dips in the light of the host star. In late 2011, Dr.Fressin and his team made the first announcement of the detection of Earth-sized exoplanets which were detected by the Kepler mission.

## What can we learn from the dips of light

It turns out that we can learn quite a lot about the exoplanet from observing the dip in light. First, by observing the time between the dips in the star’s light we can determine how long the planet takes to orbit its host star (the period of orbit). Also, by analysing the amount the host star’s light dims, we can work out the physical size of the planet as we know from the spectral type of the star what it’s physical size is.

In order to confirm that a transit event is indeed an orbiting planet we need to follow up the observation using the Doppler-shift technique to see its radial wobble. The Doppler-shift technique allows us to determine the mass of the exoplanet, because we know from the host star’s spectral type what its mass is, and so the size of the host star’s wobble is related to the ratio of the host star’s mass to the exoplanet’s mass.

By combining the two techniques we can also determine the exoplanet’s density, as the transit technique tells us its size and the Doppler-shift technique tells us its mass. The density allows us to say whether the exoplanet is gaseous or rocky.

## Is 17 billion reasonable?

So far Kepler has detected some 2,740 possible exoplanets, from the more than 150,000 stars that it is observing. As scientists only recently announced a 461 new candidates, clearly new detections are still being made. Dr. Fressin calculates that 17% of stars host a planet up to 1.25 times the size of the Earth. This figure is based on several steps of calculations – including how many of the 2,740 detections are Earth-sized planets and how many of the approximately 150,000 stars in the field of view have the correct orientation for us to see a transit event. The figure of 17% that Dr. Fressin has determined is then multiplied by the calculation that there are 100 billion stars in our Galaxy to come up with the figure of 17 billion Earth-like planets.

By anyone’s reckoning, 17 billion Earth-like planets is a lot! Even if the figure is found to be too high, it is unlikely to be out by more than a factor or 10, probably much less. This still leaves more than 2 billion Earth-like planets in our Milky Way galaxy, a very large number. Not all of these Earth-like planets would be suitable places for life to have evolved; they may be orbiting high-mass stars whose lifetimes are too brief for life to evolve, or they may be too close or too far from their host star to be suitable.

It seems to me that there is every likelyhood of not only life but intelligent life elsewhere in our Galaxy. Whether we ever make contact with extra-terrestrial civilisations is a whole different matter.

## Update

This interesting histogram has recently been produced by NASA, as their Astronomy Picture of the Day (APOD) for Saturday the 12th of January 2013.

A histogram produced by NASA showing the percentages of different types of exoplanets.

As the caption to the image on NASA’s APOD page says, these percentages are for predominantly planets in orbits close to the host star, within the equivalent of Mercury’s orbit. This is because Kepler is more likely to detect a transit event when the exoplanet is in a close orbit, because a larger range of viewing angles will still lead to our seeing a transit event.

## The Analemma (part 2)

A few weeks ago I showed a photograph an an Analemma. As the Analemma in the photograph was vertical, I explained that it must have been taken at midday. Here is the photograph again, just to remind you.

A Solar analemma. Because it is vertical, this analemma was taken at midday.

In part 1 of my series on the Analemma, I also explained how the North-South motion of the Sun in the photograph was due to the changing elevation of the Sun at midday. This is, of course, due to the tilt of the Earth’s axis, as explained in this video below.

But, what about the East-West (left-right) motion? What is this due to?

It turns out that the East-West motion is due to two effects. One is the same inclination of the Earth’s axis in its orbit around the Sun which produces the North-South variation in the Sun’s elevation at different times of the year. I will explain how this affects the East-West position of the Sun at midday in part 3 of this blog.

But, the second effect is unrelated to this, it has to do with the details of the Earth’s path around the Sun.

## The Heliocentric Universe

When Copernicus suggesed in 1547 that the Earth and the other planets went around the Sun, he argued that they would do so in perfect circles. He was, in fact, not the first to suggest that the Sun and not the Earth was at the centre of things. Aristarchus had suggested the same thing in the 3rd Century B.C., but his work had been largely ignored in preference to the teachings of Plato and Aristotle, who firmly held that the Earth was the natural centre of all things.

Building on Aristotle’s Geocentric Universe model, the Greek-Roman astronomer Ptolemy developed a sophisticated model of the Sun, Moon, planets and stars orbiting the Earth. This model was incredibly successful, and able to predict the positions of the celestial objects to a good degree of accuracy for some 1500 years.

In Copernicus’ 16th Century model, the planets orbited the Sun in circles. In the latter part of the 16th Century, the greatest observational astronomer was a Danish man, Tycho Brahe. Brahe had his own Observatory and research institute, Uraniborg, on the Danish (now Swedish) island of Hven, with a Royal patronage to fund his observing programme.

Tycho Brahe (1546-1601)

Brahe produced the most accurate observations of the planetary positions, and he found he got better agreement with Ptolemy’s geocentric model than he did with Copernicus’ heliocentric one. Towards the end of his career, a young mathematician by the name of Johannes Kepler came to work with him.

## A very particular kind of curve

After Brahe’s death, Kepler set about seeing whether he could get a heliocentric model to agree with the observations. After over a decade of trial and error, he eventually found that, if he allowed the planets to move in ellipses rather than perfect circles, that very good agreement could be obtained. As Richard Feynman once said, “an ellipse is a very particular kind of curve”. To be more precise, it is the curve obtained when one passes a string about two drawing pins (“thumb tacks” as Americans call them) and draws the ensuing locus of points.

How to draw an ellipse

Kepler’s 2nd law states that a planet will “sweep out equal areas in equal times” in its orbit. What this means is that it will speed up when near the Sun (perihelion) and slow down when further from the Sun (aphelion).

Part of the East-West motion of the Sun in the Analemma is due to Kepler’s 2nd law, the fact that the Earth changes its speed of orbit as it goes around the Sun. The Earth moves quicker when it is closer to the Sun (perihelion), and slower when it is further from the Sun (aphelion). Kepler did not know why this happens, but it is a natural consequence of Newton’s law of gravity.

## A mean Solar day

How do we measure the length of the day? It seems like a simple question. Surely, the answer is that it is the time it takes for the Earth to turn once on its axis. This is, in fact, the wrong answer. The time it takes for the Earth turn once on its axis is the sidereal day, and this is not how we measure our day. Why? The diagram below explains it.

The Earth has to turn a little bit extra for the Sun to cross the local meridian. We call this the Solar day. The sidereal day is the time for the Earth to rotate 360 degrees.

Because the Earth moves about the Sun in its orbit, the Earth has to rotate a little bit extra for the Sun to cross the local meridian on two successive occasions. This is how we define 24 hours, the solar day. But, there is an additional complication; because the Earth’s speed of orbit changes, the extra angle the Earth needs to turn to bring the Sun back over the meridian also changes. As the Earth approaches perihelion (closer to the Sun), it speeds up and so moves through a larger angle each 24 hours than when it is further from the Sun.

Of course, we cannot keep changing the length of the day, we fix it at 24 hours, which is what we call the mean solar day. This is the midday our watches will show. But, near perihelion, the Earth has to turn that little bit extra as I’ve explained, so when our watches say it is midday the Sun will still be to the East of the local meridian. So, if we were taking a photograph of where the Sun was at midday as shown by our watches, the Sun would have shifted eastwards of the mid-point.

The opposite effect happens near aphelion, when the Earth is moving more slowly in its orbit. This time, the Earth moves slightly less in 24 hours than it does in other parts of its orbit, and so the Earth does not have to turn through such a large angle to bring the Sun back over the local meridian. When our watch says midday, the Sun will have gone past the local meridian, and be to the West of it.

Hopefully, this now explains why there should be some East-West motion in the Solar Analemma. However, it turns out that there is a second effect which also causes an East-West motion, the tilt of the Earth’s axis in its orbit. I will explain this component and how it affects things in part 3 of this series.