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Posts Tagged ‘exoplanets’

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

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This story appeared recently on the Universe Today website.

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

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

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

 

 

 

 

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This story caught my attention a few months ago, so I thought I would share it in a blogpost. I’m not sure why it has taken me so long to blog about it, but it is still a newsworthy story.

Astronomers have discovered the exoplanet (which goes by the name 2MASS J21265040−8140293) with the longest period orbit, the planet takes about 1 million years to orbit its parent star. It has been found in orbit about a red dwarf star (a star with a lower mass than the Sun), which affects the calculated size of the orbit. Based on Newton’s form of Kepler’s 3rd law, which I blogged about here, the period of orbit of a planet is given by

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

where T is the period of the orbit (expressed in Earth-years), m_{1} and m_{2} are the masses of the star and the planet (expressed in terms of the mass of our Sun), and a is the size of the orbit, expressed in Astronomical Units (AUs). An astronomical unit is the average distance between the Earth and the Sun, and is just under 150 million kilometres.

As the parent star for this exoplanet has a lower mass than the Sun, and as the orbiting planet has a mass of about the mass of Jupiter, it has been calculated that its orbit is about 4,500 AUs! For comparison, Pluto orbits at about 40 AUs from our Sun. So, it is a truly huge orbit.

Astronomers have discovered a solar system where

Astronomers have discovered a solar system where

You can read the submitted paper here, it was published in Monthly Notices of the Royal Astronomical Society in

Both the parent star and the exoplanet were previously observed, this discovery has been made by looking through data archives. The parent star was discovered in 2006 as part of a programme to observe associations of stars which contained young stars. The exoplanet 2MASS J21265040−8140293 was, as the name implies, observed as part of the 2MASS project, which ran from 1997 to 2001. The exoplanet was identified from the 2MASS images in 2008.

To me this piece of news shows a few things

  1. That solar systems come in all shapes and sizes. This shouldn’t really be a surprise.
  2. Trawling through the huge amounts of archived data we have accumulated over the last few decades can lead to exciting discoveries. As most of these archives are freely available, this means that you do not necessarily need to be working at a university with lots of telescope access to make astronomical discoveries.

 

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On May 10, NASA announced that the Kepler mission had discovered and confirmed 1,284 new extra-solar planets (exoplanets). This is the largest trawl of exoplanets ever announced at one time, and takes the total of known exoplanets to over 3,000. It is sometimes hard to remember that the first exoplanets were only being discovered in the mid-1990s; we have come a long way since then.

 

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On 10 May, NASA announced that the Kepler mission had discovered and verified 1,284 new planets, taking the total of confirmed exoplanets to well over 3,000

 

Kepler discovers planets using the ‘transit technique’. This involves staring at a particular patch of the sky and looking for a dimming of particular stars. If the dimming of a particular star happens on a regular basis, it is almost certainly due to our seeing that star’s planetary system edge-one. It is a safe bet that repeated and regular dimming is caused by a planet passing across the disk of the star. This is similar to the effect Mercury would have had on the Sun during the recent Transit of Mercury (see my blogpost here about that event).

Kepler was launched in March 2009 and put into an Earth-trailing orbit. In July 2012 one of the four reaction wheels used for pointing the telescope stopped working. In May 2013 a second one failed, and in August 2013 NASA announced that they had given up trying to fix the two failed reaction wheels and Kepler ceased operation. It used the reaction wheels to keep it pointing at the same patch of the sky, a nearly square patch which covered parts of the constellations Cygnus, Lyra and Draco. This field is shown in the diagram below.

 

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The patch of sky observed by Kepler, covering parts of the constellations Draco, Cygnus and Lyra. The field of view covered 115 square degrees; the Full Moon would fit into this area over 400 times. Within this area there are over half a million stars, with about 150,000 being selected for observation.

 

Although the first exoplanets were discovered using the parallax technique (see my blog here for details of that method), Kepler has led to a huge increase in the number of known exoplanets. In fact, since its launch in 2009, Kepler has slowly become the dominant instrument for detecting new exoplanets. It took a few years for it to do this, as so much data were acquired during its mission that it has taken several years for the results to start coming out.

 

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The growth in the number of exoplanets discovered between 1995 and 2016. Since Kepler’s launch in 2009 the numbers have boomed, with Kepler being responsible for the majority of new discoveries since 2013.

 

Incredibly, in addition to this announcement of 1,284 more confirmed exoplanets, Kepler has found a further 1,327 which are more than likely to be exoplanets but require more study to be confirmed. Of the nearly 5,000 planet exoplanets found to date, more than 3,200 have been verified and 2,325 of these have been discovered by Kepler. Based on their size, nearly 550 of the newly announced 1,284 exoplanets could be rocky planets like the Earth. Nine of these 500 orbit their star in the habitable zone, the zone around a star where we believe it is possible for liquid water to exist. This means that we have discovered a total of 21 exoplanets in the habitable zone.

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Continuing with my list of the five top facts about Jupiter which BBC Five Live tweeted a few weeks ago, today I am going to talk about the fact I listed as number 2 –

all the other planets in the Solar System together would fit into it [Jupiter].

The tweet from BBC Radio 5 with the five most interesting facts about Jupiter.

The tweet from BBC Radio 5 with the five most interesting facts about Jupiter.



The list of the five facts

The list of the five facts


I think this gets across very clearly just how big Jupiter is compared to the other planets! I should add that this statement does not include Saturn’s rings, which extend a huge distance from Saturn, but if you take just the planets themselves without any rings (Saturn, Jupiter, Uranus and Neptune all have ring systems) they would all fit comfortably together into Jupiter.

In terms of the Earth, the Earth would fit across the diameter of Jupiter about 13 times, which means that it would fit into it 13^{3} \approx 2200 \text{ times }!! The great red spot, which we believe is a giant storm system, could swallow up the Earth three times. Astronomers have been seeing the great red spot ever since telescopes were good enough to see it, which is about the middle of the 1600s. Although it has changed a little in size and appearance over these 400-odd years, it has remained a prominent feature of Jupiter’s cloud tops.



The comparative sizes of Earth and Jupiter. Earth would fit into Jupiter over 2000 times!

The comparative sizes of Earth and Jupiter. Earth would fit into Jupiter over 2000 times!



Comparing Jupiter to the next largest planet Saturn, we can say that Saturn would fit into Jupiter nearly twice (1.73 times to be more precise). This is, of course, not including Saturn’s large ring system. In terms of mass, Jupiter’s mass is 318 times the mass of the Earth, whereas Saturn’s is 95 times. Both are much less dense than the Earth, because they are mainly composed of gaseous hydrogen and helium.

When we look at Jupiter, we are seeing the cloud tops. If there is a solid surface we have never seen it, but we do infer its presence from our understanding of planetary formation. A solid, rock-like core probably exists and it may be similar in size to the Earth, but 99.9 \% or so of the volume of Jupiter is mainly hydrogen and helium gas, or hydrogen and helium under high pressure in some kind of liquid metallic state. The famous bands of Jupiter are caused by gases of different temperatures. The light-coloured bands (actually called “zones”) are regions where the gases are warmer are rising, and the dark coloured bands are where the gases are cooler and are falling.

Before I finish this particular blog, I will say a brief word about exoplanets (planets around other stars). I have blogged about this topic several times before, but with regard to Jupiter’s size and mass, most of the early exoplanets found were, in fact, larger and more massive than Jupiter. This is because our early detection techniques were not sensitive to find less massive exoplanets, but this has changed a lot with the advent of the Kepler Space Observatory. This interesting graph below shows the masses of exoplanet discoveries as a function of year. Note, the y-axis (vertical axis) is in terms of mass of Jupiter, so 10^0 means the same mass as Jupiter, 10^1 means ten times more massive, 10^-1 means one tenth etc. As you can see as we come closer to the present we see that smaller and smaller mass exoplanets are being discovered, and we have even found several now which are similar in mass and size to the Earth.

A graph showing the masses of exoplanet discoveries compared to the mass of Jupiter. The y-axis is logarithmic, so 10^{0} means the same mass as Jupiter, 10 means ten times more massive, and 10^{-1} means a tenth of Jupiter's mass.

A graph showing the masses of exoplanet discoveries compared to the mass of Jupiter. The y-axis is logarithmic, so 10^0 means the same mass as Jupiter, 10^1 means ten times more massive, and 10^-1 means a tenth of Jupiter’s mass, 10^-2 means one hundredth, etc.


Next week I will talk more about Jupiter’s rotation and how long it takes to orbit the Sun.

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This story got my attention a while ago, but for some reason I forgot about it until recently. It is astounding to think that it was only in 1995 that we discovered the first extra-solar planet (exoplanet), and less than 20 years later the tally is at over 1000!



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The first few hundred exoplanets were discovered using the Doppler technique, where an orbiting planet causes its parent star to move back and forth in a rhythmic and regular manner which can be detected by shifts in the star’s spectral lines. However, in 2009 NASA launched its Kepler Space Telescope, and this led to more and more exoplanets being detected using the transit technique, and as of now most have been discovered by Kepler using this technique. You can read more about these two techniques in one my previous blogs here.

There is also a great episode of BBC Radio 4’s “In Our Time” which was broadcast last year (2013) which discusses exoplanets. The link is here, it is still available to listen to. Enjoy!



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Legend has it that it was Galileo who first noticed that the period of a pendulum’s swing does not depend on how large the swing is, but only on the length of the pendulum. The swinging back and forth of a pendulum is an example of a very important type of motion which crops up in many places in Nature, so called Simple Harmonic Motion (SHM). In this blog I will derive the basic equations of SHM, and then go on and talk about the deep connection between SHM and circular motion.

A swinging pendulum

If we start off by looking at a simple pendulum which has been displaced so that the bob is to the right of the vertical position, the angle the line of the pendulum makes with the vertical is given by \theta, and for this derivation to work \theta needs to be small.


A simple pendulum will swing back and forth, exhibiting Simple Harmonic Motion.

A simple pendulum will swing back and forth, exhibiting Simple Harmonic Motion.


The force restoring the pendulum bob back to the middle, which I have called F in the diagram above, is given by F = -T\sin(\theta) (the minus sign comes about because the the force is back towards the centre, even though the angle \theta increases as we move the bob to the right).

The restoring force, F, can be written using Newton’s 2nd law as F=ma=-T\sin(\theta). The angle \theta is measured in radians (see my blog here for a tutorial on radians). When \theta is small, \sin(\theta) \approx \theta and so we can write that



\sin(\theta) \approx \theta = \frac{x}{l}


where l is the length of the pendulum and x is the horizontal displacement of the pendulum bob.

Finally, the tension T in the pendulum chord can be written as T \approx mg where m is the mass of the bob and g is the acceleration due to gravity (9.8 m/s/s for the Earth).

Putting all of this together, we can write that the restoring force F can be written as


F = - mg\theta = -mg\frac{x}{l}

This means that the acceleration a can be written as


\vec{a} = -\frac{g}{l} \vec{x}

It is more common to write this as


\boxed{ \vec{a} = -\omega^{2} \vec{x} }


where \omega^{2} = \frac{g}{l} for the pendulum. \omega is called the angular frequency and it is related to how long the pendulum takes to complete one full swing, the period, by the equation T = \frac{2\pi}{\omega}. The frequency is just the reciprocal of the period, so we can write \nu = \frac{\omega}{2\pi} and so the angular frequency is related to the time frequency as \omega = 2\pi \nu.

Whenever the acceleration can be written as being proportional to the displacement, and in the opposite direction to the displacement, we have Simple Harmonic Motion. Other examples of SHM are an object bouncing vertically on a spring, or moving horizontally back and forth due to a spring attached at one end, even the vibrations of atoms in molecules.

Solutions to the SHM equation

What are the solutions to the second order differential equation \frac{d^{2}\vec{x}}{dt^{2}} = \vec{a} = - \omega^{2} \vec{x}? We have a displacement, \vec{x}, and it is proportional to the acceleration \vec{a}, but the acceleration acts in the opposite direction to the displacement.

We differentiate the displacement twice with respect to time to produce the acceleration (remember \vec{a} = \frac{d^2}{dx^2} \vec{x}), and for SHM, when we do this, the acceleration is proportional to the displacement and in the opposite direction.

Let us suppose we try the displacement



x = A \sin(\omega t) \text{ (remember that } \theta = \omega t)

If we differentiate this once with respect to time we get the velocity



v = \frac{dx}{dt} = \frac{d}{dt} A \sin(\omega t) = A\omega \cos(\omega t)

To get the acceleration we need to differentiate the velocity with respect to time so



a = \frac{dv}{dt} = \frac{d}{dt} A\omega \cos(\omega t) = - A \omega^{2} \sin(\omega t) = -\omega^{2}x

Loh and behold, we now have that a \propto - x, so we have shown that, if x = A \sin(\omega t) that an object which has this displacement as a function of time will display SHM.

As the acceleration is proportional to the displacement, it will be at its maximum when the displacement is maximum, so for a pendulum when the bob is at its extreme positions. The acceleration at the centre is zero, as the displacement at the centre is zero.

Conversely the velocity behaves in the opposite sense. Remember, as \vec{v} = \frac{d}{dt}\vec{x} = A\omega\cos(\omega t) it means that the velocity and displacement are 90^{\circ} out of phase with each other, when the displacement is a maximum the velocity is zero, and when the displacement is zero the velocity is a maximum. So, the bob will be travelling at its quickest when it passes through the centre, and at its extremes the velocity is (temporarily) zero.

SHM and circular motion

If an object is moving at a constant speed in a circle in the x-y plane we can write that it’s position at any time is given by



x = A \cos(\theta) \text{ and } y = A \sin(\theta)


For an object moving with a constant speed in a circle, its x-position and y-position can be written in terms of the radius A and the angular velocity.

For an object moving with a constant speed in a circle, its x-position and y-position can be written in terms of the radius A and the angular velocity \omega.


But notice that the expression for x is exactly the same as the expression which we had above, so an object moving in a circle performs SHM. But how? Surely, if it is moving with a constant speed (and hence constant angular velocity \omega), how is it also displaying SHM?

The SHM comes in when we look at the object’s x or y-position as a function of time. So, for example, if we look at the circle from below, as if we were looking along the plane of the screen from below, we would only see the x-displacement of the object, as the y-displacement would be invisible to us. The x-displacement is given by x=A \cos(\theta) = A \cos(\omega t), so the x-position will move back and forth about the central point, displaying SHM. This regular “wobble” is one of the things we look for in trying to find exoplanets – planets around other stars. If we see a regular, rhythmic wobble in the position of the parent star, it’s a pretty good bet that it has a planet going around it with the force of gravity between the host star and its planet producing the apparent SHM.


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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.



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