* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download `earthlike` and second the probability that they have suitable climate
Observational astronomy wikipedia , lookup
Space Interferometry Mission wikipedia , lookup
Discovery of Neptune wikipedia , lookup
Nebular hypothesis wikipedia , lookup
Dialogue Concerning the Two Chief World Systems wikipedia , lookup
Kepler (spacecraft) wikipedia , lookup
Formation and evolution of the Solar System wikipedia , lookup
Corvus (constellation) wikipedia , lookup
History of Solar System formation and evolution hypotheses wikipedia , lookup
Late Heavy Bombardment wikipedia , lookup
Planets in astrology wikipedia , lookup
Satellite system (astronomy) wikipedia , lookup
Astronomical naming conventions wikipedia , lookup
Aquarius (constellation) wikipedia , lookup
Planets beyond Neptune wikipedia , lookup
Circumstellar habitable zone wikipedia , lookup
IAU definition of planet wikipedia , lookup
Astrobiology wikipedia , lookup
Planetary system wikipedia , lookup
Rare Earth hypothesis wikipedia , lookup
Definition of planet wikipedia , lookup
Exoplanetology wikipedia , lookup
Extraterrestrial life wikipedia , lookup
Nciv =Ngal fstar fplanet flife Planetary Considerations The next problem is to estimate what fraction of stars with appropriate chemistry are expected to be found with planets orbiting them with climates that are suitable for life. It is useful to consider the question in two parts, first the likelihood of having planets which are ‘earthlike’ and second the probability that they have suitable climate. fplanet= fPSfhab So in algebraic terms we consider two factors, one of which is the probability fPS of getting a suitable planetary system and the other fhab is the probability that the planetary system contains an earthlike planet with a suitable (‘habitable’ climate). fPS The existence of planetary systems around other stars has been thought likely since at least the 18th century However they are very difficult to see directly with telescopes and the light reflected from such planets orbiting another star has almost never been observed. The reason it is so hard is that the light reflected from such planets is very weak in intensity compared with the intensity of the light from the star itself. Though there were several reports of observations of planetary systems, none withstood the scrutiny of the astronomical community until 1996, when indirect observations using the Doppler shift were first confirmed. The light from the planet itself is not observed. However when a planet is orbiting a star, it tugs the position of the star back and forth as it goes around. Thus the position of the star is oscillating and is moving toward the observer half the time and away from the the other half the time. Correspondingly the light from the star is blue shifted as the star moves toward the observer and red-shifted as it moves away. I will work this out for a circular orbit (with just one planet). The center of mass of the star planet system experiences zero force (to good approximation) so the center of mass Velocity is constant: Mvs+mvp=(M+m)Vcm so vs= -(m/M)vp+(1+m/M)Vcm and vs-Vcm=-(m/M)(vp-Vcm) The velocity of the planet with respect to the center of mass is given to good approximation if m<<M by Newton’s second law m(2π/T)2 R=GmM/R2 Newton: ma=F giving velocity vp= 2πR/T=(GM/R)1/2 so multiply previous by R and take square root use vs=(m/M)v The quantities on thep (with VCM=0) left are measured so we and can solve for the mass m the orbit radius if M 3/2 solve top equation for 2π/T =(GM)1/2 /R2Π/T is known. vs=(m/M) (GM/R)1/2 However the Doppler shift method only measures the part of the star's velocity which is along the line of sight betwen earth and the star. Taking account of this changes the first equation to vs=(m sin θ/M)(GM/R)1/2 where θ (sometimes called i) is the angle between the line of sight and the direction normal to the plane of the orbit. As a result one can only get a minimum value of the mass of the planet from this method unless one knows the angle. Most observed planet masses are reported as values of msin θ Transit method: The other major method for identification of exoplanets is called the transit method. It is conceptually simpler but observationally more difficult: If a planet passes between an observer on earth and the star around which it orbits, then the shadow of the planet will cause the intensity of the light from the star to dim slightly. This periodic dimming can be used to detect the presence of exoplanets. Advantages of the transit method. Better estimates of the mass are possible. Some spectroscopic information about planetary atmospheres may be accessible. Disadvantage: Only those planets with orbital planes in the line of sight or near it can be seen. The intensity variations are very small and easily obscured by effects of the earth's atmosphere as the light from the star passes through it. For the last reason, most planets discovered by the transit method have been discovered from the dedicated Kepler satellite. http://exoplanets.org Will give you a summary of recent information on the confirmed exoplanets so far discovered by all methods. (They report 2950 as of 1/30/2017 with 2504 additional unconfirmed candidates.) The Kepler satellite has identified about 3000 'candidate' exoplanetary systems by the transit method. Organization NASA Major contractors Ball Aerospace & Technologies Corp. Launch date 7 March 2009, 03:49:57.465 UTC[1] Launched from Space Launch Complex 17-B Cape Canaveral Air Force Station, Florida, U.S. Launch vehicle Delta II (7925-10L) Mission length up to 7.5 years elapsed: 4 years, 10 months, and 28 days Mass 1,052 kg (2,319 lb) Type of orbit Earth-trailing heliocentric Orbit height 1 AU Orbit period 372.5 days Wavelength 400-865 nm Diameter 0.95 m (3.1 ft) Collecting area 0.708 m2 Website kepler.nasa.gov From the enumeration of the discovered exoplanets, one can obtain an estimate of the fraction of stars which have planets. In 2002, the number of stars explored was about 1000 and the number of those stars found to have planets was 93, for a ratio of about 10-1. We will take this as our estimate of fPS. A more careful and complete statistical study was carried out in Publications of the Astronomical Society of the Pacific 12, 531 (2008). The fraction of stars having planets is similar to this estimate (48/585). More information is given about mass and period distributions. fplanet= fPSfhab OK the next question is, what fraction of those stars around which planets orbit are expected to have HABITABLE planets orbiting them? Somewhat dangerously, we will assume that an earthlike, that is solid, chemistry rich planet is required. Only a few exoplanets as low in mass as earth have yet been detected because the associated Doppler shifts are very small. To get started, we will suppose that each of planetary systems detected so far by the Doppler method also contains earthlike planets. This conjecture is supported by two facts 1) Computer simulations of the process of planet formation support it. This is a weak argument because the models of planet formation are not very reliable. 2) The mass distribution of discovered planets (which does not yet extend down to earth like planets) rises quite sharply as the mass gets smaller, suggesting that there are a lot more earthlike than Jupiter like planets. You can read here about the first detection of an earthlike exoplanet in 2007 http://en.wikipedia.org/wiki/Gliese_581_c The remaining requirement is that the earthlike planets be ‘habitable’ which we take to mean that they have a suitable climate. (They will also need suitable chemistry, but this will be nearly guaranteed by the type of star around which they formed.) What would a suitable climate be? To make an estimate, I will accept the idea that the surface of the planet will need to have an average temperature and pressure which allows liquid water to exist. This is, of course, true of earth. What arguments suggest that it be a general requirement? What is special about liquid water? To get complex chemistry, which is a minimal requirement for life we will need a medium which Is fluid, so that the constituents of reactants can move around and get in contact with each other Dissolves a large number of chemical species which will participate in the chemistry Desirably if not essentially, will hold lots of heat so that temperatures can be maintained at a uniform level. Liquid water has the needed features to a degree not matched by any other known liquid. It is an almost universal solvent, dissolving nearly everything. This property is traced to the large electric dipole moment of the water molecule as I will explain. It has an extremely high heat capacity, which can again be understood in terms of the structure of the molecules. That process tells you how to measure the specific heat (measure the work done, the mass and the change in temperature) and people have done it and tabulated the values of specific heats for many materials. 1 atmosphere= 101,325 Pa If we accept the idea that liquid water is needed for life (and, for what it’s worth, NASA makes this assumption) then we have some basis for deciding what kind of climate we need. From the phase diagram, we see that the the lower temperature limit is very well defined almost independent of surface pressure and requires that the surface temperature be, on average, larger than about 0 C, 273 K or 32 F. On the other hand, the upper temperature limit depends on what pressure we imagine a biosphere can stand. We get some hint of this in the extremes at which organisms are found to survive on earth. For example, in the ocean trenches the pressures are of order 100 Mpa (about 1000 atm) and the temperatures at which organisms survive range up to about 120 C (393 K and 240 F) Let’s suppose a temperature range of 273K < T < 500K would permit life to exist. (if the temperature were over 373 K, the pressure would have to be higher than it is on earth’s surface). To translate this requirement into a condition on the planet’s orbit, we need a discussion of how the surface temperature of a planet is determined. The surface temperature is determined by a balance of incoming electromagnetic energy from the planet’s star and outgoing energy from electromagnetic radiation which the planet emits in the infrared as it warms up. 4πr2σT4 (L/4πR2)πr2 To calculate the temperature of the surface of the earth with no greenhouse effect write A=4πr2 , take account of the fact that about 30% of the incoming radiation is reflected back out into space and equate the incoming to the out coming energy flow: Energy in/t = (1354 watts/m2)(A/4)(.7)= Energy out/t =A σ T4 σ=5.68x 10-8 watts/m2K4 If you solve for T you get about 254K =-19C=-2.2F which is lower than observed. The discrepancy arises from the greenhouse effect. Ignoring both the reflection from the clouds (cooling) and the greenhouse effect(warming) for earth, this formula predicts an average temperature on earth of 278 K which is quite close to the actual average temperature of 288K. However, this model works less well for Venus where the same calculation gives a prediction of 327K, while the observed average surface temperature is about 740K. The difference is the ‘greenhouse effect’. temperature in house goes up The greenhouse effect: Analogy with a house: insulation, less thermal energy escapes If the furnace keeps pumping in energy at the same rate and you add insulation to the walls, the temperature goes up. The greenhouse gases act like insulation for the earth: less thermal energy escapes the earth surface of the earth heats up. greenhouse gasses block escaping infrared but let incoming visible solar light in. For other planets we can use the same relation and our requirements on the temperature to get the orbital radii R which would allow liquid water to exist on the surface. Equating the input to the output and solving for T: T=(L/16πR2σ)1/4 The bounds on T (from requiring water to be liquid) thus give a range of distances R which the planet can be from a star of luminosity L in order to be ‘habitable’. Uncertainties arise because we know very little about either reflection (albedo) or the greenhouse effect on extrasolar planets. T=(L/16πR2σ)1/4 Solving the relation for the distance R one finds R=(L/16πT4σ)1/2 If we require 273K < T < 500K Then plugging values for a star with the luminosity of the sun in we find .31 AU <R< 1.09 AU This an estimate of the ‘habitable zone’ for our type of star. T=(L/16πR2σ)1/4 Solving the relation for the distance R one finds R=(L/16πT4σ)1/2 If we require 273K < T < 500K Then plugging values for a star with the luminosity of the sun in we find .31 AU <R< 1.09 AU This an estimate of the ‘habitable zone’ for our type of star. In exercise 4 you will estimate habitable zones for all these types of stars. Now with some additional assumptions we can get a rough estimate of fhab : We will assume that the probability of forming a planet in a given area of the dust disc around a planet is proportional to its area and that a Jupiter like planet is needed outside the earthlike planet to protect the lifebearing planet from meteorites. Making the somewhat arbitrary assumption that the Jupiter like planet must be at least 2AU from the star and that the star is sunlike and using the list of discovered exoplanets, we get an estimate of about 0.2 for fhab. Summary We find that fplanet =fPSfhab fPS ≈ 10-1 fhab ≈ 10-1 giving a chance of around 1 part in 100 that a star with the right chemistry will have a habitable planet orbiting it. These numbers are quite uncertain, so it might be better to say fplanet = 10-2±1