Download `earthlike` and second the probability that they have suitable climate

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 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 763 as of 1/23/2014.)
The Kepler satellite has identified more than
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.
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
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’.
For example, if we use the known
value of L/4πR2 for our sun at the
earth, known as the solar constant
and equal to 1354 watts/m2 then
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’.
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’
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