Subject: Finding Earths by M-dwarf occultation

From: Jon Thaler

Submitted: Wed, 16 Apr 2003 11:04:37 -0500

Message number: 109 (previous: 108, next: 110 up: Index)

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Finding Earths by M-dwarf occultation.
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For an Earth have a 1% effect on a star's luminosity, the star must 
have a radius about 10 Re.  This is about 0.1 Rs, so Mass ~ 0.1 Ms, L ~ 
10^-3 Ls (absolute magnitude ~12.3), and T ~ 3000K.

For a limiting magnitude m = 15.5 (see below), we can see these stars 
to a distance of about 44 pc.  There are about 40,000 stars in that 
volume, about 70% of which are M-dwarfs.  I'll assume that all of them 
are the right size.  (Most are smaller, I think.)

To estimate the probability of observing an Earth, assume that the 
radius of its orbit is 0.03 AU (5e6 km), which is in the habitable zone 
of the star.  The period of this orbit is about 6 days.  The 
probability that this orbit occults the star (ie, that we're in the 
plane) is about 0.02.

Occultation lasts for about 0.01 (half of the above) of the period, 
about 1 hour.  This means that dedicated follow-up will be required to 
confirm the signal.  If the planet is real, we'll only see it a few 
times in 300 exposures.  This is complicated by the fact that the 
return period is similar to the orbital period.

So, in one scan of half the sky, we might find (the signature is four 
low luminosities: two pairs of exposures separated by 20 minutes) about 
2e4*0.02*.01 = 4 of them, if every star has an Earth.  Because each 
scan is (nearly) independent, this implies several hundred per year.  I 
don't know enough about solar system formation to make a better 
estimate.

Here, I calculate the limiting magnitude that comes from statistical 
uncertainty.
I assume these exposure conditions:
* LSST effective aperture:      6.5 m (Area = 33.2 m^2)
* R Filter delta-lambda/lambda: 0.2
* Photon efficiency:            75%
* Seeing:                       0.5"  (Area = 0.79 "^2)
* Sky brightness                R=20.5/"^2
   Sky brightness varies.  This value yields the nominal S/N = 5 in a 10 
sec exposure
   for an R=24 star.

Detected electrons:
* R=24 Star: 54 e/sec
* Detector:   2 e/sec/pixel (0.2")
* Read-out:   7 e/pixel
* Sky:     1075 e/sec (in the "seeing" circle)

In a 10 sec exposure, the fractional uncertainty in an R=24 star's 
luminosity is sqrt(11300)/540 = 0.2, so the limiting magnitude for this 
science is larger.  To achieve 0.1% uncertainty, we need at least a 
million electrons.  An R=15.5 star gives us about 1.4e6 electrons in 10 
sec.

Comments:
* Because we're not noise limited, scanning is not optimal.  We'd be 
better off staring.
* Variable stars will be a serious background.  Current searches reduce 
this problem
   by making many measurements of a star's light curve and looking for 
the characteristic
   shape that occultation produces.  This cadence is not an option for 
the LSST.
* The well capacity of our CCDs may be about 200,000 electrons, so we 
are close
   to saturating the detector.  (The star's image covers about 20 
pixels.)  This
   introduces a systematic error that will be difficult to control.  
CMOS devices
   may fare better in this regard.

My conclusion:  Despite the possibility that the LSST has enough 
sensitivity to find Earths orbiting M-dwarfs, the systematics and 
cadence obstacles appear to be overwhelming.

Thanks to Kem Cook, Chris Stubbs, and Mike Strauss for helpful comments.


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