3
$\begingroup$

For the purposes of this question, I will define "habitable" as:

  • rocky composition like Earth
  • about the same surface gravity as Earth (about 1 G)
  • orbiting a relatively quiet star, like our Sun
  • far enough from its host star to not be tidally locked
  • receiving about the same amount of energy from its host star as we do (in the star's habitable zone).

Thus, a "habitable" exoplanet would be capable of holding on to an atmosphere conducive to life, and, if any water was present, it could exist in liquid form on the planet's surface.

Kepler data has revealed numerous planets orbiting numerous stars in only a small slice of the sky. Many of these planets have been characterized to some degree. There are compiled lists of potentially habitable exoplanets.

My question is whether the findings so far can inform a reasonable estimate of the total number of habitable (or potentially habitable) exoplanets within a given number of light-years distance from Earth, or if further sky surveys and studies would be needed. In other words, can we pin down the first few terms of the Drake Equation and apply it to our neighborhood in the galaxy?

Improve this question
$\endgroup$
1
  • $\begingroup$ Certainly on-topic both here and in Astronomy SE. Search for "Drake" here returns 14 hits, and 26 hits in Astronomy. exoplanet tag has 36 questions here, and 211 questions in Astronomy. If no good answers appear here after a while, one might consider then moving there. However it gets a little harder to move a question yourself once answers begin to appear. To me it seems that the analysis of an exoplanet database is more like something that would get an answer there than here, but you never know! $\endgroup$
    – uhoh
    Sep 19, 2018 at 4:59

2 Answers 2

Reset to default
3
$\begingroup$

We can do an estimate, but there are going to be pretty big error bars on that estimate. Kepler has given us a lot of exoplanets, but its results are biased: it's easier to detect large planets in small orbits than it is to detect small planets in large orbits. Most of Kepler's results is gas giants.

We only have one solar system which we've observed reasonably completely: our own. And our solar system is atypical, compared to the Kepler results: no gas giants in small orbits.

Improve this answer
$\endgroup$
1
  • $\begingroup$ So the formula is N(r) = 1. $\endgroup$
    – DrSheldon
    Sep 19, 2018 at 12:51
0
$\begingroup$

Over at an advanced NASA project proposal, you will find the number THREE Earth-Earth-like planets (not large super-Earths or large not-Earth-like planets) within 32 light years of us. See:

"Finding an exoplanet like the Earth requires a very large space telescope that can resolve an Earth-like planet from a Sun-like star to distances of at least 30 light years, where the maximum angular separation of the star and planet is only a few hundred thousandths of a degree. Within this volume, it is expected that there are of order a few rocky, Earth-like planets,

BUT-caveat-I have not been able to find the source document, estimate, analysis.

See: https://www.nasa.gov/directorates/spacetech/niac/2023/Diffractive_Interfero_Coronagraph_Exoplanet_Resolver/

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ "A few" is actually distinct from three; "of order a few" is even more ambiguous. $\endgroup$
    – Erin Anne
    May 27 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.