So there's a boatload of planets being discovered around stars in our solar neighborhood.

Kepler 22, 638 ly away, was discovered to have a planet Kepler 22b, calculated to be twice the size of Earth.

The observations that make these measurements aren't high resolution images but periodic changes in luminosity (the transit method).

My question is: would a similar situation with similar telescope properties (Kepler) detect our own Earth around our Sun from a similar distance? If not, then how close would you have to be to see Earth?

Presumably there is some correspondence among: distance, luminosity/size of a star, size of a planet. Any other factors that make Kepler-22 different from us?

Note: this question asks about Alpha Centauri. I'm asking specifically about distances like Kepler-22.

  • $\begingroup$ If you think this is a different question than the AC one, then please edit your question title to be specific about Kepler-22 $\endgroup$
    – user1569
    Nov 2, 2018 at 9:29

1 Answer 1


There is a lot of detail in my answer to If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect? I showed that current Doppler techniques and transit techniques would be unlikely to detect the Earth at the distance of Alpha Cen.

It does not get any easier if you are further away. Stars are point sources at 1pc or 200pc, so these spatially unresolved techniques just care what signal level is being received - further away is worse.

In the case of Doppler techniques we need an improvement in radial velocity precision and stability of a factor of a few.

With transits, Kepler was almost there. So another Kepler with a slightly bigger telescope that observed the Sun for say 5 years would do the trick (if the orbital inclination was about 90 degrees). The reason Kepler 22b was just detected is because it has twice the Earth's radius so eclipses 4 times the area and gives a deeper transit signal for a Sun-like star. It is this contrast that is the key to transit detectability.

Maybe the missing piece of information here is that there is a floor level to photometric precision that can't be bettered even if the star is close and bright. So long as you can reach this level in a time shorter than any transit, then the distance is irrelevant, all that matters is the fractional change in the brightness during transit and the number of transits you observe. The first of these grows as the square of the planet radius (for a fixed stellar radius), the second just depends on the length of your mission divided by the orbital period.

The bottom line here is that the Earth in an Earth-like orbit around a Sun-like star would not quite be detectable by Kepler in a Kepler-length mission, even if it were very close and nor if it were further away.

  • $\begingroup$ Thanks for at least responding. I've read you answer there and for whatever reason I can't extract from that the answer to my specific question here. Kepler-22 is roughly the same as our Sun, and Kepler 22b only roughly twice the size of Earth, yet we are ~700 ly apart, so what is it that makes Kepler 22b detectable, and (from all you've said here and there) Earth likely very undetectable? $\endgroup$ Oct 31, 2018 at 21:12
  • $\begingroup$ @Mitch As you say, it has twice the radius, 4 times the projected area and 4 times the transit signal. It is also in a slightly shorter period orbit, which means you get more transits in a fixed mission time. And where did I use the phrase "very undetectable"? As I said in the other answer and here, Kepler was almost good enough. $\endgroup$
    – ProfRob
    Oct 31, 2018 at 21:18
  • $\begingroup$ I don't mean to put words in your mouth. I'm just extrapolating. That answer is all about Alpha Centauri, ~4 ly away, where you say Venus might be barely detected. But this is about Kepler-22 which is ~700 ly away, presumably making things quite a bit more difficult, right? $\endgroup$ Oct 31, 2018 at 22:34
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    $\begingroup$ @Mitch Well it doesn't make it easier. It is not a straightforward function of distance. It is the contrast that is key and that doesn't change with distance. So long as you can get enough signal to reach the instrumental sensitivity limit. $\endgroup$
    – ProfRob
    Oct 31, 2018 at 22:46

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