Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm not sure how scientists determine that a planet is really an Earth-like planet. How do they do it?

share|improve this question
up vote 2 down vote accepted

The earth similarity index (ESI) is a weighted geometric mean of four similarities.

The formula documented on (as of march 23, 2014) should be adjusted a bit, since $n$ should be the weight sum instead of the number of planetary properties. With the weights provided on the site we get

$$ESI =\left( \left(1-\left | \frac{r_E - r_P}{r_E + r_P} \right |\right)^{0.57}\cdot \left(1-\left | \frac{\rho_E - \rho_P}{\rho_E + \rho_P} \right |\right)^{1.07}\cdot\\ \left(1-\left | \frac{v_E - v_P}{v_E + v_P} \right |\right)^{0.70}\cdot \left(1-\left | \frac{288\mbox{K} - \vartheta_P}{288\mbox{K} + \vartheta_P} \right |\right)^{5.58}\right)^{\frac{1}{7.92}},$$ with $r_E=6,371 \mbox{ km}~~$ Earth's radius, $r_P$ the radius of the planet, $\rho_E=5.515\mbox{ g}/\mbox{cm}^3$ Earth's bulk density, $\rho_P$ the bulk density of the planet, $v_E=11.2 \mbox{ km}/\mbox{s}~~$ the escape velocity on the surface of Earth, $v_P$ the escape velocity on the surface of the planet, and $\vartheta_P$ the surface temperature of the planet; the weight sum is $0.57+1.07+0.70+5.58=7.92$.

Mars as an example: With $r_P=0.53 r_E$, $\rho_P=0.71 \rho_E$, $v_P=0.45v_E$, $\vartheta_P=227\mbox{ K}$ we get $$ESI_s =\left( \left(1-\left | \frac{0.47}{1.53} \right |\right)^{0.57}\cdot \left(1-\left | \frac{0.29}{1.71} \right |\right)^{1.07}\cdot\\ \left(1-\left | \frac{0.55}{1.45} \right |\right)^{0.70}\cdot \left(1-\left | \frac{61\mbox{ K}}{515\mbox{ K}} \right |\right)^{5.58}\right)^{\frac{1}{7.92}}=\\ (0.811241627\cdot 0.819676889\cdot 0.716163454\cdot 0.494865663)^{\frac{1}{7.92}}= 0.833189885$$ as surface similarity. (Some of the data used from here.) Global similarity combines surface similarity with interior similarity. Global similarity of Mars with Earth is about 0.7.

Surface gravity can be calculated from radius and bulk density of a planet. The radius of an expoplanet can be estimated by the transit method, relating the estimated diameter of the star to the brightness change of the star during planet transit. The mass of the planet can be estimated by the wobble of the radial velocity of the star (using Doppler shift). By mass estimate of the star and the orbital period of the planet the distance of the planet to the star can be estimated. An estimate of the absolute brightness of the star can then be used to estimate the surface temperture of the planet. There exist more methods. The accuracy of these estimates vary with the quality of the observations.

ESI values between 0.8 and 1.0 are considered as Earth-like.

Details of the planet's atmosphere, amount of surface water, and other details are not considered in the formula. So it's just a very rough prioritization. With future spectroscopic data further refinement could be possible.

share|improve this answer

the composition of an exoplanet is what is being referred to when someone calls it earth-like. one a planet is found, scientists try to calculate the distance of the planet from the star - by studying the rotational period of the planet, wobble of the parent star and so on. Once we know the distance to the star and the type of star, we can infer if the planet is in the relevant Goldilocks zone i.e if the planet can have liquid water, tolerable temperatures and other such factors which are necessary for the sustenance of life.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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