How has the distance between sun and earth been calculated? Also what is the size of the sun?

  • $\begingroup$ Do you want to know both how the Earth-sun distance is measured and how the speed of light is measured? Those are completely different things. As I asked before, separate threads, please. $\endgroup$ Feb 15, 2011 at 8:00
  • $\begingroup$ Yes, I agree. Just that the single curiosity that how man has acquired knowledge of solar system has forked so many curiosities. And by putting them in separate threads, I didn't want to lose the context. More importantly, in 20 minutes, I can only post one question. So it will take me full day today to address all the queries separately and put them together to make holistic understanding, but yes point taken. I am isolating my questions in different threads as soon as I am permitted to start a new thread, time wise. $\endgroup$
    – xyz
    Feb 15, 2011 at 8:08
  • $\begingroup$ Okay. I edited out the part about the speed of light because it is off-topic in this question. $\endgroup$ Feb 15, 2011 at 8:11

2 Answers 2


The most precise measures of this distance are from radars in the 1960s. However, the distance has been known, though roughly, since the Ancient Times.

Aristarchus of Samos (310BC - 230BC) used the angle between the Earth-Moon axis and the Earth-Sun when the Moon is in First Quarter (elongation of the Moon, $E$ ) and then, with simple trigonometry, could deduce the distances:

$$ \cos E = \frac {distance (\text{Earth-Moon})} {distance(\text{Earth-Sun})} $$

Since he had already computed the Earth-Moon distance from the duration of lunar eclipses, he could conclude on the Earth-Sun distance. His results were false, because of too loose measure of the angle, but his method was very accurate. See Wikipedia for more details.

Another method was explored in 1672 by Cassini and Richer: they measured the parallax (i.e. the variation in angle when seen from different places) under which Mars was seen in Cayenne and Paris, at the moment of opposition. From this, they deduced the distance Earth-Mars. Then, using the Kepler law

$$\frac{a^3}{p^2}= constant$$ (where $a$ is the distance between the planet and the Sun, and $p$ the sideral time)

they could figure out what was the distance to the Sun.

  • 3
    $\begingroup$ This is a great answer, but it neglects the Venus transit method, which is basically the same thing as Mars but looking the other way. Trying to observe rare transits of Venus led to some incredible journeys around the world and makes for great stories about how far astronomers will go to get their data. $\endgroup$
    – user10851
    Apr 3, 2013 at 22:14

Another way of calculating the earth - sun distance is to look at the centrifugal and the gravitational force. This solution assumes that one already knows the mass of the sun, but thats a different problem ;-). One does only need High-School Math and Physics in order to derive a solution.

Thanks to Newton we know

$F_g = -G\frac{Mm}{r^2}$

where $G=6,674\quad10^{-11}$ is the gravitational constant. We also know the centrifugal force to be

$F_z = \frac{mv^2}{r}$

Putting these two equations together one gets:

$\frac{mv^2}{r} = G \frac{Mm}{r^2}$ $\Rightarrow r = \frac{GM}{v^2}$

Furthermore we know the duration of a year and therefore we know $v$:

$v = \omega r = 2 \pi f r = \frac{2 \pi r}{T}$


$r = \frac{GMT^2}{4 \pi^2 r^2} \Rightarrow r = \sqrt[3]{\frac{GMT^2}{4 \pi^2}} = 149,8 \quad 10^9 \; m$

Which is very close to the real value, which is varying between 147,1 Mio. and 152,1 Mio. km. According to Wikipedia the average distance is 149,6 Mio. km, so our result is actually quite good.

  • $\begingroup$ Do you mean centripetal force? $\endgroup$
    – wingerse
    Mar 9, 2017 at 18:58

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