Last: 8. The Sun as a Star  Next: 10. The Lives of Stars 
Since ancient times, many people have suggested that the stars are like the sun, but seen at such great distances that they appear faint instead of dazzlingly bright. One test of this idea is to (1) measure distances to stars, (2) calculate how bright the Sun would appear if seen at such distances and (3) compare the result with actual stars.
At this point, we have some unfinished business to complete:
To express distances between planets in terrestrial units, we need to determine the value of 1 AU.
Ancient astronomers constructed the first astronomical distance scale:
The Keplerian ModelKepler assigned each planet, including the Earth, an elliptical orbit around the Sun with a definite semimajor axis (in AU) and period.
Measuring the distance (in terrestrial units) to any planet would provide enough information to find the value of 1 AU. 
To measure the distance from the Earth (E) to a planet (P), observations are made at two different places. Suppose that one observation is made with P exactly overhead and aligned with a distant star. The other observation is made with P on the horizon; the angle between P and the distant star is θ.
The distance D from E to P can be calculated in terms of the Earth's radius, R_{⊕}:
D = 
1

R_{&oplus}^{ } 
360°

You may recognize this equation  or something like it  from an earlier lecture. Eratosthenes used it to calculate the size of the Earth. Now the same relationship is being used to calculate distances to planets!
On October 1, 1672, Mars passed in front of a moderately
bright star. Realizing that this presented an opportunity to
measure the distance to Mars, Cassini organized simultaneous
observations from Paris (Europe) and from Cayenne (South
America). 
Giovanni Cassini [Wikipedia] 
According to Kepler's model, Mars was 0.43 AU at this time. Cassini used the observed parallax and the known distance between Paris and Cayenne to to determine the distance to Mars, and find a value for 1 AU (which was 8% too small):
D_{Mars} = 6.0×10^{7} km = 0.43 AU  1 AU = 1.38×10^{8} km 
A transit of Venus happens when Venus passes between us and the Sun. Such passages take several hours, and during that time the planet can be easily observed as a small black dot moving across the face of the Sun. 
Parallax [Wikipedia] 
In 1761, 1769, 1874, and 1882, teams of astronomers scattered across the globe to observe transits of Venus. They hoped to use the parallax of Venus (relative to the Sun) to determine its distance, and thus set the scale of the solar system. However, the results were disappointing  atmospheric blurring made it difficult to determine the planet's position against the Sun.
In 1877, observations of Mars using the rotation of the Earth to carry a single observer along a WestEast baseline finally gave an accurate value:
1 AU = 1.49×10^{8} km 
For this purpose, asteroids were even better targets than Mars; their starlike appearances make parallaxes easier to measure. In 1930, observations of the nearEarth asteroid Eros gave
1 AU = 1.4960×10^{8} km 
Since 1958, radar has been used to measure distances directly by timing a radio signal bounced off another planet. The currently accepted value is
1 AU = 1.49597870×10^{8} km 
To measure stellar distances, a baseline much larger
than the Earth is required. The Earth's orbit provides
a suitable baseline for measuring distances to nearby stars;
observations made 6 months apart are separated by a baseline
of Over the course of a year, a nearby star appears to trace a small oval in the sky when compared to more distant ones. By measuring the parallax angle p (defined as half the long axis of this oval), the distance D to the star can be found:
By now this equation should seem familiar... 
Suppose we could measure stellar parallaxes from Mars instead of from the Earth. Over the course of one Mars year, nearby stars would move in ovals which are

In practice, even distances to neary stars are hard to measure, because the angles involved are so tiny. The nearest star has a parallax angles p = 0.76'', where 1'' = (1° ⁄ 3600) is a second of arc. Since stellar parallaxes are usually measured in seconds of arc, it's convenient to simplify the parallax equation by defining a new unit of distance, the parsec:
1 pc = 
360 × 3600

AU =  206264.8 AU =  3.086×10^{13} km_{ } 
With this definition, the parallax equation becomes:
D = 
1''

pc 
(It turns out that 1 pc = 3.26 ly.)
Suppose star B is twice as far as star A. Which star has a larger parallax angle?
The nearest known stars are α Centauri, a pair of sunlike stars at a distance of 1.32 pc, and their faint companion Proxima Centauri, which is marginally closer. A handfull of bright stars, including Sirius, Vega, Procyon, and Altar, lie within a sphere 10 pc in radius centered on the Sun. The same sphere also contains dozens of faint stars, most too dim to be seen without a telescope!
Parallax measurements made from Earth are limited by the effects of our atmosphere; consequently, distances greater than about 20 pc are hard to measure. More accurate measurements can be made from space; the Hipparcos satellite has measured distances to over 100000 stars within 100 pc. This animation shows the 150 nearest stars, with the Sun at the center. 
By building step by step on earlier measurements, we have now constructed a distance scale out to nearby stars:
Suppose we discovered an error in our calculation of the AU, and the correct value was 10% larger than we had thought. What impact would this have on our values for stellar distances?
Light from a central source spreads out in all directions.
Define an object's brightness as the total
amount of light a fixed area receives. Then the relationship
between brightness B, luminosity L, and distance
D is

This is called the InverseSquare law because the brightness is proportional to the inverse (that is, one over) the square of the distance.
Assuming a given value for the Sun's total energy output or luminosity, we can work out how bright it appears at a distance of 1 AU, which is where we observe it.
L_{Sun} = 3.8×10^{26} watt
D = 1 AU = 1.50×10^{8} km = 1.50×10^{11} m
B = 
L_{Sun}

= 
3.8×10^{26} watt

= 1350 watt ⁄ m^{2}_{ } 
Our result is that at our distance from the Sun, each square meter receives 1350 watt.
How bright is the Sun from other planets? We can plug in their distances and work through the calculation, but it's easier to form ratios: if the same luminosity L is seen at two distances D_{1} and D_{2}:
B_{1} : B_{2} = D_{2}^{2} : D_{1}^{2}
For example, Venus is 0.723 AU from the Sun. Compared to sunlight on Earth, sunlight on Venus is
B_{V} : B_{E} = D_{E}^{2} : D_{V}^{2} = (1 AU)^{2} : (0.723 AU)^{2} = 1 : 0.523 = 1.91 : 1
In other words, sunlight on Venus is 91% brighter than on Earth.
Now we can ask how bright the Sun would appear if seen from the same distance as a nearby star. Let's pick the distance of αCen.
D_{&alpha} = 1.32 pc = 2.7×10^{5} AU
B_{α} : B_{E} = D_{E}^{2} : D_{α}^{2} = (1 AU)^{2} : (2.7×10 AU)^{2} = 1 : 7.29×10^{10} = 1.37×10^{11} : 1
In other words, the Sun would appear 1.37×10^{11} times fainter at the distance of αCen.
At their actual distances, αCen appears 2.09×10^{11} times fainter than the Sun. So if they were viewed at the same distance αCen would appear 2.09 ⁄ 1.37 = 1.53 times brighter then the Sun, or L_{α} = 1.53 L_{Sun}.
To summarize, αCen and the Sun are not exactly the same, but they are similar in terms of brightness. This is evidence that the Sun is a star.
We can play the same game with other stars near the Sun. It turns out that some are much brighter than the Sun, some are like the Sun, and most are much fainter than the Sun. For a sample of the 150 stars nearest the Sun, this table shows how many we find in various ranges of luminosity (measured in units of L_{Sun}).

Last: 8. The Sun as a Star  Next: 10. The Lives of Stars 
Joshua E. Barnes
(barnes@ifa.hawaii.edu)
Last modified: October 19, 2006 http://www.ifa.hawaii.edu/~barnes/ast110_06/hfts.html 