Friday, 04 July 2014 00:00

Why do Stars Twinkle but Planets Don't?

By  Gideon
Rate this item
(17 votes)

When we view stars from the surface of the Earth, we a looking up through a ''sea'' of air, the atmosphere, about three hundred miles thick.

Stars are suns, like our own sun, and we see them because they emit particles of light called photons.

The closest star to our solar system, Proxima Centauri, is more than a parsec from our Sun

(1 parsec \(\approx 10 \times 10^{13}\) miles \( \approx 3 \times 10^{18}\) cm),

and most stars are much farther away (hundreds or thousands of parsecs--the Milky Way Galaxy has a radius of about eight thousand parsecs).

[To learn more about parsecs and stellar parallax, please click here.]

[To learn more about cosmic distance scales, please click here.]

Because stars are so far away their apparent angular diameters--that is, the number of degrees they cover in an arc across the celestial sphere as seen from Earth--are tiny. And so they are essentially point sources of light.

When a stream of photons from a star is scattered by the molecules in the atmosphere the photons that are scattered do not reach our eyes, and so the star appears to twinkle.

From space, above the atmosphere; or on the surface of the Moon, which has no atmosphere, stars do not appear to twinkle.

Planets are part of our solar system and so are much closer than the closest star. The Kuiper belt, of which Pluto is a member, is about \( 2 \times 10^{-4}\) parsecs away from the sun (that is, about 2 thousandths of a parsec).

Planets have angular diameters that are much larger than those of stars, and so even if some of the photons they reflect towards Earth are scattered by our atmosphere, there are plenty of other photons that still get through the atmosphere, and so planets do not appear to twinkle.


To get some idea of the differences in the apparent angular diameters of planets and stars, consider first the planet Jupiter.

Jupiter is roughly 5.2 astronomical units, or about \( 5 \times 10^{8}\) miles, or about \( 8 \times 10^{13}\) cm from the sun. (All of these numbers are very rough, but we are only interested in order of magnitude estimates.)

The distance between the Earth and Jupiter will vary according to the two planets' orbits about the Sun, but we will ignore this technical detail.

Jupiter's radius \(r\) is about \(4.4 \times 10^{4}\) miles, or about \( 7 \times 10^{9}\) cm.

If we draw a circle about the Earth with a radius approximately equal to the distance to Jupiter, then the ratio of Jupiter's angular diameter in the sky to the full angular arc of 2 \(\pi \) radians or \(360\) degrees will be approximately equal to the ratio of Jupiter's physical diameter to \(2 \pi \) times the distance \(R\) between Earth and Jupiter.


This is expressed in the diagram above, and in the equation:

\[ \frac{\delta }{2 \pi} = \frac{ d}{2\pi R} = \frac{ 2 r}{ 2 \pi R} = \frac{r}{\pi R}.\]

Where \( \delta\) is Jupiter's apparent angular diameter as seen from Earth, \(r\) is Jupiter's physical radius, \(d = 2r\) is Jupiter's physical diameter, and \(R\) is the physical distance between Earth and Jupiter.

This gives us Jupiter's apparent angular diameter in radians as:

\[ \delta \approx \frac{ 2r}{R} \approx 2 \times 10^{-4} \text{ radians}.\]

We want the value of \(\delta\) in arcseconds and so we need to convert.

There are 360 degrees in \( 2 \pi \) radians there are 60 minutes of arc in every degree, and there are 60 seconds of arc in every minute of arc.

Therefore there are \( (360)(60)(60) = 1.296 \times 10^{6} \approx 1.3 \times 10^{6}\) arcseconds in \( 2 \pi\) radians, and so the conversion factor is:

\[ \frac{ 1.3 \times 10^{6} \text { arcseconds }}{ 2 \pi \text{ radians }} \approx 2 \times 10^{5} \frac{ \text{ arcseconds }}{ \text{ radians }}.\]

And so Jupiter's apparent angular diameter is approximately:

\[ ( 2 \times 10^{-4} )( 2 \times 10^{5} ) = 40 \text{ arcseconds}. \]

(In fact Jupiter's apparent angular diameter varies from about 30 to about 50 arcseconds.)

Now let's do the same calculation for a star.

All stars are more than a parsec away from our Sun, but let's imagine that there is a star exactly one parsec away, and let's imagine that this hypothetical star has the same physical radius as our sun.

Our Sun's radius is about \( 4 \times 10^{5}\) miles, or about \(7 \times 10^{10}\) cm, or about \(2 \times 10^{-8}\) parsecs.

So a star with the same radius as our sun, at a distance of one parsec would have an apparent angular diameter in radians of about:

\[ \delta \approx \frac{2r}{R} = \frac{ 2 (2 \times 10^{-8})}{1} = 4 \times 10^{-8}.\]

In terms of arcseconds this is:

\[ (4 \times 10^{-8})( 2 \times 10^{5} ) = 8 \times 10^{-3} \text{ arcseconds}\]

or eight thousandths of an arcsecond.

And so we see that Jupiter has an apparent angular diameter about five thousand times larger than even the largest possible angular diameter of any star, and that stars are truly very close to point sources.

Read 6585 times Last modified on Thursday, 12 May 2016 14:20
More in this category: « Stellar Parallax
Login to post comments