There are two significant measures to get to grips with in astrophysics: distance and brightness.
Key Concepts
The astronomical unit (AU) is defined as the average distance from the Earth to the Sun. This is 149 597 870 700 m, or the slightly more convenient 150 million km (to 3 sf).
A light year (ly) is the distance travelled by light in one year, equivalent to 9.46 × 1015 m.
To calculate we need two quantities:
- The speed of light, 3 x 108 ms-1.
- The number of seconds in one year (365 x 24 x 60 x 60), 3.15 x 107 s.
Distance = speed x time
The parsec (pc) is the largest unit of distance and is defined by the technique of parallax.
Parallax is a technique to determine the distance to stars. You might have noticed on car or train journeys that nearby objects appear to move much faster past the window in comparison to the distant landscape.
As the Earth orbits the Sun (the process that defines the year), nearby stars move relative to the backdrop of distant stars. Using the mathematical technique of similar triangles, we can observe that the angle subtended by the Earth's movement through 1 AU is equal to that subtended by the near star relative to the distant background.
\(\tan p \approx p=\)1 AU / distance to near star
Rearranging: distance to near star = 1 AU / \(p\)
To make things simple the parsec is defined as the distance that subtends an angle of 1 arcsec.
The distance to a star in parsec =\(1\over \text{the angle in arcsec}\)
No further calculation is required.
One parsec is equal to:
- 3.09 x 1016 m
- 6.02 x 105 AU
- 3.26 ly
The luminosity of a star is equivalent to its power, the energy released in one second:
\(L= {\Delta E \over t}\)
This luminosity
is also related to the star's temperature:
\(L=\sigma AT^4\)
- \(L\) is the luminosity (W)
- \(\sigma\) is the Stefan-Boltzmann constant (= 5.67 x 10-8 Wm-2K-4)
- \(A\) is the surface area of the star (m2)
- \(T\) is the absolute temperature of the star (K)
Use flashcards to practise your recall.
Use quizzes to practise application of theory.
START QUIZ!
A telescope is lined up with a star on two nights 6 months apart. The situation is represented by the diagram below.
Distance \(X\) is:
The circle on the left hand side of the diagram represents the Earth orbitting the yellow Sun. The diameter is twice the distance between the Earth and the Sun (which is 1 astronomical unit (AU)).
A telescope is lined up with a star on two nights 6 months apart. The situation is represented by the diagram below.
If the distance from Earth to the star is 4 pc then the angle \(A\) is:
\(\text{distance in pc} ={1\over \text{angle in arcsec}}\)
\(\Rightarrow \text{angle in arcsec} ={1\over \text{distance in pc}}\)
A telescope is lined up with a star on two nights 6 months apart. The situation is represented by the diagram below.
If the angle \(A\) is 0.1 arcsec, the distance from Earth to the star is:
\(\text{distance in pc} ={1\over \text{angle in arcsec}}\)
This photograph shows the stars in the constellation Orion.
From this image alone we can deduce that the stars are of different...
In fact, all of these quantities are different. However, the others can't be deduced from the photograph (e.g. one could be more luminous but further away).
This photograph shows the stars in the constellation Orion.
Alnilam has a higher luminosity than Betelgeuse. From the photo we can deduce that Alnilam is:
Alnilam is less bright to observers here at Earth and so must be further away. The other options are irrelevant - only distance could affect the observed brightness for a fixed luminosity.
This photograph shows the stars in the constellation Orion.
Betelgeuse and Rigel are a similar distance from the Earth, they also have similar...
They have a similar brightness to observers in the same location. Therefore, they must have similar luminosity.
Star X has the same radius as the Sun and temperature 12 000 K.
The ratio \(\text{luminosity}_\text{X}\over \text{luminosity}_\text{sun}\) is equal to:
The surface temperature of the Sun is approximately 6000 K. HINT: It's good to learn this!
\(\Rightarrow T_\text{X}=2T_\text{sun}\)
\(L \propto T^4\) so \({L_\text{X}\over L_\text{sun}}=2^4\)
Star Y has twice the temperature of the Sun but half the radius.
The ratio \(\text{luminosity}_\text{Y}\over \text{luminosity} _\text{sun}\) is equal to:
\(L = σAT^4\) and \(A = 4πr^2\)
\(\Rightarrow L\propto r^2T^4\)
\({L_\text{Y}\over L_\text{sun}}=({1\over2})^2\times 2^4\)
Star Z has the same luminosity as the Sun but half the radius.
The ratio \(T_\text{Z}\over T_\text{sun}\) =
\(L = σAT^4\) and \(A = 4πr^2\)
\(T^4\propto{1\over r^2}\) so \(T\propto{1\over \sqrt r}\)
Since the radius has halved, the temperature is increased by a factor \(\sqrt 2\).
The units of the Stefan-Boltzmann constant are:
\(L = σAT^4\)
\(σ ={L\over AT^4}\)
How much of Stellar quantities have you understood?
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