Those luminous heavenly bodies which have their own light are called stars. We can see near about 2500 stars on a clear night with our naked eye. But in actual, total no of stars are countless. The star which is nearest to earth is sun. It is the brightest star of all. The second nearest star except sun is Alpha-Centauri.
Basic properties of stars are discussed below:
1. Stellar Spectra: Spectrum of all stars is continuous. If we observe the spectrum of stars then we will find some dark lines in it. Stars emit radiations having different wavelength. When these spectrum lines pass through some outermost region having low temperature then some of the radiations are absorbed. This occurs on the basis of the elements that are present in the region having low temperature. This low temperature region is also called as ‘Reversing Layer’. So, due to absorbing of radiations dark lines are produces in the spectrum. After studying the spectrum carefully we can find the composition of various elements present in the reversing layer.
Stars may or may not be of same color. The spectrum lines generated depends upon the star color. This is the main reason that the different stars have different spectrums. After the observations of Stellar Spectra, it was found that the spectrum lines are divided into main seven categories. These categories are named with particular letters .i.e. O, B, A, F, G, K and M.
As the time passed a famous Indian scientist Megh Nath Saha put forward his submissions. He said that the difference found in the spectrum lines of stars is due to the different temperatures of the stars. But in the above discussion we have discussed that the difference in the spectrum lines is due to the different colors of stars.
So, a relation was found which relates the color and temperature of the star.
T= b
Or we can write the above equation as
T = b / ——— (1)
In the above equation b is the Wien’s constant. The value of b is .2898 cmK. Electrostatic figure 4.11 is the wavelength and T is the temperature of the black body. In the above equation T and 1 are inversely proportional to each other. If the temperature will be increased then the wavelength will be decreased. The common colors which appear in the visible region are Violet, Indigo, blue, green, yellow, orange and red. It means the star which will be of blue color will have extremely high temperature. On the other hand the stars having red color will have very low temperature. In the different classes of the stellar spectra the stars are ordered in the decreasing order of the temperature. The above equation can be used to find out the temperature of star by calculating the wavelength at point having maximum no of radiations. Among the seven classes of Steller spectra, our sun falls under the G class. The table shown below is showing the all the details of the different spectra classes.
| Color | Spectra Type | Description of absorption spectrum | Surface temperature in K |
| White | A | Lines of Hydrogen | 9500 to 11000 |
| Blue | B | Lines of neutral helium |
15000 to 23000 |
| Very Blue | O | Lines of ionized helium | 30000 to 40000 |
| Yellow | G | Iron and band of carbon, Lines of ionized calcium | 5800 |
| Orange | K | Bands due to hydrocarbons | 4500 |
| Red | M | Bands of Titanium oxide | 3500 |
| Green | F | Lines of ionized metals and hydrogen | 6500 to 7500 |
2. Brightness of stars: The brightness of a star can be explained with the help of a simple system known as the system of magnitude. Now let’s understand the term Magnitude. The magnitude is the calculation of the brightness of a star when it is watched from earth. But if we talk about the absolute brightness of a star then, it is the calculation of the brightness of a star from a distance of near about 10 parsec’s. The value of 1Parsec is equal to 3.26 light years. The stars which we observe in the sky on a clear night seem to be the same. But it is the fact that the brightness of all the stars is not same. On the basis of brightness of stars, they are divided into six main magnitude classes. These categories were made by a Greek Astronomer called ‘Hipparchus’. Stars which are brightest are kept in the first magnitude class. The other stars which have some fainter look or having less brightness are placed successively in second, third, fourth, etc classes. The approximate difference in the brightness of one magnitude class to that of the other successive magnitude class is approximately 2.512 i.e. [(100) 1/5]. It means that the stars present in one magnitude class are [(100) 1/5] times brighter than the other successive magnitude class. So, the equation formed is shown below:
Brightness of star present in first magnitude class / Brightness of star present in second magnitude class
(2.512)5 = 100
Suppose the brightness values of two stars be l1 and l2, their magnitudes be m1 and m2. So the relation formed will be
l1 / l2 = 100(m2 – m 1) 1/5
On taking the logarithms on both sides of the above equation we get:
m2 – m 1) / 5 log10 100 = log10 (l1 / l 2)
= – log10 (l2 / l 1)
Or m2 – m 1) * 2 / 5 = – log10 (l2 / l 1)
Or m2 – m 1 = -5/2 log10 (l2 / l 1)
= -2.5 log10 (l2 / l 1)
Suppose l1 = l0. Think that if the value of the brightness will be zero for first star. Then m1=0.
After substituting the values in the above equation we get.
m2 = -2.5 log10 (l2 / l 1)
If we write the general form of the equation then it will be like this:
m = -2.5 log10 (l2 / l 1)
Using the above equation we can easily find the value of magnitude.
Most of the stars have negative magnitude. If the stars will be brighter than the zero magnitude then we can say that the stars have negative magnitudes. If any star has a magnitude of -5 then it is said that its brightness will be 100 times more than that of a star having zero magnitude.
Magnitude of some of the objects is given below:
| Object | Magnitude |
| Sun | -26.5 |
| Venus | -4.0 |
| Sirius | -1.5 |
| Jupiter | -2.0 |
3. Chemical composition of a star: The chemical composition of almost all stars is same. The space which is present between stars contains some dust particles and gases. These dust particles and the gases are also made up of the same material as that of the stars. As we know that everything is made up of atoms. So, from the observations it is cleared that nearly 88% of atoms present in stars are hydrogen atoms. The remaining percentage is of helium atoms i.e. 11.8% and 0.2% are some other atoms.
4. Size of stars: The value of luminosity and surface temperature can be used to find out the size of the stars. Lets suppose that the star as a perfectly black body. Take its surface temperature as T.s. Then by the formula of the Stefan’s law:
E=
T4. Here E is the energy emitted by the star per unit area per second.
In the above equation is the Stefan’s constant.
Now form other equation by supposing its radius as R. So, the surface area will be:
= 4 
R2
The Equation for finding the total energy emission of star per second is shown below:
To calculate the radius of the star we need the values of T, L and the value of the Stefan’s constant. Divide the equation 2 by the sun’s radius i.e. 6.95 x 108m. Then the resultant equation after division will be:
The basic range in which all the radii of stars fall is .02 to 200 times than the radius of sun. Stars have different names depending upon their sizes. Some are called giant stars and some are called supergiant stars. Stars which are nearly similar to that of sun but smaller than that of giants are called dwarfs. Stars which are smaller in size as compared to sun are known as small dwarfs.
5. Mass of stars: If the circular motion of two stars is around the common centre of mass then they will form a gravitationally bound system. This system formed by two stars is known as Binary star system.
By checking the motion of the binary star system the masses of two stars can be calculated.
Consider two stars having masses M1 and M2. Distance between their centre’s is a. Then the equation will be:
In the above equation T is the common time period of the revolution around the centre of mass of the binary system. G is the Gravitational constant.
If we take T i.e. the time period of revolution in years and the separation distance of two stars from their centre’s i.e. a is taken in astronomical unit, then the masses of the stars will become solar masses. When these conditions will be applied to the above equation the resultant equation becomes:
M1 and M2 = a3 / T2 —— (2)
If the location of the centre of masses of the binary system are found then we can easily calculate the two distances a1 and a2. We know that
a1 + a2 = a
So, M1a1 = M2a2
Then the estimations of the masses of stars M1 and M1 can be done.
Lets take an example of a ‘Sirius’ star. The masses M1 and M2 of ‘Sirius’ star are calculated and founded that both the value are 1. This value was 2.4 times greater than the value of solar mass.
6. Density of a star: The basic formula by which we can calculate the density of a star is shown below:
= Mass / Volume = 
In the above equation M is the mass and R represents the radius of the star.
From the observations it is clear that the density increases as we move towards the centre of stars. The variation in the density in case of the stars having low temperature is 50000Kgm-3 and for stars having high temperatures is near about 10Kgm-3
