Luminosity refers to the total amount of energy produced by different celestial bodies(stars, galaxy) per unit time and it is basically measured in joules per second or watts in SI units. Values of luminosity are given in terms of the luminosity of the sun or in terms of magnitude which is called the absolute bolometric magnitude of an object is the measure of the total energy emission rate. A bolometer can be used to measure the radiant energy by the method of absorption and measurement of heating.

Luminosity generally depends upon two factors:

**The size of the star**– The larger a star is, the more energy it puts out and the more luminous it is.

**The temperature of the star**– If two stars are of the same size but have different temperatures, then the star with higher temperature will be more luminous than the star with lower temperature.

**The Brightness of a Star:**

The brightness of a star which appears to our naked eyes mainly depends upon two factors:

- Luminosity of the star
- Distance of the star from the Earth

The Apparent Brightness of a source is a consequence of geometry. As light rays emerge from a source, they spread out in an area. According to Inverse Square Law of brightness, the Apparent Brightness (B) of a source is inversely proportional to the square of its distance (d).

Mathematically,

\(B\propto \frac{1}{d^{2}}\)

**Luminosity Theory:**

- Luminosity depends on the surface area of the star. If radius of a star is R then,

**Surface area of the star = 4PR****2**

- Two stars having the same temperature, one with radius 2R will have 4 times greater luminosity than a star with radius R.

- The luminosity of a star also depends upon its temperature. When considering a star to be a completely black body, the radiation emitted per second will be according to the Stefan- Boltzmann law.

** Energy emitted per second (E) = sAT****4**

Where,

s= Stefan’s constant with a value of \(5.7\times10^{-8}Wm^{-2}K^{-4}\)

A= Surface Area of the Star

T = absolute temperature of the star

- Calculating the energy output for a star which is of the same size as the sun.

R = 6.96×108m and T = 6000K

So, surface area of sun = surface area of star = A = \(4\pi r^{2}\)

= \(4\times \pi \times ( 6.96\times 10^{8})\\ \\ =6.08\times 10^{18}\)

Therefore \(E=5.7\times10^{-8}\times6.08\times10^{18}\times6000^{4}=4.5\times10^{26}J\)

A star having twice the temperature of Sun would have 16 times greater energy output.

**Example** – The Sun having a surface temperature of 6000K produces radiation with lmax = 420 nm. Find out the temperature of the Sirius if l max of Sirius is 72nm.

**Answer** – \(l_{max1}T_{1}=l_{max2}T_{2}\)

Therefore,

\(420\times10^{-9}\times6000=72\times10^{-9}\times T_{2}\)

\(T_{2}=35000K\)