Classification Of Stars

Question 1

Luminosity

Answer 1

rate of light energy released / power output of a star

Question 2

Intensity

Answer 2

this is the ​power received from a star (luminosity) per unit area​ (W m-​2).​
intensity of a star follows the ​inverse square law​

Question 3

Apparent magnitude

Answer 3

how bright the object appears​ in the sky, therefore this will depend on a star’s luminosity and distance from the Earth.

Question 4

Hipparcos scale

Answer 4

classifies astronomical objects by their apparent magnitudes, with brightest ​stars given apparent magnitude of ​1​, and faintest ​stars given apparent magnitude of ​6​.
​intensity of a magnitude 1 star is 100 greater than a magnitude 6 star​.
The Hipparcos scale is ​logarithmic​, as the magnitude changes by 1, the intensity changes with a ratio of ​2.51​.

Question 5

Absolute magnitude

Answer 5

what its ​apparent magnitude would be if it were placed 10 parsecs away from the Earth.

Question 6

Equation for magnitudes

Answer 6

m − M = 5log( d/10)
d is the distance in parsecs

Question 7

Parallax error

Answer 7

apparent change of position of a nearer star in comparison to distant stars​ in the background, ​as a result of the orbit of the Earth around the Sun.​
property is measured by the ​angle of parallax (θ)​
The greater the angle of parallax, the closer the star is to the Earth.

Question 8

Astronomical Unit (AU)​

Answer 8

average distance between the centre of the Earth and the centre of the Sun.
1 AU = 1.50 × 10^11 m

Question 9

Parsec (pc)​

Answer 9

The distance at which the ​angle of parallax is 1 arcsecond (1/3600th of a degree)​.
This distance can also be described as ​the distance at which 1 AU subtends an angle of 1 arcsecond​.
1 pc = 2.06 × 105 AU = 3.08 × 1016 m = 3.26 ly

Question 10

Light year (ly)

Answer 10

The distance that an EM waves travels in a year in a vacuum. 1 ly = 9.46 × 1015 m

Question 11

To find d in magnitude equation

Answer 11

To find ​d ​use trigonometry. tan θ = opp → tan θ = r → d = θr ​As tan θ ​≈ ​θ for small θ
Where d and r are in metres and θ is in radians.
Using the equation above you can find d in parsecs,.
d = 1/θ
Where d is in parsecs and θ is in arcseconds.

Question 12

Parallax diagram

Answer 12

Question 13

Black body radiator

Answer 13

is a ​perfect emitter and absorber of all possible wavelengths of radiation​.
➔ Stars can be approximated as black bodies.

Question 14

Stefans law

Answer 14

the power output (luminosity/P) of a black body radiator is ​directly proportional​ to its ​surface area (A) ​and its​ (absolute temperature)4​ .​
P =σAT^4
Where T is the absolute temperature and σ is the ​Stefan constant​ (5.67 x 10-​ 8​ W m-​ 2 ​K-​ 4)​ .

Question 15

Weins displacement law

Answer 15

​the peak wavelength (λm​ax)​ of emitted radiation is ​inversely proportional​ to the ​absolute temperature (T)​ of the object.
The peak wavelength (λm​ ax)​ is the wavelength of light released at maximum intensity.
λmaxT = constant = 2.9 × 10−3m K Where the unit mk is ​metres-Kelvin​s

Question 16

Weins displacement law explanation

Answer 16

Wein’s law shows that the peak wavelength of a black body ​decreases​ as it gets hotter meaning frequency increases so energy of wave increases​
law can be used to ​estimate ​temperature of black-body sources.

Question 17

Black body curves graph

Answer 17

follows Wein’s law, as the temperature of the body increases, the peak wavelength decreases.
intensity of light emitted by a star follows the ​inverse square law​, meaning ​intensity is inversely proportional to the distance between the star and the observer.
shown in the following formula:
I=P /4πd^2
P is the power output by the star and d is the distance from the star.

Question 18

Stellar spectral classes

Answer 18

Stars classified into ​spectral classes​ based on the strength of absorption lines.
absorption lines dependent on the temperature​ of the star.
This is because energy of particles which make up star is dependent on its temperature.

Question 19

Hydrogen balmer lines

Answer 19

absorption lines that are found in the spectra of ​O, B and A type stars​.
-caused by excitation of hydrogen atoms from the ​n = 2 state​ to higher/lower energy levels.
-if temperature of a star is ​too high​, majority of hydrogen atoms will become excited to higher levels than n=2​ or electrons might even become ionised, so hydrogen balmer lines will not be present.
-if temperature of star too low​, hydrogen atoms unlikely to become excited,​ or may not be present at all, so hydrogen balmer lines will not be present.
-therefore intensity of hydrogen balmer lines is dependent on temperature​.

Question 20

Spectral class table

Answer 20