Astrophysics

Question 1

Define Luminosity:

Answer 1

The total energy emitted in the form of em radiation each second.

Question 2

Define luminosity of stars in terms of luminosity of the sun:

Answer 2

Most stars have luminosity about a million times more than the sun.

Question 3

Define intensity of an object:

Answer 3

The power received per unit area at earth

Question 4

What does the brightness of a star depend on?

Answer 4

Its luminosity and its distance from earth.

Question 5

Describe Hipparchus scale:

Answer 5

Apparent magnitude 1 = very bright, 6 is very dim.

1 is 100 times more intense than 6.

Question 6

How did the Hipparchus scale change?

Answer 6

Changed to logarithmic scale so difference in apparent magnitude 1 and 2 is about 2.51 more bright. The range was extended in both directions e.g. the sun is around -26 apparent magnitude and the limit of the naked eye is around +6 and limit of marge large telescopes is AM +20.

Question 7

How do you calculate brightness/intensity ratio between 2 stars?

Answer 7

I2/I1 =2.51^m1-m2

I is the intensity and m is the apparent magnitude.

Question 8

How is brightness different to intensity?

Answer 8

Brightness is subjective and intensity is an actual value.

Question 9

What is apparent magnitude, m?

Answer 9

Based on how bright things are from earth.

Question 10

Explain parallax:

Answer 10

The apparent change in position e.g. moving fast in a car means things in background seem to move slower than things closer.

It is measured in angle of parallax, the greater the angle the closer it is to you. The angle is the angle between the line between the sun and the star and the line between the earth and the star.

This can be used to measure distance to nearby stars by measuring how fast they move relative to very distant stars when the earth is in different parts of its orbit.

Question 11

Explain parsecs:

Answer 11

If a star is 1 parsec away, then the angle of parallax is 1 arc second = 1/3600 of a degree.

Roughly 3.08 * 10^16

Question 12

Explain absolute magnitude, M:

Answer 12

Based on the luminosity of the star.

It is based on what the apparent magnitude would be if it was 10 parsecs away from earth.

Relationship is m - M = 5log(d/10)
d is the distance in parsecs.

Question 13

Why is absolute magnitude useful?

Answer 13

You can use it to calculate the distance from earth if you know m and M.

Especially handy since the distance of most stars is too distant to measure using parallax.

Question 14

What is a standard candle?

Answer 14

Objects you can measure the luminosity of directly e.g. type 1a supernovae

Question 15

Define AU:

Answer 15

mean distance between earth and sun (mean because distance is not constant as orbit is not perfectly spherical)

1.5 * 10^11m

measured when Venus passed between the earth and sun in 1769.

Question 16

Define light year:

Answer 16

The distance that em waves travel in 1 year.

The further away we see an object, the further into the past it is.

Around 9.46 * 10^15.

Question 17

How many light years in 1 parsec?

Answer 17

1pc = 3.26 ly

Question 18

What is a black body?

Answer 18

A body that absorbs and emits all wavelengths of EM radiation which is what makes it pure black.

Question 19

What objects supposedly act as black bodies?

Answer 19

Stars as they are very good emitters.

Question 20

Describe graph of intensity against wavelength:

Answer 20

The shorter the wavelength the lower the temperature as the graph varies with temperature.

The peak wavelength gives the temperature which is Wiens displacement law = max wavelength * temperature = 2.0 *10^-3 mK (metre kelvins)

Question 21

What is power output related to?

Answer 21

The temperature and the surface area of the star.

this is proportional to the fourth power of the stars temperature.
It is also directly proportional to the stars surface area therefore P = kAT^4 where k is Stefan’s constant.

Question 22

What are the units of Stefan’s constant?

Answer 22

Wm^-2K^-4

Question 23

What does the power of a star obey?

Answer 23

Inverse square law as the radiation dissipates so it lands over a larger area I = P/(4pi*d^2)

Question 24

Why are measurements inaccurate on earth?

Answer 24

Dust and light pollution affect it so observatories are at high altitudes and away from cities and at low humidity.

Only lets some waves through atmosphere.

Detectors are sometimes more sensitive depending on the material. Glass absorbs UV better but is transparent to VL.

Question 25

Which waves does the atmosphere let through?

Answer 25

VL, most radio waves, very near infrared and some UV.

Question 26

What is the Balmer Series?

Answer 26

Classified by colour - wavelengths correspond to energies moving from higher levels to first excitation level or n = 2 (n=1 is ground state)

Question 27

What are the requirements for the Balmer series?

Answer 27

Electrons in hydrogen already need to be in n=2 state which happens when there are high temperatures so collisions between electrons give them extra energy.

The energy can’t be too high otherwise the electrons will reach ionisation e.g. above 10000K.

The light is emitted in all directions so there may be a gap in the absorption of light by the observer.

Question 28

Principles of the use of spectral classes:

Answer 28

When light created from a star passes through its ‘atmosphere’, some wavelengths are absorbed which leads to absorption spectra.

Question 29

Describe spectral class system:

Answer 29

O - blue, 25000-50000K and has absorption lines He and H (atoms that are most commonly seen in the star) - has weak Balmer lines as most hydrogen is ionised.

M- red, is less than 3500K - weak Balmer lines as too little hydrogen and too cool to be excited.

goes in order ‘Oh To Be a Fine Guy/Girl Kiss Me’ -mnemonic