Question

Astrophysics Question

Answer 1

A star lies at a distance of 50pc, has a bolometric flux of 6.02×10−16Wm−2, and
its spectrum peaks at 290nm. Calculate:
(a) The temperature of the star [3 Marks]
(b) The radius of the star in Solar Radii [7 Marks]
(c) The mean speed of the gas particles (you may assume m = mneutron = 1.67×10−27kg)
[3 Marks]

Answer 2

I'm totally lost on this one. I have no idea about how to relate the variables.

Answer 3

I'm sorry, I'm not good with astrophysics

Answer 4

Np, thanks for looking.

Answer 5

Do i use Wien's displacement law for (a)?
\[\lambda_{\max}T =0.0029\]

Answer 6

so \[T =\frac{ 0.0029 }{ 290\times10^{-9} }=10,000\]

Answer 7

@Astrophysics

Answer 8

They would use parsecs eh..

Answer 9

\[50 pc = 163.1 ly\] right?

Answer 10

yh

Answer 11

Ok lets see..Wein's law should work, \[163.1 ly = 1.543 \times 10 ^{18}m\]\[\lambda_{\max} = 290 \times 10^{-9}m\]\[I = 6.02 \times 10 ^{-16}\frac{ W }{ m^2 }\]\[\lambda _{\max} T = 2.898 \times 10^{-3}m K\] solve for T

Answer 12

We will apply Stefan - Boltzman Law for the radius

Answer 13

ok

Answer 14

I got T=9993K

Answer 15

What is the intensity? It's not 10^(-16) right?

Answer 16

no, intensity and flux are different.

Answer 17

It's the same units, I think that's suppose to be intensity of radiation, I'm not sure what bolometric flux is

Answer 18

It's the first time I've come across it that's why I'm struggling

Answer 19

Mhm interesting, it has same units as intensity and since we are applying these same formulas I think we can go with it, as I'm not too sure what exactly it is

Answer 20

yh, I guess we could.

Answer 21

Unless google says something different

Answer 22

lol

Answer 23

Haha it's been some years since I've dealt with these kinds of questions, but in any case lets see, so we have stefan boltzmann law which is \[R = \int\limits_{0}^{\infty} I(\lambda)d \lambda \] which becomes...\[R = \sigma T^{4}\]

Answer 24

\[\sigma = 5.6705 \times 10 ^{-8} W/m^2 K^4\]

Answer 25

is R the flux?

Answer 26

ok

Answer 27

Note that the R is the distance to the start

Answer 28

We can definitely use the total power P of the star to find the radius though

Answer 29

Distance to the star*

Answer 30

\[P = 4 \pi r^2 \sigma T^4\] right, that would be the total power?

Answer 31

yes I get that

Answer 32

Now we may apply the inverse square law and get \[P = 4 \pi R^2 I\]

Answer 33

Simply equate the two now and solve for r :)

Answer 34

ok so the 'r' in the first power equation will be the radius of the star?

Answer 35

R = distance to the star
r = radius of the star
sorry about that

Answer 36

Np, i got it

Answer 37

You good from here? I have to go for a while right now

Answer 38

yh, thanks for the help. really appreciate it.

Answer 39

Np

Answer 40

astrophysics question answered by astrophysics

Answer 41

Can anyone please help me with part (c)?

Answer 42

The other two were blackbody radiation related, the third I'm not entirely sure about but I believe this could work: https://en.wikipedia.org/wiki/Root-mean-square_speed \[v = \sqrt{\frac{ 3RT }{ M_m }}\] read up on the variables

Answer 43

Is it the same as\[v=\sqrt{\frac{ 2kT }{ m }}\]

Answer 44

Correct, using maxwell distribution

Answer 45

in the link you gave it's '3kT' instead of '2kT', why the difference?

Answer 46

Here it shows 3 http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html

Answer 47

It's 3, not 2 I think you meant to put energy then there's a 2

Answer 48

\[v = \sqrt{\frac{ 2E_k }{ m }}\] is equivalent to \[v = \sqrt{\frac{ 3kT }{ m }}\]

Answer 49

https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution
I think the reason why there's a 2 here is because it calculates the 'most probable value' whereas we want the mean value, so 3 is correct.

Answer 50

Ok, that's it for this question. Thanks again for your help. Hope you have a great day.

Answer 51

I think you're right, and np :) take care!