Question

Problem 16.66


A star whose surface temperature is (50) kK radiates (4.0×10^27) W.

If the star behaves like a blackbody, what is its radius?

Direct answer: 3.0 x 10^7

How do you get to the answer (inc. formulas)?

Answer 1

do you know the stefan's law?

Answer 2

according to stefans law
\[E=AT ^{4} \]
use this E=total radient energy emitted
A=surface area

Answer 3

\[r = \sqrt{\frac{ A }{ 4\pi }}\] A being surface area of an sphere.

I do not get 3*10^7 when I put \[A = \frac{ E }{ T ^{4} }\] in place of A to get the radius.

Further help much appreciated!

Answer 4

I assume T is the (50 000) K though

Answer 5

i'm getting 2.8*10^7

Answer 6

My numbers input;
E = (4*10^27)
T^4 = (50000^4)

\[r =\sqrt{\frac{ \frac{ E }{ T ^{4} } }{ 4\pi }}\]

r = 7136,496465

What am I doing wrong so far?

Answer 7

probably the pre given answer is a bit wrong!

Answer 8

godd you guys missed the stefan's constant!!! O.O

Answer 9

The answer should be close to 3.0 * 10^7 , 7136 is far from it :P

Answer 10

You guys used the wrong equation. Do a dimensional analysis of the equation you used, and you'll find the units are wrong.

The correct equation is:\[L=\frac{ P }{ A }=\sigma T ^{4}\]where L is luminance; P is power; A is area; T is temperature; and sigma is the Stefan-Boltzmann constant.

Answer 11

L is radiance, not luminance.

Answer 12

Using the correct equation, I get:\[r=2.997\times 10^{7}m \approx 3\times10^{7}m\]

Answer 13

That's it.

Note that dimensional analysis is a powerful tool in physics. You can use it to be sure your answer and your equations are correct; and you can use it to gain insight on what factors influence a physical phenomenon. If you get to the point in physics were you are doing derivations and proofs of equations, dimensional analysis is invaluable.

Answer 14

Thanks a lot to all contributors! Finally managed to get from A to Z.

Using all information given to me, this is how I got to the answer. :)

----------
L = radiance [W * sr^−1 * m^−2]
P = power [W]
A = surface area [m]
sigma = Stefan-Boltzmann constant = 5.6703x10^-8 [W * m^-2 * K^-4]
----------

\[L = \frac{ P }{ A } = \sigma T ^{4}\]

Turned it into
\[A = \frac{ P }{ \sigma T ^{4} }\]

put A above into
\[r =\sqrt{\frac{ A }{ 4\pi }}\]

which would be (when putting in all numbers)
\[r = \sqrt{\frac{ \frac{ 4.0 \times 10^{27} }{ 5.6703 \times 10^{-8} \times (50000)^{4} } }{ 4\pi }} = 2.99697 \times 10^{7} \approx 3.0 \times 10^{7}\]