0
$\begingroup$

I think that the energy flux is given by:

$$\text{G}=\mathcal{k}\cdot\text{T}^4\cdot\left(\frac{\text{R}}{\text{D}}\right)^2\tag1$$

Where $\mathcal{k}$ is the Boltzmann constant, $\text{T}$ is the surface temprature of the sun, $\text{R}$ is the radius of the sun and $\text{D}$ is the distance from the sun to the earth.

Now, for $\text{D}$ we know that is changes over a year because the earth makes a elliptical orbit around the sun.

Question: Is my formula right?

||cite|improve this question
$\endgroup$
  • $\begingroup$ $\frac{\pi R^2}{\pi D^2}$ is fine, however, the flux is not $k T^4$. Try to take total energy produced by the Sun instead of $k T^4$. Or include Sun surface. $\endgroup$ – jaromrax Mar 3 '17 at 14:54
  • 1
    $\begingroup$ Why don't you consult a textbook (Stefan-Boltzmann Law) before posting such a question here? $\endgroup$ – freecharly Mar 3 '17 at 16:58
0
$\begingroup$

Your formula for sure has wrong units. Usually, energy flux has units of $\text{W} = \frac{\text{J}}{\text{s}}$, maybe $\frac{\text{W}}{\text{m}^2} = \frac{\text{J}}{\text{m}^2\text{s}}$.

The $G$ you define on the other hand has units of $\text{J}\text{K}^3$, which doesn't fit the standard unit definition.

You may be thinking about the Stefan-Boltzmann law, where for a blackbody we have $$ P = \sigma \cdot A\cdot T^4 $$ with $P$ the radiated power, $A$ the surface area, $T$ the temperature and $\sigma ={\frac {2\pi ^{5}k^{4}}{15c^{2}h^{3}}}=5.670373\times 10^{{-8}}\,{\mathrm {W\,m^{{-2}}K^{{-4}}}}$.

||cite|improve this answer
$\endgroup$
  • $\begingroup$ The units I need are $\text{W}/\text{m}^2$ $\endgroup$ – treq Mar 3 '17 at 14:40
  • $\begingroup$ @treq What are you trying to calculate? You are mentioning the sun and earth, so I assume you want to get to the Solar constant (en.wikipedia.org/wiki/Solar_constant). In that case, just divide by $A$ in the Stefan-Boltzmann law. $\endgroup$ – Wojciech Morawiec Mar 3 '17 at 14:51
  • $\begingroup$ The thing I mean is that: $$\text{G}=\sigma\cdot\text{T}^4\cdot\left(\frac{\text{R}_\text{sun}}{\frac{\text{p}}{1+\epsilon\cos\left(\theta\right)}}\right)^2$$ $\endgroup$ – treq Mar 3 '17 at 14:53
  • $\begingroup$ Where I used kepler's law, is that correct? I try to calculate the sunconstant $\endgroup$ – treq Mar 3 '17 at 14:53
  • $\begingroup$ @treq Yes, that is correct. Do you see that in your question you have written $k$ where it should be $\sigma$? The $\sigma$ is correct, as you can see by my and your formula. $\endgroup$ – Wojciech Morawiec Mar 3 '17 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.