# Energy flux formula

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?

• $\frac{\pi R^2}{\pi D^2}$ 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. – jaromrax Mar 3 '17 at 14:54
• Why don't you consult a textbook (Stefan-Boltzmann Law) before posting such a question here? – freecharly Mar 3 '17 at 16:58

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 $\sigma ={\frac {2\pi ^{5}k^{4}}{15c^{2}h^{3}}}=5.670373\times 10^{{-8}}\,{\mathrm {W\,m^{{-2}}K^{{-4}}}}$.
• The units I need are $\text{W}/\text{m}^2$ – treq Mar 3 '17 at 14:40
• @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. – Wojciech Morawiec Mar 3 '17 at 14:51
• The thing I mean is that: $$\text{G}=\sigma\cdot\text{T}^4\cdot\left(\frac{\text{R}_\text{sun}}{\frac{\text{p}}{1+\epsilon\cos\left(\theta\right)}}\right)^2$$ – treq Mar 3 '17 at 14:53
• @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. – Wojciech Morawiec Mar 3 '17 at 14:54