<p>You can do a rough 0th order estimate:</p>
<p>The Sun has about <span class="math-container">$1360 W/m^2$</span> on the Earth's surface and a brightness of <span class="math-container">$\approx -26$</span> magnitudes. Every 5 magnitudes corresponds to a factor 100 in brightness or every magnitude corresponds to a factor of <span class="math-container">$^5\sqrt{100}$</span>. The moon has <span class="math-container">$\approx -13$</span> magnitudes. Thus it is about 13 magnitudes less bright than the Sun:</p>
<p><span class="math-container">$$P_{moon} = \frac{1}{^5\sqrt{100}^{13}}P_{Sun} = P_{Sun}\cdot 6.3\cdot10^{-6} = 8.5mW/m^2$$</span></p>
<p>If you do that for a planet with -2 magnitudes (so about Venus or Jupiter give or take a magnitude), you are left with <span class="math-container">$340nW/m^2$</span> and for the most distant planet, Neptune with <span class="math-container">$\approx 7$</span>mag only <span class="math-container">$8.5\cdot10^{-11} W/m^2$</span>.</p>
<p>All values above are the power per square meter. For a telescope with 5 inch of unobstructed aperture you have about 1% of these values (<span class="math-container">$\pi\cdot (0.0254\cdot5)^2/4$</span>), neglecting losses in the optics and reduced aperture for the sake of secondary mirror mounts etc.</p>

CC BY-SA 4.0
2023-09-23T08:43:39.557Z
Thank you very much. Is there way to estimate the intensity only for the infrared range?
Tumpa
CC BY-SA 4.0
2023-09-23T11:54:36.233Z
Integrate the Planck function for the Sun for the wavelength range of interest and the Sun's surface temperature. Do that for the whole function as well, and take the corresponding ratio. The 1360W already include the absorption in the Earth's atmosphere.
planetmaker
CC BY-SA 4.0
2023-09-24T23:43:01.850Z
@Tumpa why not go ahead and post a new, separate question about infrared. You'll need to specify 1) your telescope type (reflector is much better than refractor because various glass types have different IR cutoffs) 2) your approximate elevation (atmospheric absorption will be very different, especially at longer wavelengths) and then mention again that you have  a silicon(?) or other semiconductor photodiode and whether it has a glass window over it. Ideally a link to a data sheet for the specific photodiode. Ping me here when you do post your question!
uhoh
CC BY-SA 4.0
2023-09-23T07:12:04.023Z
2023-09-23T07:22:52.950Z
2023-09-23T07:22:52.950Z
planetmaker
<p>I have a photodiode that can measure intensity in picowatt to milliwatt range.</p>
<p>I have a telescope of 5 inch diameter. I want to calculate the intensity of moonlight received by this area? This will help me to decide whether I can use the photodiode for my measurements. I want to do similar calculations for other planets as well. How can I do this?</p>

CC BY-SA 4.0
2023-09-25T11:15:09.640Z
Related: https://astronomy.stackexchange.com/q/49335/34443
ksousa
<amateur-observing><data-analysis><spectroscopy>
Calculating intensity of moonlight per unit area
Tumpa
# Calculating intensity of moonlight per unit area

(Asked by: username_0 on 2023-09-23)

I have a photodiode that can measure intensity in picowatt to milliwatt range.

I have a telescope of 5 inch diameter. I want to calculate the intensity of moonlight received by this area? This will help me to decide whether I can use the photodiode for my measurements. I want to do similar calculations for other planets as well. How can I do this?

**Tags:** amateur-observing, data-analysis, spectroscopy
> **Comments:**
> **username_1**: Related: https://astronomy.stackexchange.com/q/49335/34443  

---

## 1 answers

### Answer by username_2 on 2023-09-23. Score: 2

You can do a rough 0th order estimate:

The Sun has about $1360 W/m^2$ on the Earth's surface and a brightness of $\approx -26$ magnitudes. Every 5 magnitudes corresponds to a factor 100 in brightness or every magnitude corresponds to a factor of $^5\sqrt{100}$. The moon has $\approx -13$ magnitudes. Thus it is about 13 magnitudes less bright than the Sun:

$$P\_{moon} = \frac{1}{^5\sqrt{100}^{13}}P\_{Sun} = P\_{Sun}\cdot 6.3\cdot10^{-6} = 8.5mW/m^2$$

If you do that for a planet with -2 magnitudes (so about Venus or Jupiter give or take a magnitude), you are left with $340nW/m^2$ and for the most distant planet, Neptune with $\approx 7$mag only $8.5\cdot10^{-11} W/m^2$.

All values above are the power per square meter. For a telescope with 5 inch of unobstructed aperture you have about 1% of these values ($\pi\cdot (0.0254\cdot5)^2/4$), neglecting losses in the optics and reduced aperture for the sake of secondary mirror mounts etc.

> **Comments:**
> **username_0**: Thank you very much. Is there way to estimate the intensity only for the infrared range?  
> **username_2**: Integrate the Planck function for the Sun for the wavelength range of interest and the Sun's surface temperature. Do that for the whole function as well, and take the corresponding ratio. The 1360W already include the absorption in the Earth's atmosphere.  
> **username_3**: @Tumpa why not go ahead and post a new, separate question about infrared. You'll need to specify 1) your telescope type (reflector is much better than refractor because various glass types have different IR cutoffs) 2) your approximate elevation (atmospheric absorption will be very different, especially at longer wavelengths) and then mention again that you have  a silicon(?) or other semiconductor photodiode and whether it has a glass window over it. Ideally a link to a data sheet for the specific photodiode. Ping me here when you do post your question!  

---

