1
$\begingroup$

My planet (I'll call it $P$ for practical reasons) is 2 AU from the Sun, a distance in which it receives 25 % of the sunlight that Earth receives. This can be estimated with the Inverse Square Law:

\begin{gather*} E=\frac{1}{d^{2}} \end{gather*}

For $P$:

\begin{gather*} E=\frac{1}{2^{2}}=0.25 \end{gather*}

With the following formula you can find the illuminance (in lx) on the Earth if you know the apparent visual magnitude of the Sun:

\begin{gather*} E=10^{\frac{-14.18-V}{2.5}} \end{gather*}

For the Earth:

\begin{gather*} E=10^{\frac{-14.18-(-26.75)}{2.5}}\approx 106660 \end{gather*}

Then, by multiplying the illuminance on Earth by 0.25, the illuminance on $P$ is found, being 26 665 lx.

My question is, what angle must the Sun form with the Earth's surface for this illuminance to occur, and at what time of day does this happen?


I found this formula to find the illuminance (in lx) knowing the distance to the light source (in m), its intensity (in cd) and the angle of incidence:

\begin{gather*} E=\frac{I}{d^{2}}\cos\theta \end{gather*}

Solving for $I$:

\begin{gather*} I=\frac{Ed^{2}}{\cos\theta} \end{gather*}

For the Sun from Earth at noon:

\begin{gather*} I=\frac{106660\cdot 149598000000^{2}}{\cos\ 0°}\approx 2.387\times 10^{27} \end{gather*}

The intensity of the light source (the Sun) is this, in cd.

Then, solving for $θ$:

\begin{gather*} θ=\arccos\frac{Ed^{2}}{I} \end{gather*}

And finally replacing the corresponding values:

\begin{gather*} θ=\arccos\frac{26665\cdot 149598000000^{2}}{2.387\times 10^{27}}\approx 1.32° \end{gather*}

Here, the value of $E$ corresponds to that of $P$, but the distance to the light source corresponds to that of the Earth and its intensity remains the same. Then, the angle of the Sun ($θ$) should be 1.32° for the expected illuminance to occur, but this value is not convincing. Did I do something wrong with this procedure? I'm not an expert in photometry.

$\endgroup$
  • $\begingroup$ Is the "Sun" a G-type main sequence star of 1solar mass and about as old as the Sol? $\endgroup$ – TheDyingOfLight Sep 22 '19 at 18:51
  • $\begingroup$ In fact, it is our own Sun. $\endgroup$ – URIZEN Sep 23 '19 at 1:07
1
$\begingroup$
  1. Your calculator gave you the value of arccos in radians. For example, the function ACOS of Excel or LibreOffice Calc does this. To get degrees, multiply by 180 / π = 57.29577951.

  2. You want one quarter of the illuminance we have at noon. No need to calculate the absolute values; just notice that illuminance is proportional to the cosinus of the angle with the normal. Then arccos 0.25 is 1.32 radians, that is, 75°31′.

    Since the formula assumes that 0° is at noon, the 75° correspond with the sun being 15° above the horizon, which, at the equinox (or on the Equator), would be one hour after sunrise or one hour before sunset.

| improve this answer | |
$\endgroup$

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.