<h2>Setting up the problem</h2>
<p>Here is a little picture of our geometry (nothing to scale):
<a href="https://i.sstatic.net/BI4wP.png" rel="noreferrer"><img src="https://i.sstatic.net/BI4wP.png" alt="enter image description here" /></a></p>
<p>Here, <span class="math-container">$i$</span> is the phase angle, <span class="math-container">$v$</span> is the distance from the Sun to Venus, <span class="math-container">$e$</span> is the distance from the Sun to Earth and <span class="math-container">$d$</span> is the distance from Venus to Earth. Please pretend the line segments originate from the center of each body.</p>
<p>The brightness <span class="math-container">$b$</span> of Venus will be proportional to <span class="math-container">$p$</span> the percentage illuminated, given by Greg Miller in the notes as <span class="math-container">$p=\frac{1}{2}(1+\cos i)$</span>.  It will be inversely proportional to the square of the distance <span class="math-container">$d$</span> between Venus and Earth.  We can express this as:</p>
<p><span class="math-container">$$b \propto \frac{p}{d^2}$$</span></p>
<p>We just need to define <span class="math-container">$d$</span> as a function of <span class="math-container">$i$</span> and then we can find where <span class="math-container">$b$</span> is maximized.</p>
<h2>Finding the distance as a function of Phase Angle</h2>
<p>From the law of cosines, we have <span class="math-container">$$e^2=d^2+v^2-2dv\cos i$$</span>  We can write this as a quadratic polynomial in <span class="math-container">$d$</span>: <span class="math-container">$$d^2-(2v\cos i)d-(e^2-v^2)=0$$</span>  We can solve for <span class="math-container">$d$</span> using the quadratic formula to get two solutions:
<span class="math-container">$$d=\frac{2v\cos i \pm \sqrt{(2v\cos i)^2+4(e^2-v^2)}}{2}$$</span></p>
<p>We will get negative distance if we subtract the radical, so we can consider only the positive cases with a single solution of <span class="math-container">$d$</span> in terms of <span class="math-container">$i$</span> as:</p>
<p><span class="math-container">$$d=\frac{2v\cos i + \sqrt{(2v\cos i)^2+4(e^2-v^2)}}{2}$$</span></p>
<h2>Defining brightness as a function of phase angle</h2>
<p>Plugging <span class="math-container">$d$</span> and <span class="math-container">$p$</span> back in to <span class="math-container">$b$</span> and simplifying, we get</p>
<p><span class="math-container">$$b \propto \frac{p}{d^2} = \frac{2(1+\cos i)}{(2v\cos i + \sqrt{(2v\cos i)^2+4(e^2-v^2)})^2}$$</span></p>
<p>This now looks like a classic calculus problem, where we can take the derivative of <span class="math-container">$b$</span> with respect to <span class="math-container">$i$</span>, set the result equal to zero and solve to find a value for <span class="math-container">$i$</span> which maximizes <span class="math-container">$b$</span>.  But why do calculus when you have a computer?</p>
<h2>What phase angle yields maximum brightness?</h2>
<p>Here is a graph of the (proportional) value of b through all phase angles in one degree increments:</p>
<p><a href="https://i.sstatic.net/wSVxw.png" rel="noreferrer"><img src="https://i.sstatic.net/wSVxw.png" alt="enter image description here" /></a></p>
<p>Zero degrees is fully illuminated Venus at maximum distance from Earth.  180 degrees is fully shadowed Venus at minimum distance.  The maximum brightness is at 119 degrees phase angle <span class="math-container">$i$</span>.  That is at about 26% illumination.</p>
<p><strong>Note:</strong></p>
<ol>
<li>We make the additional assumption that the Earth and Venus orbits are coplanar.</li>
<li>I never make misnakes, but please feel free to check my work.</li>
</ol>

CC BY-SA 4.0
2022-10-29T03:10:58.213Z
"I never make _misnakes_, but please feel free to check my work." talk about irony ;D
Nilay Ghosh
CC BY-SA 4.0
2022-10-28T18:43:24.717Z
2022-10-28T18:43:24.717Z
Connor Garcia
<p>How can I find (in terms of the angle) the moment when in the system Sun-Earth-Venus, Venus can be seen the most bright if its brightness (flow received in Earth) is proportional to the size projected in its illuminated side? ¿What percentage of Venus's surface can be seen illuminated from Earth at that moment? We have to assume circular orbits for the Earth and Venus around the Sun and the distances between those planets and the Sun are known (1 UA and 0,723 AU)</p>

CC BY-SA 4.0
2022-10-27T22:52:06.383Z
The brightness of Venus situation is really interesting! Because it is only a slim crescent when closest and full disk when furthest, these tend to cancel and the overall brightness is fairly constant. See for example answers to [Venus' magnitude during inferior conjunction](https://astronomy.stackexchange.com/q/33797/7982) and [Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths' side?](https://astronomy.stackexchange.com/q/36232/7982)
uhoh
CC BY-SA 4.0
2022-10-28T13:38:16.020Z
JPL Horizons can compute this in actual terms rather than hypothetical.  Select "Phase angle" and "Visual Mag" from the Observer Table Settings.  It only computes data for a given instant, so you'll want have it compute it at daily intervals for a year.
https://ssd.jpl.nasa.gov/horizons
Greg Miller
CC BY-SA 4.0
2022-10-28T13:44:21.793Z
Just realized you asked for illuminated percentage, that can be computed from the phase angle as $ k = \frac{1+\cos i}{2} $, where $ i $ is the phase angle.
Greg Miller
<the-sun><orbit><earth><venus><brightness>
Does anybody know how to solve this Earth-Venus-Sun problem?
Miguel Gutiérrez
# Does anybody know how to solve this Earth-Venus-Sun problem?

(Asked by: username_0 on 2022-10-27)

How can I find (in terms of the angle) the moment when in the system Sun-Earth-Venus, Venus can be seen the most bright if its brightness (flow received in Earth) is proportional to the size projected in its illuminated side? ¿What percentage of Venus's surface can be seen illuminated from Earth at that moment? We have to assume circular orbits for the Earth and Venus around the Sun and the distances between those planets and the Sun are known (1 UA and 0,723 AU)

**Tags:** the-sun, orbit, earth, venus, brightness
> **Comments:**
> **username_1**: The brightness of Venus situation is really interesting! Because it is only a slim crescent when closest and full disk when furthest, these tend to cancel and the overall brightness is fairly constant. See for example answers to [Venus' magnitude during inferior conjunction](https://astronomy.stackexchange.com/q/33797/7982) and [Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths' side?](https://astronomy.stackexchange.com/q/36232/7982)  
> **username_2**: JPL Horizons can compute this in actual terms rather than hypothetical.  Select "Phase angle" and "Visual Mag" from the Observer Table Settings.  It only computes data for a given instant, so you'll want have it compute it at daily intervals for a year. https://ssd.jpl.nasa.gov/horizons  
> **username_2**: Just realized you asked for illuminated percentage, that can be computed from the phase angle as $ k = \frac{1+\cos i}{2} $, where $ i $ is the phase angle.  

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## 1 answers

### Answer by username_3 on 2022-10-28. Score: 8

## Setting up the problem

Here is a little picture of our geometry (nothing to scale):
[![enter image description here](https://i.sstatic.net/BI4wP.png)](https://i.sstatic.net/BI4wP.png)

Here, $i$ is the phase angle, $v$ is the distance from the Sun to Venus, $e$ is the distance from the Sun to Earth and $d$ is the distance from Venus to Earth. Please pretend the line segments originate from the center of each body.

The brightness $b$ of Venus will be proportional to $p$ the percentage illuminated, given by Greg Miller in the notes as $p=\frac{1}{2}(1+\cos i)$. It will be inversely proportional to the square of the distance $d$ between Venus and Earth. We can express this as:

$$b \propto \frac{p}{d^2}$$

We just need to define $d$ as a function of $i$ and then we can find where $b$ is maximized.

## Finding the distance as a function of Phase Angle

From the law of cosines, we have $$e^2=d^2+v^2-2dv\cos i$$ We can write this as a quadratic polynomial in $d$: $$d^2-(2v\cos i)d-(e^2-v^2)=0$$ We can solve for $d$ using the quadratic formula to get two solutions:
$$d=\frac{2v\cos i \pm \sqrt{(2v\cos i)^2+4(e^2-v^2)}}{2}$$

We will get negative distance if we subtract the radical, so we can consider only the positive cases with a single solution of $d$ in terms of $i$ as:

$$d=\frac{2v\cos i + \sqrt{(2v\cos i)^2+4(e^2-v^2)}}{2}$$

## Defining brightness as a function of phase angle

Plugging $d$ and $p$ back in to $b$ and simplifying, we get

$$b \propto \frac{p}{d^2} = \frac{2(1+\cos i)}{(2v\cos i + \sqrt{(2v\cos i)^2+4(e^2-v^2)})^2}$$

This now looks like a classic calculus problem, where we can take the derivative of $b$ with respect to $i$, set the result equal to zero and solve to find a value for $i$ which maximizes $b$. But why do calculus when you have a computer?

## What phase angle yields maximum brightness?

Here is a graph of the (proportional) value of b through all phase angles in one degree increments:

[![enter image description here](https://i.sstatic.net/wSVxw.png)](https://i.sstatic.net/wSVxw.png)

Zero degrees is fully illuminated Venus at maximum distance from Earth. 180 degrees is fully shadowed Venus at minimum distance. The maximum brightness is at 119 degrees phase angle $i$. That is at about 26% illumination.

**Note:**

1. We make the additional assumption that the Earth and Venus orbits are coplanar.
2. I never make misnakes, but please feel free to check my work.

> **Comments:**
> **username_4**: "I never make _misnakes_, but please feel free to check my work." talk about irony ;D  

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