Nowadays Venus is very bright. I can spot it during broad daylight without problem. It's because it's near Earth and appear as a crescent.

This made me think: as it's reaching the inferior conjunction the crescent becomes slimmer, so I expect that it will fade a bit. So there should be an optimal phase angle for optimal brightness.

So my question is at which phase angle Venus is the brightest when viewed from the Earth?

  • $\begingroup$ Would astronomy.stackexchange.com be a better home for this question? $\endgroup$
    – Qmechanic
    Dec 7, 2013 at 13:30
  • $\begingroup$ @Qmechanic no, I think this is fine here. $\endgroup$
    – David Z
    Dec 18, 2013 at 17:02

1 Answer 1


There is indeed an optimal angle where Venus is brightest. Have a look at the following figure:

enter image description here

The distance between Earth and Sun is $\Delta$, between Sun and Venus is $r$, and between Earth and Venus is $\rho$. The amount of light that Venus receives from the Sun is $$ f \sim \frac{\pi a^2}{r^2}, $$ where $a$ is the radius of Venus. When viewed from Earth, see don't see the full disc $\pi a^2$, but rather an illuminated surface $$ \sigma = \frac{1}{2}(1 + \cos\psi)\pi a^2, $$ where $\psi$ is the phase angle, as shown in the figure. This can be understood as follows: one half is half a disc (with surface $\pi a^2/2$), and the other half is half an ellipse (the projection of a circle), with semi-major axis $a$ and semi-minor axis $a\cos\psi$ (the length of the orange line in the figure) and thus a surface $(\cos\psi)\pi a^2/2$. When $\cos\psi$ is negative, Venus will appear as a crescent.

Combining these results, the flux that we receive from Venus is $$ F = K\pi a^2\frac{1 + \cos\psi}{2r^2\rho^2}, $$ with $K$ a constant that depends on the albedo of Venus. Now, with some trigonometry we find $$ \Delta^2 = r^2 + \rho^2 - 2r\rho\cos\psi, $$ so that $$ F = K\pi a^2\frac{2r\rho + r^2 + \rho^2 - \Delta^2}{4r^3\rho^3} = K\pi a^2\frac{(r + \rho)^2 - \Delta^2}{4r^3\rho^3}. $$ Now, this flux will be at a maximum when $dF/d\rho=0$, so $$ \frac{dF}{d\rho} = K\pi a^2\left(\frac{2(r+\rho)\rho - 3[(r + \rho)^2 - \Delta^2]}{4r^3\rho^4}\right)=0, $$ which is true when the numerator is zero, thus $$ \rho^2 + 4r\rho + 3r^2 - 3\Delta^2 = 0, $$ with solution $$ \rho = -2r + \sqrt{r^2 + 3\Delta^2}. $$ Now, $\Delta=1\,\text{AU}$ and $r = 0.723\,\text{AU}$, so that $$ \begin{align} \rho &= 0.431,\\ \psi &= 118^\circ,\\ \frac{1}{2}(1+\cos\psi) &= 0.267, \end{align} $$ which means that 26.7% of Venus' disc is illuminated at maximum brightness, in agreement with this site. Also, from $$ r^2 = \Delta^2 + \rho^2 - 2\Delta\rho\cos\theta, $$ we find that the elongation $\theta$ between Sun and Venus is $39.7^\circ$. The corresponding maximum flux is $$ F_\text{max} = 2.752\,K\pi a^2. $$ For comparison, when Venus is at opposition, we have $\psi=0^\circ$ and $\rho=r+\Delta$, so that $$ F_\text{opp} = \frac{K\pi a^2}{r^2(r+\Delta)^2} = 0.644\,K\pi a^2. $$ According to wikipedia, Venus has a magnitude $-3.82$ at opposition, so from the magnitude-flux relation $$ m_\text{max} - m_\text{opp} = -2.5\log\left(\frac{F_\text{max}}{F_\text{opp}}\right), $$ it follows that $m_\text{max} = -5.39$. This is somewhat different from the true value $-4.89$. The main reason is that my analysis doesn't actually calculate maximum brightness, but rather greatest brilliancy: the difference between them is that the latter refers to the moment when the apparent illuminated surface of Venus is at its maximum, which I derived.

However, the surface brightness of Venus isn't the same at every point: the outer edges are somewhat dimmer than the central parts due to Lambert's law, so I overestimated the brightness. It also means that the true maximum brightness occurs when the planet has a slightly larger phase, but it would take us too far to calculate this.


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.