# When Venus is the brightest when viewed from Earth?

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?

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

## 1 Answer

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