Does Venus have to be brighter when its bright side is larger?
How much light you see and how bright it looks are two different questions.
Planets and Small Solar System Objects
A light becomes brighter when it is close to you, and dims when it is far away. Planets have one more factor: How much of the side illuminated by the sun is facing the Earth? Considering distance and phase together, you can understand why "rounder" is not necessarily "brighter".
Guess first
Does seeing a larger proportion of the bright side ensure that the planet looks brighter?
Show answer and explanation
No, it depends on distance and other factors — The illumination ratio only describes the disk area, and the brightness is also affected by the heliocentric/geocentric distance, reflection characteristics, etc.
Then connect experience and science
First see which half faces the sun
The sun illuminates half of the planet, but the entire illuminated surface is not necessarily visible from Earth. i is the angle between the sun and the earth as viewed from the planet.
cos i = (r²+Δ²−R²)/(2rΔ)
Convert angles to disk proportions
The proportion of bright area seen is given by the phase angle. It only describes the area ratio of the bright part of the disk.
k = (1+cos i)/2
The brightness also depends on the distance between the two
Planets farther from the sun receive less sunlight; planets farther from the Earth appear dimmer. The empirical equation considers distance and phase together.
m = constant + 5log₁₀(rΔ) + phase term
Don’t take the decimals of empirical formulas as the truth
Different formulas will give slightly different predictions. Saturn also has to consider how open its rings are; actual surface and atmospheric changes can also change brightness.
The smaller the magnitude, the brighter it is; reported as 0.1 magnitude
r is the heliocentric distance, Δ is the geocentric distance, and R is the sun-to-earth distance. The three use the same units. First, find the phase angle i according to the cosine theorem, and then find the illumination ratio k; the apparent magnitude must also choose the phase polynomial of each planet.
give it a try
First move Venus and compare the changes in illumination ratio and magnitude. The smaller the magnitude number, the brighter it becomes. Note that the distance also changes. Compare the differences between the two empirical formulas.
No need to memorize symbols
- r / Δ / R
- Heliocentric planet distance / Geocentric planet distance / Sun-Earth distance, all in AU
- i
- Phase angle, vertex at planet
- k
- Visible disk lighting ratio
- m
- Visual magnitude, the smaller it is, the brighter it is
- B / ΔU
- Observed geometric parameters of Saturn’s rings, derived from Chapter 45
Algorithm, accuracy and scope of application
Realize the day-to-Earth triangle phase angle and illumination ratio of Chapter 41 41.1/41.2, Venus fast approximation, Muller and Astronomical Almanac two sets of eight-planet empirical magnitude formulas (Muller's original table does not have Pluto, and the return is not provided). Saturn includes terms |ΔU| and ring opening angle B, but does not calculate the ring geometry from dates, and does not include unmodeled effects such as the dark side of the ring. The main interaction is an imaginary circular orbit with the earth R=1AU and each planet having a fixed heliocentric radius, not the actual date ephemeris; Δ/i/k is strictly calculated from triangles to avoid independently adjusting the distances of three impossible triangles. The original book Venus 41.a/c covers the real given r=.724604, R=.983824, Δ=.910947. The magnitude is only reported as 0.1, reminding that the empirical fitting is affected by clouds, surface, dust storms and extremely thin phases; the bright surface ratio is not regarded as brightness. The heliocentric or rectangular input equivalent path of 41.3/41.4 has been treated as a pure function and verified. No independent interface has been set up. The position angle of the bright edge of the moon is left to Chapter 48.
my understanding check
Which one is brighter, −4 or −2?
−4 — The smaller the magnitude value, the higher the brightness.
Where is the peak of the phase angle?
planet center — It is the angle between the sun-planet-Earth.
Muller Saturn requires ring opening angle, why?
Rings will affect the observed brightness — The extent to which the ring opens towards us affects the reflected light contribution, but there are still cases that the simplified formula does not take into account.
The bright surface ratio answers "how much bright surface is seen", and the apparent magnitude answers "how bright it looks"; phase, distance and reflection characteristics should be looked at together.
Jean Meeus · Astronomical Algorithms, 2nd ed. · No. 41 chapter
Planets and Small Solar System Objects · All interactive courses