A planet that crosses in front of its star is a tiny eclipsing binary in which one member is dark. You met the idea in Finding Planets Around Other Stars, and the geometry of eclipsing binaries carries over. NASA's Kepler and TESS missions found thousands of planets this way by watching stars dim on a schedule.
💡 Why it works
The star's disk has area $\pi R_\star^2$ and the planet covers $\pi R_p^2$ of it. If the disk were uniformly bright, the fraction of light lost would be the ratio of areas: $$\delta = \frac{\pi R_p^2}{\pi R_\star^2} = \left(\frac{R_p}{R_\star}\right)^{2}.$$ The repeat time is the orbital period, and Kepler's third law turns it into the orbit's size: $a^3 = M_\star P^2$ (AU, years, solar masses). During a central transit the planet moves a distance $2R_\star$ at its orbital speed $2\pi a/P$, which gives the transit's length.
🔑 Key idea
Depth $\delta = (R_p/R_\star)^2$, so $R_p = R_\star\sqrt{\delta}$. Period gives $a = (M_\star P^2)^{1/3}$, and a central transit lasts about $t \approx PR_\star/(\pi a)$.
A dark planet with one-tenth of the star's radius blocks one-hundredth of its light: $(0.1)^2 = 1\%$.
Depths an observer far away would see. Earth-size planets around Sun-like stars dim them by less than one part in 10,000.
Planet crossing the Sun
$R_p/R_\odot$
Depth
Jupiter
$0.103$
about 1%
Neptune
$0.035$
about 0.13%
Earth
$0.0092$
about 84 ppm
Worked examples
✏️ Worked example: An Earth-size planet
A star of $0.90\,R_\odot$ dims by 100 parts per million during each transit. How big is the planet? ($R_\odot = 109\,R_\oplus$.)
A planet transits a $1.0\,M_\odot$, $1.0\,R_\odot$ star every 3.5 days. Find its orbital radius and how long a central transit lasts ($1\text{ AU} = 215\,R_\odot$).
In solar radii, $a = 0.0451 \times 215 = 9.70\,R_\odot$.
$t \approx PR_\star/(\pi a) = (84\text{ h})(1.0)/(\pi \times 9.70) = 2.8$ h.
Answer: $a \approx 0.045$ AU (a "hot Jupiter" distance, far inside Mercury's 0.39 AU), and each transit lasts about 2.8 hours.
✏️ Worked example: What is it made of?
Another transiting planet has a radius of $1.5\,R_\oplus$, and the wobble it gives its star yields a mass of $5.0\,M_\oplus$. Find its mean density. (Earth's mean density is 5.51 g/cm³.)
Density scales as mass over radius cubed: $\rho = 5.51 \times 5.0/1.5^3$ g/cm³.
$1.5^3 = 3.375$, so $\rho = 5.51 \times 1.48$.
Answer: About 8.2 g/cm³, denser than Earth: a rocky planet, probably with a large iron core. Combining transits (size) with wobbles (mass) is how astronomers tell rocky worlds from gas-rich ones.
⚠️ Common mistake
Taking the depth to be $R_p/R_\star$ instead of its square. A dip of 1% does not mean the planet is 1% of the star's radius; it means $R_p/R_\star = \sqrt{0.01} = 0.1$, ten times larger. The light lost depends on the area covered. Check: Jupiter, about one-tenth of the Sun's radius, would cause a 1% dip, not a 10% one.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.