A star's luminosity is the power it gives off. What we see from Earth is its apparent brightness: the power arriving on each square meter. The two are linked by distance, which parallax can measure.
Why it works
Picture all the light leaving a star at one instant as a growing bubble. By the time it reaches distance $d$ it is spread evenly over a sphere of area $4\pi d^2$. The total power $L$ has not changed, so each square meter receives $$b = \frac{L}{4\pi d^2}.$$ Twice as far, the sphere has $2^2 = 4$ times the area, so the same light is spread four times thinner. This inverse-square law also works backward: measure $b$ and $d$, and you get $L = 4\pi d^2 b$.
Key idea
Apparent brightness $b = \dfrac{L}{4\pi d^2}$: it falls off as the square of the distance. Twice as far means a quarter as bright.
Worked examples
Worked example: Sunlight at Earth
The Sun gives off $3.83 \times 10^{26}$ W. How much power falls on each square meter at Earth's distance, $1.496 \times 10^{11}$ m?
- Sphere area: $4\pi (1.496 \times 10^{11})^2 = 2.81 \times 10^{23}\ \text{m}^2$.
- $b = \frac{3.83 \times 10^{26}}{2.81 \times 10^{23}} = 1360$ W/m².
- This is the "solar constant" measured by satellites, about $1361$ W/m², the power behind Earth's weather and climate.
Answer: About $1360$ W/m².
Worked example: Equally bright, not equally far
Two stars look equally bright, but star A is $25$ times as luminous as star B. How do their distances compare?
- Equal $b$: $\frac{L_A}{d_A^2} = \frac{L_B}{d_B^2}$, so $\frac{d_A}{d_B} = \sqrt{\frac{L_A}{L_B}}$.
- $\sqrt{25} = 5$: star A is $5$ times farther away.
Answer: Star A is $5$ times as far.
The same law sets how much sunlight each planet gets. Jupiter, at $5.2$ AU, receives $\frac{1}{5.2^2} = \frac{1}{27}$ of the sunlight Earth does, one reason its icy moons stay frozen. Later, standard candles use this law in reverse to measure distances to galaxies.
Common mistake
A natural guess is that a star twice as far away looks half as bright. But light spreads in all directions, over an area that grows as $d^2$, so the star looks a quarter as bright. Check: Mars is $1.52$ AU from the Sun and gets $\frac{1}{1.52^2} = 0.43$ of the sunlight Earth gets, not $\frac{1}{1.52} = 0.66$.