# Why do some of the planets in our solar system appear to be brighter than some stars in the night sky?

Dec 1, 2015

How bright objects appear depends much more on the distance to the object than on its intrinsic brightness, and planets are much closer than stars.

#### Explanation:

In astronomy, there are two things that determine how bright an object appears to be here on Earth;

1. Total amount of light the object gives off (luminosity).
2. How far away the object is.

We should expect that if we increase the luminosity of an object, we will also increase the brightness seen here on Earth, and if we move an object farther away, it should appear dimmer. Astronomers often use flux as a measurement of how much light they receive from an object. Flux is the amount of light that an area receives from a light source. The equation for flux is;

$F = \frac{L}{4 \pi {d}^{2}}$

Where $L$ is the luminosity and $d$ is the distance to the object. Flux follows our two rules above, it increases with luminosity and decreases with distance. Notice that if the distance has a greater effect on the flux than luminosity does.

So even though stars are much brighter than planets, they are so far away, that we only get a tiny amount of light from them. Lets consider the flux of Sirius, the brightest star in sky. Sirius has a luminosity of ${10}^{28} \text{W}$ which is $25$ times as bright as the sun, while its distance from Earth is $8 \cdot {10}^{16} \text{m}$. That yields a flux of;

$F = {10}^{28} / \left(4 \pi {\left(8 \cdot {10}^{16}\right)}^{2}\right) = {10}^{-} 7 {\text{W/m}}^{2}$

Now lets look at one of the planets, Jupiter. Jupiter itself does not generate its own visible light, but it does reflect light from the sun. We can use the flux equation to figure out how much light Jupiter gets.

${F}_{\text{sun"=L_"sun"/(4 pi d_"J"^2)= (4*10^26 "W")/(4 pi (8*10^11 "m")^2) =50 "W/m}}^{2}$

Now we can calculate the effective luminosity for Jupiter, as though it was a star, using the flux from the sun and the radius of Jupiter, ${R}_{\text{J}}$.

${L}_{\text{J" = 4 pi R_"J"^2 F_"sun" = 4 pi (7*10^7 "m")^2(50 "W/m"^2) = 3*10^18 "W/m}}^{2}$

Now we can find the flux from Jupiter here on Earth.

${F}_{\text{J" = L_"J"/(4pi (d_"J" - d_"E")^2) = (3*10^18 "W/m"^2)/(4 pi (6*10^11 "m")^2) = 6*10^-7 "W/m}}^{2}$

Jupiter is not perfectly reflective, though, so lets assume that only half of the light that reaches Jupiter is reflected. That's still $3 \cdot {10}^{-} 7 {\text{W/m}}^{2}$, or $3$ times as bright as Sirius!