VY Canis Majoris planetary system

This is actually pretty simple. We just have to calculate the stellar flux, which is the luminosity of the star divided by the surface area of a shell with the radius of the planet's orbit. In other words: $$F=\frac{L}{4\pi r^2}$$ Now, let's denote the Sun's luminosity as $L_{\odot }$, the current radius of Earth's orbit as $r_i$, the luminosity of VY Canis Majoris as $L_{\text{VY}}$, and the new radius of Earth's orbit as $r_f$. We need to set the fluxes equal, in order to keep conditions on Earth the same. Therefore, $$F_{\odot }=F_{\text{VY}}\to \frac{L_{\odot }}{4\pi r_i^2}=\frac{L_{\text{VY}}}{4\pi r_f^2}$$ The $4\pi$s cancel, and, given that Wikipedia gives VY Canis Majoris's luminosity as about 270,000 times that of the Sun, we get $$\frac{L_{\odot }}{r_i^2}=\frac{270,000}{r_f^2}\to r_f=\sqrt{270,000}{r}_i\simeq 520r_i$$ Therefore, Earth would have to be 520 times as far away - 520 AU - for conditions to be such that we could "function normally", as you put it.

posted almost 9 years ago

HDE 226868‭

601 reputation 61 463 80 49