Finding the Luminosity of the Sun

 Worksheet 2, Question 1

When the back of your hand is about 8cm away from a 100-Watt incandescent light bulb, the bulb feels like the Sun on a sunny spring day.  The Earth-Sun distance is 1 AU or $a = 1.5 \times 10^{13}$ cm. Estimate the luminosity of the Sun ($L_o$) in units of ergs/second (1 Watt is $10^7$ ergs/sec)

Since the bulb feels like the sun on a sunny spring day, we can assume that the Flux of the sun ($F_o$) is equal to the flux of the bulb ($F_b$).

$F_o=F_b$

Flux = $\frac {Energy} {area \times time}$

Flux of the bulb is equal to power of the bulb ($P_b$) over area of the bulb ($A_b$) and Flux of the sun is equal to luminosity (power) of the sun ($L_o$) over area of the sun ($A_o$).

$\frac{P_b}{A_b}$ = $\frac{L_o}{A_o}$

$ A_b = 4 \pi x^2$

$A_o= 4 \pi a^2$

$\frac{P_b}{4 \pi x^2}$ = $\frac{L_o}{4 \pi a^2}$

We can now solve for the luminosity of the sun!

(x and a are defined in the photo)

$L_o$ = $\frac {P_ba^2}{x^2}$

Plug in our values

$L_o$ = $\frac  {10^9 erg/s \times (1.5 \times 10^{13}cm)^2}{(8 cm)^2}$

Our answer

$L_o$ =$3.5 \times 10^{33} erg/s$