Worksheet 6, Problem 1: The Light of a Single Sun

1. Use a lamp fitted with a 100 Watt incandescent light bulb to estimate the luminosity of the Sun. Do this with the knowledge that:

• The Earth-Sun distance is 1 astronomical unit (AU), or a 1.5 x 10^13 cm
• An estimate of how far from the bulb you have to hold your hand in order for it to feel like a sunny Spring day;
• The fact that incandescent light bulbs convert electrical power into radiant energy with an efficiency of only about 3%.

As we dive further into the realm of black bodies, we begin to realize and appreciate their simplicity in characterizing the objects that we are dealing with in our exploration of space. In fact, in a simple experiment, we can roughly calculate the luminosity of the sun, with a few basic numbers and a 100 Watt incandescent light bulb. 

To approach this, we began covering our eyes and stepping up to our lit incandescent bulb. Carefully, we adjusted the distance of our hand away from the bulb, until the temperature resembled that of a warm spring day. As we've learned, our distance from the sun varies throughout the seasons, with the spring providing a rough average for the course of the year. Our group reached a consensus distance away from the bulb of 20cm. 

This value compares to the distance of 1 AU or \(1.5 \times 10^{13}\) cm that the earth is away from the sun on a warm spring day. So, you may ask, why is this information valuable? Essentially, what we have done here is matched the amount of radiance that reaches our body from the sun on a spring day to the amount of radiance that reached our hand from the bulb. Since this represents energy per area, we can consider this quantity to be the flux of the black bodies and as we've stated, we can set them equal to each other as follows: \[F_{sun} = F_{bulb}\] Since we know that flux is equal to luminosity per area, we can rewrite the above relationship as:\[\frac{L_{sun}}{SA_{sun-earth}}=\frac{L_{bulb}}{SA_{bulb-hand}}\] Since we know that black bodies radiate equally in all directions, the surface area that we are interested in is that of the sphere with a radius corresponding to the distance the object is away from the black body. The surface area of a sphere is: \(SA = 4\pi r^2\), where r is equal to the distance, d. Updating our relation above, we now have: \[\frac{L_{sun}}{4\pi (d_{sun-earth})^2}=\frac{L_{bulb}}{4\pi (d_{bulb-hand})^2}\] In trying to solve for the luminosity of the sun, the last remaining unknown is the luminosity of the bulb. Fortunately, we can easily calculate that based on the power output and efficiency of the incandescent bulb. Given that the bulb converts its 100 watts of electrical power to 3% radiant light, we can assume that the other 97\% is radiated as heat. \[L_{bulb} = 100 \, watts \times 97\% = 97 \, watts\] We can now solve for the luminosity of the sun: \[L_{sun} = \frac{L_{bulb}}{4\pi (d_{bulb-hand})^2}4\pi (d_{sun-earth})^2= \frac{L_{bulb}}{(d_{bulb-hand})^2} (d_{sun-earth})^2\] \[L_{bulb} = \frac{97\, watts}{ (20\,cm)^2} (1.5 \times 10^{13}cm)^2 \approx 5 \times 10^{25}\, watts\] Finally, we need to convert to ergs per second, and we have our answer! \[5 \times 10^{25}\, watts \, \times \frac{10^7 \,erg/s}{1\, watt} = 5 \times 10^{32}\, erg/s\]

Acknowledgements: Johnathan Budd & Willie Pirc