Estimating the Luminosity of the Sun using a 60W Lightbulb

In this post I'll be doing problem #2 from this week's worksheet on stellar properties. I chose this problem because it used a crucial concept of physics to answer a fun and extremely entertaining question. Here's the problem:

Use a 60 Watt light bulb to estimate the luminosity of the Sun. How does your value compare to the actual value?

That's the entire question. Use a light bulb to estimate the luminosity of the Sun. Where would one even begin?

It's actually quite simple. You need to only use a light bulb to find the Sun's luminosity - that means you need to measure some quantity that is the same for the light bulb and the Sun. That quantity turns out to be the power absorbed by your hand.

Think about it like this: on a hot summer day, you can walk outside and stretch out your hand. The rays of light from the Sun will be absorbed into your hand, and you will feel a "heat" that is caused by the Sun, not by the air temperature. If you hold your hand close to a light bulb, you will feel a similar sensation. This sensation is the power, or energy absorbed by your hand per second. So, we have:

Now, we're given the power output of the bulb. How can we relate that to the power output (luminosity) of the Sun? We need to somehow relate the power output of the bulb with the power absorbed by your hand. We can use Gauss's Law for light intensity to find this. Because of conservation of energy, the amount of energy outputted by a source of light per time (i.e. power) is the same on any surface enclosing the source. In other words, if you surround a light bulb with a shell of radius R, that shell will absorb the same amount of energy as a shell of radius 2R.

For the purposes of this problem, it will be convenient to surround our light bulb with an imaginary spherical shell with a radius equal to the distance from your hand to the center of the light bulb (the distance at where you feel the same heat on your hand as you do on a hot summer day). Because a sphere is rotationally symmetric, the power absorbed at any point on the sphere is the same. So, we can express the power absorbed by your hand as a fraction of the power output of the light bulb. Thus, we can write:

We know from before that the power absorbed by your hand is the same for the Sun's case. Except in this scenario, the distance from your hand to the Sun is an astronomical unit (AU)! That's about 1.5 x 10^13 cm. So for the Sun:

Equating the two, we have:

The distance between your hand and the bulb will be around 10cm, so:

The actual luminosity of the sun is 3.839x10^33 ergs/second, so we're within an order of magnitude to the correct answer!