# How many electrons are there in the Solar System?

I was asked this question in an interview some twenty years ago. This is a typical Fermi problem. Solving it means making a good estimate with little data. Of course, there are many possible approaches; my answer went something like this…

Because

1. the Sun makes almost all of the Solar System’s mass (more than 99%) and
2. the Sun consists mostly of hydrogen (around three quarters is hydrogen and the rest is mostly helium; this is a good enough approximation)

the number of electrons roughly equals the mass of the Sun $$m_S$$ divided by the mass of the hydrogen atom $$m_H$$. But I didn’t know $$m_S$$ and $$m_H$$ off the top of my head, so I had to improvise.

Let’s start with $$m_H$$. The Avogadro number is much easier to memorize than the mass of the hydrogen atom—there are 6.0 · 1023 atoms in one gram of hydrogen. Therefore

$$m_H \approx 2 \cdot 10^{-27} \, \mathrm{kg} .$$

The mass of the Sun was more challenging. My starting point was the gravitational force

$$F = G \frac{m_S m_E}{r^2}$$

between the Sun and Earth, which is also the centripetal force

$$F = m_E r \omega^2$$

that makes Earth circle around the Sun. To calculate the Sun’s mass, we only need the gravitational constant $$G$$, the distance between the Sun and Earth $$r$$, and the angular velocity $$\omega$$:

$$m_S = \frac{r^3 \omega^2}{G} .$$

Luckily, back then I knew the value of the gravitational constant

$$G = 6.7 \cdot 10^{-11} \, \mathrm{m^3 kg^{-1} s^{-2}} .$$

Next, the angular velocity is 2π radians over 365 days, i.e.,

$$\omega = 2.0 \cdot 10^{-7} .$$

Also, it takes a little bit more than eight minutes for light to travel from the Sun to Earth at the speed of 3.0 · 108 m/s, so

$$r = 1.5 \cdot 10^{11} \, \mathrm{m} .$$

Therefore

$$m_S \approx 2 \cdot 10^{30} \, \mathrm{kg} .$$

Finally, there are around 1057 electrons in the Solar System.