What are the chances of getting struck by lightning while in Earth orbit?

Question

Getting struck by lighting in Earths upper atmosphere is a real possibility. It is often found in the 100km range, and can reach 400 km (250 mi) in diameter. The ISS has an orbit of 400km, and as pointed out below it appears that electrical disturbances can travel light years in space.

According to Popular Science

Galaxy 3C303, Keeping Current Generated from a Very Large Array image, this image shows the huge jet of current stretching for 150,000 light years across galaxy 3C303. Philipp P. Kronberg, Richard V.E. Lovelace, Giovanni Lapenta, Stirling A. Colgate via arXiv

According to Boeing Commercial airplanes have some some protection built into them, this page also indicates that most lighting strikes to planes occur while the plane is a cloud and between altitudes of 5,000 feet (1,524 meters) and 15,000 feet (4,572 meters).

So what are the odds of being struck by lightning while in Earth Orbit? Are some orbits at more risk then others? Are lightning protection devices installed in satellites or the ISS?

For comparison; according to National Geographic

The odds of becoming a lightning victim in the U.S. in any one year is 1 in 700,000. The odds of being struck in your lifetime is 1 in 3,000.

Answer 1

The probability is, simply, zero.

Lightning is the heavy discharge between two electrically charged bodies that have enough electrostatic potential to ionize that medium.

Where lightning occurs:

• Within the clouds

• Between two different clouds with different charge

• Clouds to earth

Why cloud to earth not cloud to space?

Wikipedia article states:

In order for an electrostatic discharge to occur, two things are necessary:

1. a sufficiently high electric potential between two regions of space must exist; and
   
2. a high-resistance medium must obstruct the free, unimpeded equalization of the opposite charges.

So let us make few assumptions; Let the height of the cloud be 85 kilometres (which is the maximum height at which the clouds are present), the cloud develops a negative charge, the spacecraft is at 300 kilometres with respect to the cloud, the tree on the ground at 85 kilometres with respect to the cloud, and the spacecraft and the tree on the ground both develop equal and opposite charge in respect to the cloud.

For lightning to occur, it requires a breakdown voltage.

Paschen's law states that a breakdown voltage is described by the equation:

$${V={a*p*d}\over{Ln(p*d)+b}}$$

• $a$ and $b$ are gas composition constants
• $p$ is pressure (in Atmospheres or Bar)
• $d$ is the gap distance (in meters)

For the cloud to the Earth, the values are:

$$a=4.36*10^7, b=12.5, p=1, d=85*1000$$

On substituting it, this becomes: $${{4.36*{10^7}*{1}*85*1000}\over Ln({1}*85*1000) + 12.8} =1.5345497371054913*10^{11}$$

This is for vacuum (I couldn't find the constant values), but the breakdown voltage should be very high, and the nature of the lightning is always choosing the simplest and the closest path.

For lightning to occur and sustain its voltage, it depends on the levels of ionization, which in turn depends upon the electrons being able to hit other electrons (avalanche breakdown), and that probability is given by: $$P={N \sigma \over A}={x\over mean\ free\ path}$$

• $N\sigma$ is the number of electrons
• $A$ is the area
• $x$ distance travelled by the electron

The probability is inversely proportional to the mean free path, but in vacuum the mean free path is very large and hence very little ionization occurs (since the probability decreases). Hence, even though the discharge can occur towards the space, its intensity is very small (since the probability of electron collisions is very small), and the lightning dies out before reaching the spacecraft (since the electrons lose their energy as they travel). So the spacecraft are safe from lightning, but us on earth aren't. ;)