Do International Space Station astronauts have to perform spacewalks only while its not facing the Sun directly? If so, I'm curious about the reason behind it. Is it only because they would be shielded from solar winds and radiation? Can't the spacesuits protect them completely? Also, what is the average speed of charged particles during solar proton events when they reach the station?


2 Answers 2


No. Most International Space Station (ISS) spacewalks last in the order of several hours, often up to 8 hours during more difficult and time-consuming repairs and installations, and some nearly 9 hours*. And the station, in its roughly 93 minutes orbit is never in Earth's shadow for that long.

There are both advantages as well as disadvantages to spacewalking in either sunlight or complete darkness, but that's already discussed by two answers in How are EVA's in LEO affected by being at the night side of Earth? You'll also find answers to your other questions on our site, if you search for them.

As for your last question, while that should really be a new question, because it's rather difficult to answer and we prefer that standalone questions are asked separately, here's a few interesting tidbits that might perhaps help you frame it; Most of solar proton events are harmless in LEO except the most violent ones, since ISS orbits below Van Allen belts. But its orbit does cross South Atlantic Anomaly during some orbits for a short period and it's less protected there. And while highest velocity that these particles reach is in polar regions due to acceleration of Sun's and Earth's magnetic reconnection, ISS' orbital inclination doesn't take it over such high latitudes.

*Refer to Wikipedia's List of spacewalks and moonwalks 1965–1999 and List of spacewalks 2000—2014 that both include their duration.

  • $\begingroup$ Could you please explain that formula you linked? Or post a link to explanation because it looks interesting and I am afraid my school-level physics knowledge (which kinda deterorated over the years) is far from being able to decipher it myself. $\endgroup$
    – Maurycy
    Jul 11, 2015 at 16:26
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    $\begingroup$ @MaurycyZarzycki That formula is derived from 3rd Kepler's law of planetary motion and simplified for orbital period of orbiting bodies with insignificant mass fraction compared to the primary and in circular orbit around it, which is $T = 2\pi \sqrt{a^3/\mu}$ where $\mu$ is standard gravitational parameter ($GM$), and $a$ is semi-major axis (for circular orbit, it's just radius of Earth plus orbital altitude). $\endgroup$
    – TildalWave
    Jul 11, 2015 at 16:43
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    $\begingroup$ @MaurycyZarzycki Note that this formula is a first order approximation and there are second order perturbations that ISS' orbit is a subject to, such as nodal precession due to Earth not being a perfect sphere (technically it's an oblate spheroid), and it also doesn't include atmospheric drag which also varies due to atmospheric tides, and so on. But it's accurate to within a few seconds for a single ISS' orbit, so good enough. $\endgroup$
    – TildalWave
    Jul 11, 2015 at 16:49
  • $\begingroup$ This is the linked formula: sqrt((412.5 km + 6378.1 km)^3*4*pi^2/((6.673×10^−11 N*(m/kg)^2) * 5.97219×10^24 kg)). It is $T = 2 \pi \sqrt{a^3/ \mu}$, with GM substituted for $\mu$ - the gravitational constant times the mass of the Earth. So 412.5 km + 6378.1 is $a$, because it is the average altitude at which the ISS orbits plus the average radius of the Earth. Instead of multiplying by $2 \pi$, $4 \pi^2$ was placed within the 'sqrt' (square root). $\endgroup$
    – kim holder
    Jul 11, 2015 at 19:13

The ISS orbit is 93 minutes, and space walks last for hours.

Thus the difficulty in space suit design. It needs to handle being in direct sunlight for 45 or so minutes, then almost immediatley into deep darkness. (Though the reflection off Earth helps a lot in low earth orbit. Which makes a difference for space suits used Beyond LEO).


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