If a ladder attached to the International Space Station were oriented "vertically", that is, perpendicular to the Earth's surface, how much force would an astronaut with a mass of 100 kg climbing the ladder have to exert to take each step upward?


1 Answer 1


If the ladder is mounted to the top of the station: None, he actually needs force to climb down!

The center of gravity of the station is in free fall around the Earth. At this point, centrifugal force and gravity cancel out precisely. Any other point of the station experiences a small residual force (hence the term "microgravity"). A point above the center of mass is in an orbit with the same angular velocity, but further away from Earth. An orbit at this height would have a angular velocity that is slightly lower, so he is too fast and feels and outwards force.

We can assume that gravity is constant on this small scale and centrifugal force scales as $F = m \omega^2 r$. ISS is in an orbit with about $r=6800km$ while the astronaut is in an orbit $r'=6800km + x$. The difference in centrifugal force is now just the relative difference of these radii. If the ladder is e.g. 68m long, the astronaut feels about 1/100000 of his natural weight pushing him outwards, about 1/10 N.

  • $\begingroup$ Another way to look at this force (or another force?) is as tidal force: the gravity pull on the astronaut is slightly weaker than on the center of ISS because the astronaut is slightly farther away from Earth. In a less handwavy form: $$ F_\text{tidal} \approx 2\Delta r G M m / R^3 \approx 2 \Delta r m g / R $$ , which differs from your calculation by a factor of two. Any ideas about the physical meaning of this difference? Do the two expressions describe the same effect, or are they different things that should be added together or something? $\endgroup$
    – al13n
    Dec 31, 2016 at 0:34
  • $\begingroup$ This is exactly the same thing. The factor of two comes from the definition of the tidal force formula: It describes the difference in force between two parts of the object on opposite sides of the object, not between one side and the center of gravity. $\endgroup$
    – asdfex
    Dec 31, 2016 at 12:52
  • $\begingroup$ I believe $\Delta r$ is the distance from the center of mass, not the opposite end. The formula with the factor of two is easily derived from the $F=GMm/R^2$ with assumption that $\Delta r$ is measured from center of mass. Am I doing it wrong? (It does seem intuitive to me that the tidal and centrifugal force should be exactly the same thing, but I'm still failing to explain the factor of two.) $\endgroup$
    – al13n
    Jan 1, 2017 at 4:07
  • $\begingroup$ Sorry, I have to correct myself: The two forces are not the same. Essentially we are looking at (a) the derivative of the centrifugal force with respect to r and (b) the derivative of the gravitational pull with respect to r. As gravity scales with $r^{-2}$ and centrifugal force scales with r, we get a factor of 2 into the game. The different exponents on r don't matter as this is just a constant in our calculation. In the center of the station both forces are equal, and so the residual forces just differ by the factor of 2 from differentiation. We have to add up both forces to get the total. $\endgroup$
    – asdfex
    Jan 1, 2017 at 13:25

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