# Climbing a ladder attached to an extended object while in Earth Orbit

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?

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.
• 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? Dec 31, 2016 at 0:34
• 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.) Jan 1, 2017 at 4:07
• 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. Jan 1, 2017 at 13:25