The ISS and weightlessness (free fall)

Problem

The International Space Station orbits at $\sim 400\ \text{km}$ above the Earth. The astronauts on board are “weightless”. Does this mean Earth’s gravity doesn’t act there? What are the astronauts actually experiencing?

Solution

Gravity is still there, and it’s strong. At $400\ \text{km}$ the orbital radius is $r = R_T + h \approx 6{,}77\cdot 10^6\ \text{m}$. The gravitational acceleration is: $g(r) = \frac{GM_T}{r^2} = \frac{3{,}98\cdot 10^{14}}{(6{,}77\cdot 10^6)^2} \approx \frac{3{,}98\cdot 10^{14}}{4{,}58\cdot 10^{13}} \approx 8{,}7\ \text{m/s}^2$

Comparing with the surface value: $\frac{g(r)}{g_0} = \frac{8{,}7}{9{,}8} \approx 0{,}89 = 89\%$

Gravity on board is still almost $90\%$ of the surface value: it is not absent at all.

What they actually experience: free fall. The station and the astronauts orbit together, i.e. they fall together towards the Earth with the same acceleration $g(r)$. Because they fall in unison, there is no contact force between the astronaut and the floor: the normal reaction that we “feel” as weight on the ground is missing. The sensation of weightlessness is therefore microgravity due to free fall, not a lack of gravity.

$\ev{\text{Gravity acts (}\sim 89\%\text{); the astronauts are in free fall, hence with no floor reaction: microgravity.}}$

Links

Topics: Gravitation Concepts: Weight force · Law of universal gravitation · Orbits Objects: Satellite