True vs apparent weightlessness
True weightlessness is when the resultant gravitational force on a body is zero (at infinity, or at a neutral point). Apparent weightlessness is when a body exerts no contact force on its support, which happens when body and support have the same acceleration g, as in free fall or in orbit.
In orbit: mg − N = ma with a = g ⇒ N = 0What this actually means
An astronaut in the ISS is NOT beyond gravity. At 400 km up, g is about 8.7 N kg⁻¹, roughly 89% of the surface value. If gravity had switched off, the station would fly off in a straight line instead of orbiting.
The real explanation is a two-line Newton's-second-law argument. For the whole vehicle, gravity is the only external force, so a = g. For the astronaut alone, mg − N = ma, and since a = g, N = 0. By Newton's third law, the astronaut pushes on the balance with zero force, so the balance reads zero.
So apparent weightlessness means no CONTACT force, not no weight. What a weighing scale reads is always the normal contact force, not your weight, and the two only agree when you are not accelerating.
True weightlessness is much rarer: it needs the resultant gravitational force itself to be zero. That happens at infinity, or at the neutral point between two bodies where the fields cancel exactly.
The same logic explains the lift questions from Dynamics and the equator-versus-pole reading. At the equator part of your weight goes into providing centripetal force for the Earth's spin, so N = mg − F_c and the scale reads slightly low.
Explaining weightlessness by saying there is no gravity in orbit, rather than that there is no contact force.
Prove it — watch it be true
- Set a low Earth orbit and read the g value at that radius, confirming it is still close to the surface value.
- Confirm the only force arrow drawn on the satellite is gravity, with nothing supporting it.
- Compare the satellite's centripetal acceleration readout with g at that radius and see they are equal, which is why N would be zero.