Apparent weight in free fall When a body falls freely solely under gravity (ignoring air resistance), what apparent weight does it possess on a scale attached to it?

Introduction / Context:
Apparent weight is the normal reaction a supporting surface exerts on a body. It can differ from true weight depending on acceleration of the reference frame. A classic case is free fall, which underpins weightlessness sensations in elevators and spacecraft in orbit.

Given Data / Assumptions:

• Body is in free fall under gravity g.
• No air resistance; body and attached scale accelerate together.
• Apparent weight is the scale reading (normal force N).

Concept / Approach:
For vertical motion with acceleration a, Newton’s second law on the body with upward positive gives: N − W = −m a (if downward acceleration). In free fall, a = g downward, so N = 0. The body is locally weightless although gravitational force W = m g still acts; what vanishes is the contact force that registers on a scale.

Step-by-Step Solution:

Verification / Alternative check:
In a falling elevator with broken cable (idealized), occupants and the floor accelerate together at g; hence no normal reaction is developed. Astronauts in orbit are in continuous free fall around Earth, experiencing microgravity (practically weightlessness).

Why Other Options Are Wrong:

• minimum (but not zero): incorrect; it is exactly zero for ideal free fall.
• maximum / equal to true weight: apply to accelerating upward frames, not free fall.

Common Pitfalls:
Confusing true weight (m g) with apparent weight (normal force). Only the latter vanishes in free fall.

Final Answer:
no