Consider a person of mass m standing on a scale inside an elevator (or spaceship) that has vertical acceleration a (taking up as positive).
Step 1 — List the forces on the person.
Only two: gravity mg downward, normal force N upward.
Why this step? The scale reads N, the force it exerts on you. By Newton's 3rd law you push on it equally — that's the number it displays.
Step 2 — Apply Newton's 2nd law along the vertical.N−mg=ma
Why this step? Net force = mass × acceleration. The person accelerates with the elevator, so their acceleration is also a.
Step 3 — Solve for apparent weight.N=m(g+a)
The punchline: set a=−g (the elevator is in free fall). Then N=0. Gravity (mg) is still there, but there's no contact force left, so you feel weightless even though g is unchanged.
A common myth: "Astronauts float because there's no gravity in space." At the ISS altitude (~400 km), g≈8.7m/s2 — about 89% of surface gravity! Plenty of gravity.
Quick check with the formula: a circular orbit has centripetal acceleration a=g directed inward; the astronaut accelerates with the station, so relative to the station N=0. Same physics as the falling elevator, just curved.
What physical quantity does a bathroom scale actually report? → the ==normal force N==.
Formula for apparent weight (up positive)? → N=m(g+a).
Value of a that produces weightlessness? → a=−g (free fall).
Is g zero at the ISS? → No, g≈8.7m/s2.
You feel "heavy." Is the elevator's acceleration up or down? → up (a>0).
What is "apparent weight"?
The magnitude of the normal/support force N a body experiences — what a scale reads and what you "feel."
Formula for apparent weight in an accelerating elevator (up positive)?
N=m(g+a).
What condition gives weightlessness?
When apparent weight (normal force) is zero, N=0, which happens at a=−g (free fall).
Why do astronauts float in the ISS?
They and the station are both in free fall (same acceleration g), so N=0 — apparent weightlessness, NOT zero gravity.
Is gravity zero at ISS altitude (~400 km)?
No, g≈8.7m/s2, about 89% of surface gravity.
Elevator descends at constant velocity — what does the scale read?
mg (normal weight), since a=0.
You feel heavier in an elevator. Direction of its acceleration?
Upward (a>0, so N=m(g+a)>mg).
True vs apparent weightlessness?
True: g≈0 (deep space). Apparent: g=0 but body and support share acceleration so N=0 (free fall/orbit).
Recall Feynman: explain to a 12-year-old
Imagine standing on a bathroom scale in a lift. The number on the scale isn't really "how much gravity pulls you" — it's "how hard the floor is pushing your feet." Now imagine the lift's rope snaps and the whole lift, you, and the scale all drop together. The floor isn't pushing your feet anymore (it's falling just as fast as you), so the scale shows zero. You'd float inside the lift! Gravity is still pulling you — that's why you're falling — but because nothing is pushing back, you feel weightless. Astronauts feel the same: they're forever falling around the Earth, always missing the ground.
Dekho, sabse important baat: tum gravity ko directly "feel" nahi karte. Jo tum mehsoos karte ho woh hai normal force — yaani floor ya scale jo tumhare upar push kar raha hai. Isiliye weightlessness ka matlab "gravity gayab ho gayi" nahi hai, balki matlab hai ki normal force zero ho gaya (N=0). Bathroom scale bhi yehi N dikhata hai, tumhara real weight mg nahi.
Formula simple hai: lift me khade ho aur lift ka acceleration a hai (up ko positive lo), toh Newton ka second law lagao — N−mg=ma, jisse N=m(g+a) aata hai. Agar lift upar accelerate kare toh N>mg (bhaari feel hota hai), niche accelerate kare toh halka. Aur agar rope toot jaye aur lift free fall me chali jaye, toh a=−g, isse N=m(g−g)=0 — bilkul weightless! Par yaad rakho, gravity tab bhi puri ki puri kaam kar rahi hai, isiliye toh tum gir rahe ho.
Astronauts ke baare me sabse bada myth: "space me gravity nahi hoti." Galat! ISS pe (~400 km upar) g abhi bhi lagbhag 8.7m/s2 hai — surface ka 89%. Toh woh float kyun karte hain? Kyunki station aur astronaut dono ek saath free fall kar rahe hain Earth ki taraf, bas itni tezi se side me ja rahe hain ki Earth ko "miss" karte rehte hain. Dono ka acceleration same hai, isliye ek doosre pe push nahi karte → N=0 → apparent weightlessness.
Exam tip: velocity ki direction matter nahi karti, sirf acceleration matter karta hai. Lift niche constant speed se ja rahi ho toh a=0 aur scale normal mg hi dikhayega. Sign convention ek baar fix karo (up = +) aur usi pe tiko — falling lift ka acceleration −g likhna, +g nahi.
