Weightlessness — true (free fall) vs apparent: Question bank

1.2.25 · D5 · Physics › Newton's Laws & Dynamics › Weightlessness — true (free fall) vs apparent

Definition Yeh page jin words aur symbols ka use karti hai (pehle padho)

True weight $W = m g$ — kisi bhi mass $m$ wale body par actual gravitational pull, jahan $g$ local gravitational acceleration hai. Yeh sirf tabhi zero hota hai jab $g \to 0$ .
Apparent weight $W_{a pp} = N$ — normal (support) force ki magnitude jo surroundings tumhare upar push karte hain. Yahi scale read karta hai aur yahi tumhe "feel" hota hai.
$a$ — support ka acceleration (elevator, ship, ya station jo tumhe carry kar raha hai). Person support ke saath accelerate karta hai, isliye $a$ person ka bhi acceleration hai.
Sign convention (is poori page par yahi use hoga): upar positive hai. Toh $g > 0$ hamesha (yeh multiply hokar downward pull deta hai), jabki $a > 0$ ka matlab support upar accelerate kar raha hai aur $a < 0$ ka matlab neeche accelerate kar raha hai.

True or false — justify

An astronaut in the ISS experiences almost zero gravity.

False — ~400 km par,

g \approx 8.7 m/s^{2}

, surface gravity ka lagbhag 89%; woh float karte hain kyunki woh free fall mein hain, toh

N = 0

A body in free fall has zero weight.

Mixed — iska apparent weight (

N

) zero hai, lekin iska true weight

m g

unchanged hai; gravity poori tarah on hai, aur yahi wajah hai ki yeh girta hai.

A scale in an elevator always reads

m g

False — scale normal force

N = m (g + a)

read karta hai; yeh

m g

ke barabar sirf tabhi hota hai jab support ka acceleration

a = 0

ho.

If you move downward you become lighter.

False — sirf acceleration

N

ko affect karta hai, velocity nahi; constant speed se neeche jaane par

a = 0

hota hai aur

N = m g

(bilkul normal).

Weightlessness requires the surroundings to fall at the same rate as you.

True — agar support aur body ek hi acceleration share karein, toh dono ek doosre par press nahi karte, isliye

N = 0

In deep space far from all masses, a body is weightless.

True — lekin yeh true weightlessness hai (

g \to 0

), free-fall (apparent) weightlessness se alag cause hai jahan

g \neq = 0

Feeling "heavy" in a lift means gravity got stronger.

False — true weight

m g

unchanged hai; lift upar accelerate kar raha hai (

a > 0

), toh

N = m (g + a) > m g

aur tumhare paon par extra push extra weight jaisa feel hota hai.

Two bodies of different mass in the same free-falling lift both read

N = 0

True —

N = m (g - g) = 0

chahe

m

kuch bhi ho, kyunki mass factor out ho jaata hai; dono ek hi acceleration

g

se girte hain.

Spot the error

"The cable snapped, the lift falls, so

a = + g

and

N = m (2 g)

Sign galat hai — upar positive convention ke saath, neeche ka acceleration

a = - g

hota hai, jo deta hai

N = m (g - g) = 0

. Up-positive convention lo aur usi par tikay raho (dekho Newton's Second Law).

"Astronauts float because the ISS is beyond Earth's gravitational field."

Earth ka field indefinitely extend karta hai; ISS altitude par

g \approx 8.7 m/s^{2}

. Woh continuous free fall ki wajah se float karte hain, gravity ki absence ki wajah se nahi.

"The scale measures how hard Earth pulls you, so it reads

m g

Scale sirf woh contact force sense kar sakta hai jo woh tumpe exert karta hai, yaani

N

(apparent weight); Newton's Third Law ke hisaab se tum equally push back karte ho, aur

N = m g

sirf equilibrium mein hota hai.

"In free fall gravity is cancelled by an upward force, so net force is zero."

Kuch bhi gravity ko cancel nahi karta — gravity hi net force hai, isliye tum

g

se accelerate karte ho. Bas koi contact force nahi bachta, toh tum kuch feel nahi karte.

"Weightlessness means the body has no acceleration."

Ulta hai — free-fall mein weightless body ka maximum downward acceleration

g

hota hai; contact force vanish hota hai, acceleration nahi.

"A lift decelerating as it goes down makes you lighter because it's going down."

Neeche jaate hue decelerate karna matlab acceleration upar point karta hai (

a > 0

), toh

N = m (g + a) > m g

— tum heavier feel karte ho, lighter nahi.

"Orbit needs no gravity, otherwise the station would fall."

Station gir hi raha hai — gravity centripetal acceleration

a = g

inward provide karta hai; iska sideways speed bas ise zameen miss karta rehta hai.

Why questions

Why does a bathroom scale, not gravity, decide what you "feel"?

Tumhare nerves pressure/contact forces sense karte hain; gravity har atom par uniformly act karta hai bina kuch push karne ke, isliye sirf support force

N

(apparent weight) "weight" ki tarah register hota hai.

Why does mass cancel out when checking weightlessness?

N = m (g + a)

mein,

a = - g

set karne par

N = m \cdot 0 = 0

kisi bhi

m

ke liye; har mass ek hi acceleration se free-fall karta hai, toh koi bhi apne support par press nahi karta.

Why is orbital weightlessness the "same physics" as a falling lift?

Dono mein, body aur uska support ek acceleration share karte hain (

g

neeche lift mein,

g

inward orbit mein), toh relative normal force zero hai — bas seedha girna versus curved girna.

Why does velocity direction never enter

N = m (g + a)

Newton's second law force ko acceleration

a

se relate karta hai, velocity se nahi; ek body kisi bhi direction mein constant speed se move kar sakti hai aur phir bhi

a = 0

ho sakta hai, hence

N = m g

Why can we treat the falling lift as a place where "gravity vanishes"?

Lift ke accelerating (non-inertial) frame mein ek pseudo-force

- m g

appear karta hai jo real gravity ko cancel karta hai — dekho Non-inertial Frames & Pseudo-forces; andar, sab kuch aise behave karta hai jaise

g = 0

ho.

Why does an upward acceleration (

a > 0

) increase the scale reading?

Tumhe upar accelerate karne ke liye, floor ko gravity se zyada push karna padta hai, toh

N > m g

; wahi extra push hai jo scale report karta hai aur tumhari legs feel karti hain.

Edge cases

Lift at rest on the ground — apparent weight?

a = 0

, toh

N = m (g + 0) = m g

; apparent weight exactly true weight ke barabar, yahi ek case hai jahan dono coincide karte hain.

Lift accelerating downward faster than

g

(say

a = - 1.5 g

, rocket-driven).

N = m (g - 1.5 g) = - 0.5 m g

, yaani formally negative; floor tumhe neeche pull nahi kar sakta, toh tum floor se lift off ho jaoge aur ceiling ke against press karne lagoge.

A person floating mid-air inside a free-falling lift, not touching anything.

Woh weightless hain

N = 0

ke saath trivially (no contact), aur phir bhi

g

se neeche accelerate kar rahe hain — gravity act kari, kuch push back nahi kiya.

At the exact instant the cable snaps, before any speed builds up.

N

immediately

0

par drop ho jaata hai, kyunki yeh acceleration par depend karta hai (

a = - g

turant), na ki kitni velocity accumulate hui hai.

A body in deep space (

g \approx 0

) sitting on a scale that isn't accelerating.

True weightlessness:

N = m (0 + 0) = 0

kyunki

g

khud zero hai — free fall jaisi hi reading lekin fundamentally different cause.

Astronaut at ISS altitude but momentarily at rest relative to Earth (not orbiting).

Woh seedha neeche girenge

a = g \approx 8.7 m/s^{2}

par (support-free) aur phir bhi weightless rahenge (

N = 0

); orbiting sirf woh sideways speed add karta hai jo impact prevent karta hai.

Standing in an elevator moving up at constant velocity.

a = 0

, toh

N = m g

— bilkul normal weight; kisi bhi direction mein constant velocity rest se indistinguishable hai scale ke liye.

Connections

Newton's Second Law — upar ke har answer ka ek application $N = m (g + a)$ ka hai.
Normal Force — woh force jo scale actually report karta hai.
Newton's Third Law — kyun scale tumhara push us par read karta hai.
Non-inertial Frames & Pseudo-forces — girte hue lift ke andar "gravity cancelled" wala view.
Circular Motion & Centripetal Acceleration — orbital free fall.
Gravitation — Variation of g with altitude — kyun ISS par $g \approx 8.7$ hai.
Free Fall Kinematics — support chale jaane ke baad ki motion.