[Physics] Trouble understanding the concept of true and apparent weight

Question

I need help understanding the concept of true weight vs apparent weight. I understand this much: if someone is standing in an elevator on a scale, the further up they go the less the reading on the scale becomes. But why is this? Is it that distance affects the force of gravity? The further away the object goes [from the Earth's surface] the less the attractive force? Also, if on some other planet with radius $r$ an object is some distance $d$ away from the surface and is 1% less than its true weight on surface, what is the ratio $d/r$?

Answer 1

Okay. Firstly, I would like to point out that you are mixing two very different concepts here:
(1) Variation in the value of gravity $g$ as the distance from the surface of the earth changes.
(2) True and apparent weight

(1) Variation in the value of gravity
Alright. Variation in gravity. Firstly, lets get clear on the value of $g$. What exactly is $g$? It's like this: Suppose you are somewhere. Maybe sitting somewhere having pizza or flying in the sky. The earth applies a force on you. Let's call this force $F$. Then the value of $g$ is simply defined as $F/m$. That's it.
Now suppose the radius of earth is $R$ and you are at distance $d$ from the surface. (Note, from surface of the earth, not the center.) The force applied on you by the earth is $$F = \cfrac{GM_em}{(R+d)^2}$$ So, now, $$g = F/m = \cfrac{GM_e}{(R+d)^2}$$ Have a look at it. The value of $g$ indeed depends on $d$, your distance from the surface of the earth. But, near the surface of the earth, $d<<R$, so we can approximate the above expression to $$g = F/m = \cfrac{GM_e}{R^2}$$ which is independent of $d$. But note that it is valid only for small values of $d$.

(2) True and apparent weight
Okay. Answer to the next part of the question. True and apparent weight. True weight is simply weight. What is your true weight? It's simply $mg$. Mass multiplied by gravity. End of story.
Now, Apparent weight. I'll denote it by $W_A$. It's defined as $$W_A = N$$ where $N$ is the normal force in the direction opposite to the direction of gravity. That is away from the center of the earth. You may be standing and someone may be trying to push you horizontally. That normal reaction force doesn't count. Only the vertical Normal Force counts.
So suppose you jump from the top of the building because your dog died. You are falling. Your '(True) Weight' is simply $mg$. Your Apparent weight is $0$. Because there is no normal force applied on you currently. (Offcourse the ground will apply one hell of a normal force when you finally reach it.)
Now suppose you are standing in an elevator at rest. True weight, offcourse is $mg$. But Apparent weight is also $mg$. Because you are at rest, $N = mg$.
Elevator moving with constant speed: $N = mg$
Suppose the magnitude of elevator's acceleration is $|a|$.
Elevator moving upwards, and slowing down: $N = mg - m|a|$
Elevator moving upwards, and increasing speed: $N = mg + m|a|$
Elevator moving downwards, and slowing down: $N = mg + m|a|$
Elevator moving downwards, and increasing speed: $N = mg - m|a|$
So, why did they introduce the concept of Apparent Weight. Apparent weight is the weight you 'feel'. Think about it! When you are falling, you feel weightlessness. Hence Apparent Weight is $0$. When in an elevator with moving upwards with increasing speed, you feel heavier. Hence more is the Apparent Weight!

(3) $d/R$ ratio
The ratio $d/R$ where weight would be 1% lesser:
$$\cfrac{GM_e}{(R+d_1)^2} = 0.99 \cfrac{GM_e}{R^2}$$ Solve it for $d_1$. That's your answer!