Physics · Gravitation · NEET
For a small height h above the surface, weight becomes W' = W(1 - 2h/R). For a depth d below the surface, weight becomes W' = W(1 - d/R). The height formula has the factor 2h/R (from the exact form g(h) = g/(1 + h/R)^2 expanded for small h). The depth formula g(d) = g(1 - d/R) is exact for a uniform earth, with no 2. Mixing these two is the most common NEET mistake.
No. W = W(1 - 2h/R) is only an approximation for h much smaller than R. For a large height (like h = R/3 or h = R) you must use the exact form: W' = W / (1 + h/R)^2, which comes from g(h) = GM/(R+h)^2. NEET 2025 used h = R/3 and the exact form gives (3/4)^2 = 9/16, not the 2h/R shortcut.
Going up, g follows the inverse-square law g ∝ 1/(R+h)^2, and for small h this drops at the rate 2h/R. Going down, only the inner sphere of radius (R-d) pulls you, and mass falls off, so g ∝ (R-d), dropping at the slower rate d/R. Same distance moved, but height loses weight about twice as fast as depth.
Set the two equal: g(1 - 2h/R) = g(1 - d/R). Cancel g and R: 2h = d. So the depth that matches a given height is d = 2h. This is exactly the NEET 2017 question: height 1 km matches depth d = 2 km.
Yes, at the centre of the earth (d = R). Put d = R in W' = W(1 - d/R) to get W' = W(1 - 1) = 0. At infinite height weight also approaches zero, but it never becomes exactly zero at any finite height above the surface.
The acceleration due to gravity at a height 1 km above the earth equals that at a depth d below the surface. Then:
A body weighs 48 N on the surface of the earth. The gravitational force on it at a height equal to one-third the radius of the earth from the surface is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For small h: W' = W(1 - 2h/R). For any h: W' = W/(1 + h/R)^2, where R is the earth's radius and W = mg is the surface weight.
W' = W(1 - d/R). This is exact for a uniform earth. At d = R (the centre), weight is zero.
Set 1 - d/R = 1/2, giving d/R = 1/2, so d = R/2. At half the earth's radius deep, g and weight are halved.
On the earth's surface. Weight decreases whether you go up (height) or down (depth), so the surface is the maximum.
No. Mass m is constant everywhere. Only g changes, so weight W = mg changes while mass stays the same. This is a common NEET conceptual trap.
