Physics · Gravitation · NEET
When you go to depth d, the shell of Earth ABOVE you no longer pulls you (its gravity cancels out inside). Only the smaller inner sphere of radius (R − d) pulls you. Less mass acting on you means smaller g. So g_d = g(1 − d/R): the deeper you go, the smaller the effective mass below, and the smaller g becomes.
At the centre, d = R, so g_d = g(1 − R/R) = g(1 − 1) = 0. Physically, at the centre you are surrounded equally by mass on all sides, so every pull cancels out. There is no net force, so g = 0. A body there would be weightless even though it is not moving.
Assume Earth has uniform density ρ. At the surface: g = (4/3)πGρR. At depth d, only the inner sphere of radius (R − d) pulls you, so g_d = (4/3)πGρ(R − d). Divide: g_d/g = (R − d)/R = 1 − d/R. Therefore g_d = g(1 − d/R). This step-by-step ratio is the fastest way to remember it.
For SMALL height h, g_h = g(1 − 2h/R) — note the factor of 2. For depth d, g_d = g(1 − d/R) — no factor of 2. So near the surface, g falls TWICE as fast per metre of height as per metre of depth. That is exactly what the 2017 NEET PYQ tests: g at height 1 km = g at depth 2 km.
In real deep mines g can slightly rise at first because Earth's density is NOT uniform — the dense iron core makes the inner sphere heavier than the simple model assumes. But for NEET, always use the uniform-density result g_d = g(1 − d/R), where g decreases steadily and reaches zero at the centre. NEET questions assume uniform density unless stated otherwise.
A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth?
The acceleration due to gravity at a height 1 km above the earth equals that at a depth d below the surface. Then d equals:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
g_d = g(1 − d/R), where g is surface gravity, d is the depth, and R is Earth's radius. It is a linear decrease, reaching zero at the centre (d = R).
Yes. At the centre d = R, so g_d = g(1 − 1) = 0. Mass surrounds you equally on all sides, so all gravitational pulls cancel and net g = 0.
They come from different physics. Depth uses the shrinking inner mass giving g(1 − d/R). Height uses the inverse-square law approximated for small h, giving g(1 − 2h/R). Do not mix them.
With height. For the same small distance, g falls twice as fast going up as going down, because of the factor of 2 in the height formula.
For a uniform-density model, g is maximum at the surface. It decreases both above (with height) and below (with depth), so the surface is the peak.
