Physics · Gravitation · NEET
At the centre, mass is spread equally in every direction around you. Every pull from one side is cancelled by an equal pull from the opposite side. The net gravitational force becomes zero, so g = 0 at r = 0. This is why the graph starts from the origin.
It decreases as you go deeper. Only the mass in the sphere below your feet pulls you; the outer shell contributes nothing. As you go down, less mass is below you, so g falls. Inside the Earth g = (GM/R³)·r, so g is directly proportional to r (a straight line). At depth it becomes smaller, reaching zero at the centre.
Inside the Earth, the effective mass that pulls you grows as r³ (bigger sphere), and gravity has 1/r², so g ∝ r³/r² = r, a straight line. Outside, the full mass M is fixed while distance grows, so g ∝ M/r² = 1/r², a curve. Two different rules because inside the pulling mass changes, outside it stays fixed.
g is maximum exactly at the surface, r = R (about 6400 km from the centre). This is the peak of the graph — the meeting point of the rising straight line and the falling 1/r² curve. Both the inside formula and the outside formula give the same value g = 9.8 m/s² here.
No. It has two clearly different parts joined at the surface. From centre to surface it is a straight line rising through the origin. From surface outward it is a smooth curve dropping as 1/r². They meet at the peak (r = R), but the shapes are different — straight then curved.
Starting from the centre of the earth (radius R), the variation of g (acceleration due to gravity) with distance r is best shown by which graph?
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?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Inside the Earth at distance r from the centre, g = (GM/R³)·r, or equivalently g = g_surface·(r/R). It shows g is directly proportional to r, giving the straight-line part of the graph.
Outside the Earth at distance r from the centre, g = GM/r². Here g is inversely proportional to r², giving the curved 1/r² part that falls slowly as you move away.
g is maximum at the Earth's surface, where r = R (about 6400 km from the centre). Its value there is about 9.8 m/s². Both going deeper and going higher make g smaller.
It tests whether you truly understand the two different behaviours — linear inside (g ∝ r) versus 1/r² outside — in a single picture. Many students wrongly assume g keeps rising toward the centre, so it is a common trap worth one guaranteed mark.
In the ideal uniform-density model used for NEET, yes, g = 0 at the centre. The real Earth has a denser core, so g does not fall perfectly linearly, but for NEET always use the uniform model: g rises linearly inside and is zero at the centre.
