Simulation · Physics · Class 11

g inside and outside the earth

From the lesson g above and below the surface in Gravitation. Change the values and watch what happens.

The idea behind it

NCERT §7.6

At height h the distance from the centre is R_E + h, so g(h) = G M_E / (R_E + h)² = g (R_E/(R_E + h))².
For h much smaller than R_E, the binomial approximation gives g(h) ≈ g (1 − 2h/R_E).
At depth d, the shell of thickness d above pulls with zero net force; only the inner sphere of radius R_E − d, of mass M_E (R_E − d)³/R_E³, pulls.
So g(d) = g (1 − d/R_E) exactly, for a uniform earth. At the centre (d = R_E) g is zero.
g is greatest on the surface and falls whether you go up or down. For the same small distance, going up reduces g about twice as much as going down: 32 km up gives about 9.70 m/s², 32 km down about 9.75 m/s².
Weight changes with g, mass does not. A body weighing 63 N on the surface weighs 63/(1.5)² = 28 N at a height R_E/2; one weighing 250 N weighs 125 N halfway to the centre.
Use the exact form (R_E/(R_E + h))² whenever h is comparable to R_E; 1 − 2h/R_E gives nonsense there (zero at h = R_E/2).

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