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Gravitational intensity below the Earth's surface

Field inside the Earth Consider a point inside the Earth at distance r from the centre (r< R). Let the field strength at that point be g. Therefore:

g = GM'/2 where M' is the mass of the Earth within radius r. But M' = r3M/R3. Therefore mg = GmMr/R and so:

Gravitational intensity (g) at radius r = go[r/R]

This means that theoretically the gravitational field intensity decreases linearly inside the Earth; however, this is only true if we assume that the Earth has a uniform density.


In fact the density increases with depth, the density of the Earth's crust being about 2.8x103 kg m-3 while that of the surface of the core is 9.7x103 kg m-3 .

This actually results in an increase in g for a short distance below the surface. The theoretical and actual variations are shown above. If R r = h, then the theoretical reduction in gravitational intensity at a depth h below the surface is given by:

go g = hgo/R

Gravitational intensity above the Earth's surface

Consider a point a distance r from the centre of the Earth where r > R.

g = GM/r2 but go = GM/R2. Therefore:

g = go[R2/r2]

For a height h above the Earth's surface, r = R + h. Using the result above, and assuming that h is small compared with the radius of the Earth R, we can say that:

go g = [2h/R]go

Thus the value of g at a depth h below the Earth's surface is greater than that at the same distance above the surface.

Example problem
Find the gravitational intensity at a point 1000 m above mean sea level.
Take R = 6400 km and go = 9.81 ms-2.

go g = [2x103go]/6.4x106 = 0.31x10-3go = 0.003 ms-2

© Keith Gibbs 2020