Gravity inside Earth.?

Question

Can anyone show or derive an equation so that i can compute for the acceleration due to gravity when a person is inside the earth (or when R is less than Rearth).

Answer 1

If you imagine a spherical shell (centered at the earth's center)
inside the earth, the gravitational force at the surface of that
sphere is equal to the force that would be exerted by a point mass at
the center, with mass equal to that contained within the whole shell.
In particular, all the mass outside the shell has a total contribution
of ZERO. Assuming (somewhat unrealistically) that the mass density is
uniform throughout the interior of the earth, the mass contained
within a spherical shell of radius r is proportional to r^3, while the
force varies with distance by 1/r^2, so altogether the force varies as
r^3/r^2 = r. Therefore, at R/2 from the center, the force is 1/2 that
at the surface.

Answer 2

It can be shown that, at any point within the earth, the net gravity is equal to that of the mass inside the sphere centered at the earth's center and containing the point in question. In other words,the mass under your feet counts, and the mass over your head doesn't. The density of the earth deep down is a bit higher than the density at the surface, but for mud-on-the-wall approximations, this can be ignored, and the result is that the gravitational force decreases linearly with increasing depth, reaching zero at the center. You can use this result to predict what would happen to a rock dropped down a hole bored all the way through the earth; the rock would reach the other end of the hole in about 45 minutes, and get back to your end 45 minutes later. It is not a coincidence that the period of a satellite in low earth orbit is about 90 minutes.

Answer 3

This cannot be done in general without knowing the run of mass (and/or density) with radius inside the Earth.

As Newton showed in one of his celebrated results, in a spherically symmetric distribution of matter, the matter "outside" your radial position has NO NET gravitational attraction on you, while all the matter "inside" your radial position attracts you gravitationally as though it were all concentrated at the centre. (This is a very specific, geometrical result that is ONLY true for spherical symmetry!)

Thus if the mass contained in a sphere out to radial distance r (< R, the Earth's radius) is M(r), the gravitational acceleration experienced would be:

g(r) = G M(r) / r^2.

The only simple, if highly idealized case to consider is that of UNIFORM DENSITY. In that case, M(r) is proportional to r^3, so that g(r) becomes a LINEAR FUNCTION of r. In that case,

g(r) = (r / R)*g(R)

where g(R) is the value of the gravitational acceleration at the surface of the Earth, of course.

Live long and prosper.

P.S. The actual number for the half-period of an oscillation through the Earth under that idealized assumption is very close to 42 minutes. This is true for an ARBITRARILY DIRECTED FRICTION-FREE TUNNEL, i.e. one going from any point on such an Earth to any other point on Earth. (That is WHY Douglas Addams asserted that 42 was the answer to almost anything!) Correspondingly, the period of an Earth-grazing satellite around an atmosphere-free Earth would be about 84 minutes.

Answer 4

Assuming the Earth to be homogeneous (which it isn't) and spherical,
g = (gs)r/R
where
gs = surface gravity (9.80665 m/s^2)
r is distance from the center, and
R is the radius of Earth
r and R must be in the same units, of course.