analytical solution to center of earth problem

Theresa Wiegert theresaw at
Mon Mar 22 22:00:28 CET 2004

I'm afraid the ontopicness of this message tends now to become rather
farfetched, so for those who are not physics nerds as some of us, you
can skip the following.

Anyhow: According to Newton's shell theorem #2 A uniformly dense spherical
shell exerts no gravitational force on a particle (or duck in these
circumstances, the size of a duck compared to the earth is more or less
that of a particle...) located anywhere inside it.
(There's a proof for it also, but I don't have time/energy to write it
down. see e. g. Physics vol 1 by Halliday, Resnick and Crane if you're
interested.) (there's a shell theorem #1 as well: A uniformly dense
spherical shell attracts an external particle (duck) as if all the mass of
the shell were concentrated at its centre)

ANd then to the 'problem':
Suppose a hole is created through the Earth from one side (Duckburg) to
the other (I'm assuming now that the hole goes through so our poor
test-duck has somewhere to get out in case his speed is grave when he
reached the middle...). A duck of mass m is is dropped into the tunnel
from rest at the surface. a) what is the force exerted on the duck when it
is a distance r from the centre? b) What is the speed of the duck when at
distance r from centre? and c) speed at r=0? Neglect all frictional forces
and assume that the Earth has a uniform density.

Solution: From the theorem #2 above, we conclude that the gravitational
force on the particle only comes from the portion of the Earth that lies
inside the sphere of radius r, and we can also consider this mass to be
concentrated in the centre of the Earth (shell theorem #1). Let the mass
inside r be M, and total mass of the Earth M_E (radius R). Then the
fraction of the mass inside radius r is the same as the fraction of the
volume inside radius r (true only if density is uniform):
M/M_E=(4/3Pi*r^3)/(4/3Pi*R^3) or M=M_E*r^3/R^3
Regarding this mass as concentrated at the centre, we find the
gravitational force on mass m to be proportional to r:

Now it will be hard, I have to introduce vectors in ascii... Hm. Let rr be
a vector r from the center of the earth to the duck m, then we note that
the force F (now vector F=FF) acting on the duck is opposite in direction
to rr (r in vector form) -> FF=-(GmM_E/R^3)*rr (answer to question a).
Since we got a minus here, the form of the force looks very much like that
of a spring force, F=-kx!
Ok, so given the similarity with the spring force we can represent the
potential energy U of the system (earth and falling duck) as 1/2*kr^2,
taking U=0 in the center (NB k here is the constant in the force law:
k=GmM_E/R^3). AppÃlying conservation of energy at the surface and at
radius r we have K_s+U_s=K_r+U_r or 0+1/2*kR^2=1/2*mv^2+1/2*kr^2. Solve
for v (velocity) yields:
v=SQRT(k/m*(R^2-r^2))=SQRT(GM_e/R^3*(R2-r^2)) and at the center (r=0) the
value becomes v=SQRT(GM_E/R) (look in tables for the values) = 7910km/s
So... it is a good idea to dig the hole out on the other side as well to
avoid flat ducks. (probably worse than flat, but let's not go into gross

Now, it's definitely time to stop procrastinating, no matter how fun it

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