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a hole in a shell



REPOSTING WHAT WAS LOST

Here is nice problem for a calculus based physics class. Consider
an infinitely thin spherical shell whose mass and radius are equal
to those of our planet. We all know that g should be 9.8 m/s^2
at the outer surface and zero anywhere inside the shell.

Now suppose a circular hole is punched in the shell. The half
angle of that hole, as seen from the center, is 1 degree. The mass
which was removed, 2.28*10^-6 of the total, becomes a point object
attracted by the punched shell. It moves along the axis of the hole,
toward the center. How does the value of g change with the distance?

It turns out to be always less than 9.8. And it changes gradually from
a maximum value close to 9.0 to zero. The value of g at the “sea
level” is only 4.88 m/s^2. That value would be obtained if the entire
mass of the punched shell were concentrated at a distance sqrt(2)*R
from the hole (as previously stated by Bob Sciamanda, in the electric
context).

At the height of one hole radius (55.76 km) g=8.206 m/s^2 and at
the “depth of one hole radius” (inside the sphere) it is 1.44 m/s^2.
The agreement with g=G*M/r^2 is already very reasonable (1%)
at 5 hole radii above the sea level. At the depth of 10 hole radii
the value of g is already 0.02 m/s^2.

Nothing profound, only an exercise on numerical integration.
I did this for electric charges and used the program in the context
of gravitation. This helped me to find a bug, as posted several
minutes ago.
Ludwik Kowalski
P.S.
It is interesting to speculate on the "like charges repel" versus
"like masses attract". Or on the "holes as free carriers" in some
electric materials. A hole is not an antiparticle (as a positron is)
but it is treated as a positive electric carrier. The absence of the
mass (in the punched region of the gravitational shell) acts as a
"small negative mass" inserted into the body of the shell; it can
be said to be responsible for g<9.8 above and g>9.8 below. Silly?