12.4 - Shell theorem: inside the sphere
Another section discussed how to calculate the force of gravity exerted on an object on
the surface of a sphere (a groundhog on the Earth). Now imagine a groundhog
burrowing halfway to the center of the Earth, as shown to the right. For the purposes of
calculating the force of gravity, what is the distance between the groundhog and the
Earth? And what mass should be used for the Earth in the equation?
The first question is easier: The distance used to calculate gravity’s force remains the
distance between the groundhog and the Earth’s center. Determining what mass to use
is trickier. We use the sphere defined by the groundhog’s position, as shown to the
right. The mass inside this new sphere, and the mass of the groundhog, are used to
calculate the gravitational force. (Again, we assume that the Earth’s mass is
symmetrically distributed.)
If the groundhog is 10 meters from the Earth’s center, the mass enclosed in a sphere
with a radius of 10 meters is used in Newton’s equation. If the animal moves to the
center of the Earth, then the radius of the sphere is zero. At the center, no mass is
enclosed, meaning there is no net force of gravity. The groundhog is perhaps feeling a
little claustrophobic and warm, but is effectively weightless at the Earth’s center.
The volume of a sphere is proportional to the cube of the radius, as the equation to the
right shows. If the groundhog burrows halfway to the center of the Earth, then the
sphere encloses one-eighth the volume of the Earth and one-eighth the Earth’s mass.
Let’s place the groundhog at the Earth’s center and have him burrow back to the
planet’s surface. The gravitational force on him increases linearly as he moves back out
to the Earth’s surface. Why? The force increases proportionally to the mass enclosed
by the sphere, which means it increases as the cube of his distance from the center.
But the force also decreases as the square of the distance. When the cube of a quantity
is divided by its square, the result is a linear relationship.
If the groundhog moves back to the Earth’s surface and then somehow moves above
the surface (perhaps he boards a plane and flies to an altitude of 10,000 meters), the
force again is inversely proportional to the square of the groundhog’s distance from the
Earth’s center. Since the mass of the sphere defined by his position no longer varies,
the force is computed using the full mass of the Earth, the mass of the groundhog, and
the distance between their centers.
Inside a sphere
To calculate gravitational force
·Use mass inside the new shell
·r is distance between object, sphere’s
centerVolume of a sphere
V = volume
r = radius
12.5 - Sample problem: gravitational force inside the Earth
Assume the Earth’s density is uniform. The Earth’s mass and radius are given in the table of variables below.
Variables
What is the gravitational force on the
groundhog after it has burrowed
halfway to the center of the Earth?
gravitational force F
mass of Earth ME = 5.97×10^24 kg
radius of Earth rE = 6.38×10^6 m
mass of inner sphere Ms
distance between groundhog and Earth’s
centerrs= 3.19×10^6 m
mass of groundhog m = 5.00 kg
gravitational constant G = 6.67×10í^11 N·m^2 /kg^2