
Theoretically speaking, if you were to travel to
or near the core of the Earth, would you be
heavier because the radius would decrease between
you and the core; therefore, decreasing the
radius in the formula
Fg=Gm_{1}m_{2}/r^{2} and
increasing the Fg?
  Answer 1:
Hi, you're halfway there with that reasoning,
but you've forgotten that one of the other
variables in that equation also changes when you
move closer to the center of the Earth: the mass
of the object pulling on you, which I'll call
M_{1}. More specifically, in the
formula F_{gravity} = G * M_{1}
* M_{you} / r^{2}
,
M_{1} is the mass enclosed within
a sphere that has a surface at radius r. So, when
you're on the surface of the Earth, M_{1}
= M_{Earth}. But when you're closer to
the center of the Earth, M_{1} is less
than M_{Earth}. And when you're very
close to the center, you can approximate the mass
density of Earth, rho, as constant, so that the
mass contained within a sphere of radius r is just
the volume (4/3*pi*r^{3}) times the
density: M_{1} = rho * 4/3 * pi *
r^{3} .
Plugging that into the
force equation gives
F = G * 4/3 * pi *
M_{you} * rho * r .Since rho is just
some number (somewhere between 1 and 10 grams per
cubic centimeter), and M_{you} stays
constant, the force goes to 0 as r goes to 0. And
that makes sense because at the exact center of
the Earth, which direction would gravity pull on
you? All directions look the same, so there can't
be a force due to gravity! And so the force near
the center must be nearly zero. Physicists
call that a "proof by symmetry", which can be very
useful while involving no effort, both of which
are appealing to physicists! Another example that
comes to mind as I drink my coffee: the next time
you're holding a glass of liquid, put it down with
a slight bump on the table, keeping the bottom as
flat as you can. Or think back to the movie
"Jurassic Park", when the kid has his water on the
car's dashboard and the T. Rex is stomping closer.
What do the waves on the surface look like?
Better yet, what do they HAVE to look like?
Circles, because there aren't any preferred
directions in this setup! The waves only "know"
about the symmetries of the cup, which are
circular. Symmetries in all aspects of
physics can take you a long, long distance, in
ways that still blow my mind. Good luck with your
studies!
  Answer 2:
Actually, the acceleration due to gravity is
found from the following formula g= G
M(r)/r^{2} where M(r) is the mass of earth
between the center of earth and distance r from
the center. So, given how the mass of the
earth between the center and some distance r from
center varies, it turns of the g is almost
constant from the surface down to a depth of about
3000 km. thereafter g decreases until it would be
zero at the center of the earth . If you
google this site: gravity you
can see a graph of gravity (actually the value of
little g) vs depth.
  Answer 3:
No, actually you would get lighter. If you
are inside of a hollow, massive sphere, you would
experience no gravity. This is because the sides
of the sphere are pulling on you equally in all
directions. Even if you are closer to one edge of
the sphere, there is more material on the other
side even if it's farther away. You experience no
gravity. Now, the Earth is not a hollow
sphere, but what this does mean is that as you
burrow into the Earth, the mass of the planet is
divided into two parts: (1) the mass that is above
your current depth, and (2) the mass still inward
toward the center. The mass of (1) doesn't matter,
but you do feel gravity from mass (2). However, as
you continue going inward, the mass of (2) is
progressively getting smaller because the radius
out to you is getting smaller. Mass is
proportional to volume, which is in turn
proportional to the cube of the radius.
Gravitational force, alternatively, is inversely
proportional to the square of the radius.
r^{3}/r^{2} = r, which means that
the amount of gravity you experience is
proportional to the radius, and the farther out
from the center you are, the more gravity there
is, until, of course, you break out again through
the surface of the Earth. Now, in this
calculation, I assumed that the Earth has the same
density throughout. It doesn't; the core is a lot
denser than the mantle or the crust. Because of
this, it may be that for a while you would get
heavier as you bored in for exactly the reason you
suggested in the first place because you're
getting closer to the surface of the denser, more
massive core. You would have to do the calculation
(density of the mantle is about 2.5 thousand
kg/m^{3}, density of the core is about 10
thousand kg/m^{3}, the Earth has a radius
of 6.4 million meters, and the radius of the core
is I think about 3 million meters). The
only time in which you could bore into something
and get heavier as you go farther in is if all of
the mass is concentrated at the very center 
something which only happens in a black hole.
There, your weight really does become infinite as
you approach the center of the hole.
  Answer 4:
You would be lighter because all of earth's
mass would be distributed equally all around you.
All of it would be pulling on you, but the pull
would be equal from every direction, so it would
all cancel out. On the other hand, if you squeezed
Earth to a smaller radius, so it had the same mass
but you were still outside it, then your weight
(gravitational force) would go up just like you
predict from the Fg equation.
Click Here to return to the search form.





Copyright © 2015 The Regents of the University of California,
All Rights Reserved.
UCSB Terms of Use


