Gravity at the Earth's Center
(Two questions with the same answer)
(1) My students had a couple of questions
that I thought were interesting. I told them I'd ask ya'll.
- What would you weigh is you were at the exact center of the earth?
- What would you weigh is you were 3 meters from the center of the earth?
Please include supporting evidence for your answers
I am confused about Newton's discussion of the force on a particle within a
sphere in the Principia. In one place, he says that the force would be zero,
since the attraction of all the particles in the sphere would cancel each
Just a little further on, he says that the force would be directly
proportional to the particle's distance from the center of the sphere.
Can you clarify these two seemingly conflicting statements?
Thank you for any light that you can shed.
Dear Teacher (and this also answers the student of Newton):
Yours is an old question, first tackled by Newton, as the "Hollow Earth Paradox." If the Earth where a hollow sphere (inner and outer
surfaces spherical) and someone dug a hole that reached the hollow interior,
and then stepped into it--what would that person experience?
Newton's answer--there would be no gravity inside the hollow. Any object
thrown into it--say, a stone--would continue in a straight line with
constant velocity (ignoring air resistance).
Newton's argument was roughly as follows. Take an object at a point in
space anywhere in the cavity and draw from it a double cone (like a teepee,
extending to both sides). Each side of it will cut part of the sphere,
and the gravity of the two parts will tend to pull the object in opposite
directions (make a drawing and you will see).
Newton showed that the pulls of both part cancel each other: one part may
be closer, but then it will also be smaller. Since all directions can be
covered by a collection of such cones, the total force is zero. Today we
get this result much more quickly by the theory of the potential, but that
takes three-dimensional calculus, which Newton did not have.
Now: Imagine you are somehow in the middle of the solid Earth--by some
magic, not crushed by the rocks, suffocated or incinerated. In your mind
you can divide all matter on earth into two parts: a smaller sphere
containing everything that is CLOSER than you to the Earth's center, and
a hollow sphere containing everything that is MORE DISTANT.
By Newton, the hollow sphere exerts no pull, while the interior sphere,
like the Earth, pulls as if all its mass were concentrated in the middle
(that's another thing easily shown from potential theory). If you are
halfway to the center, and the density everywhere is the same (actually,
matter gets compressed towards the middle) then only 1/8 of the Earth
mass is pulling you, but at half the distance, the pull is 4 times stronger
("inverse squares law"), so the final result is 1/2 of the gravity on the surface. At 1/N times the radius, the pull of gravity is just 1/N the pull at the surface.
As you get deeper and deeper, the inside sphere gets smaller and its pull
is weaker, so gravity too weakens. At the center, it is zero. At 3 meters
from the center, it is the pull of a 3-meter sphere of rock, experienced
on its surface--the pull of a tiny asteroid.
Please note--that is just the pull of gravity on YOU. The rocks above you
are also all pulled down, all the way to the surface of the Earth, and
their weight is likely to crush you before you get very far. There may
perhaps exist a cave a mile deep, but if so, none is much deeper, because
there is too much weight piled on top.
(a) Radiation hazard in space--1
I am working at the University of Arkansas School of Architecture along with members of the Habitability team at NASA for the manned Mars mission.
It has been explained to me that radiation will be a big issue in the
design of a Mars habitat. I was wondering how feasible it would be to use
nuclear power to produce a eletromagnetic field around the habitat to
reduce or deflect the radiation. Is it possible to create a magnetic
field strong enough to provide radiation protection? And if so, how much
energy would it require?
I have not calculated the field needed, but it is probably very strong, too expensive to set up, too much mass and energy are needed, and a strong
magnetic field would affect instruments.
The cheap and simple way is to build a shelter--especially since the dangerous events are the ones of solar outbursts, which are rare and
last a day at most. You can calculate the shielding, but 20-50 cm rock
should do a pretty good job (remember gravity is weaker, too, they will
weigh less than on Earth).
Have you read Ben Bova's "Mars"? It's fanciful science fiction, but his physics seems OK. The Mars astronauts are hit by a solar outburst halfway to Mars and wait it out, huddled in a special shielded area of