Further Note on "Gravitation Inside A Uniform Hollow Sphere"
This article is
written in response to questions received from Claus Olesen. Here is
his e-mail to me:
I saw your page
http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/.
Interesting...[I have a question] related to "Gravitation Inside
A Uniform Hollow Sphere."
Imagine a straight
pipe going all the way through the earth, e.g. being a diameter of the
Earth. If the pipe is filled with water, then what is the pressure in
the water as function of distance from the center of the Earth? In the
Mariana Trench is it roughly 11000m/(10m/1atmosphere) = 1100 atmospheres?
If the pipe is empty,
e.g. contains vacuum, then what is the traveling time from rest from
one end of the pipe to the other?
Thank you.
Claus Olesen
Dear Claus,
Permit me to replace the assumption of a water-filled pipe with another
assumption that will produce the desired result: Let's assume that the
earth is a uniform sphere of some hypothetical liquid having an overall
radius of 6,400 km. Let's assign this liquid some additional properties:
We know that the
average density of the earth is about 5.5 times that of water, i.e.,
5,500 kg/m3. We will assume that our hypothetical liquid
has the same density. We will also assume that our liquid is incompressible
so that the density remains uniform throughout the sphere.
OK. We wish to calculate
the pressure at some depth below the surface. We will use the differential
form:
dp(r)
= -g dr
where,
· r is distance measured from the center of the sphere,
· dp(r) is differential pressure as a function of the distance
r,
· is density
(assumed constant),
· g is gravitational acceleration, and
· dr is an increment of distance measured from the earth's surface
downward.
We recognize that
g inside the earth must be a function of r, i.e., g = g(r). Let's determine
what that function actually is before we proceed.
First of all, we
know from Newton's Law of Universal Gravitation that:
g(r)
= GM/r2
where G is Newton's
universal constant and M is the field generating mass. Let a be a point
in the interior of our liquid sphere at a distance ra from the center.
Let's find the value of g at a:
· We know
that a is on the surface of a sphere of radius ra < 6,400
km.
· We also know that there is a hollow spherical shell of thickness
6,400 km - ra above a.
By my previous article,
"Gravitation Inside A Uniform Hollow Sphere," we know that
the mass contained in this hollow shell contributes nothing to the value
of g at a. Thus, the only non-zero contribution must come from the mass
inside the sphere of radius ra. This contribution is:
g
= GM/ ra2.
Since our liquid
sphere is assumed to have a constant density, we may further write the
field generating mass as:
M
= (4
ra3/3).
Combining these
last two results, we discover that g inside the earth is a decreasing
linear function of the distance r as we approach the center. Dropping
the subscript a, we have:
g
= G(4
r3/3) / r2 = 4Gr/3.
Finally:
dp(r) = -g dr
dp(r)
= -(4Gr/3)
dr = -(4G2r/3)
dr.
Integrating the
right-hand side between the limits of 6,400,000 m and r yields the interior
pressure of the earth as a function of r:
p(r)
= 2G2(6,400,0002
- r2)/3
where the earth's
radius has been expressed in units of meters for dimensional consistency.
As a point of interest,
setting r = 0 gives an estimate of the hydrostatic pressure at the center
of the earth:
p
= 2G2(6,400,000
m)2/3 = 1.7 x 1011 nt/m2.
Let's check this
result using your information on the Mariana Trench. You estimate the
pressure to be 1,100 atm at 11,000 m. We will use the expression obtained
above to calculate the pressure at a depth of 11,000 m in our liquid
sphere.
Recall that:
p(r)
= 2G2(6,400,0002
- r2)/3
where the earth's
radius is expressed in m for dimensional consistency. To obtain an estimate
of the pressure at the bottom of the Mariana Trench, we need an estimate
of the average density r near the earth's surface. Since the earth's
surface is about 70% ocean and 30% land, let's use:
r = 0.7 x (1,000
kg/m3) + 0.3 x (5,500 kg/m3) = 2,400 kg/m3.
If Mariana is 11,000
m in depth, then the corresponding value of r is:
6,400,000
m - 11,000 m = 6,389,000 m ~ 6,390,000 m.
The pressure at
the bottom of the trench may then be estimated as:
p(r)
= 2G2(6,400,0002
- 6,390,0002)/3 = 1.03 x 108 nt/m2
= 1,030 atm.
The difference error
is (1,100 - 1,030)/1,100 = 6%, which is, overall, not bad!
OK, now to the evacuated
pipe.
An evacuated shaft
implies that we may ignore air resistance, so we are dealing with an
idealized case. Using our previous results, let's begin with the expression:
g = lr, where l
= 4G/3.
Assuming that the
projectile starts at the earth's surface with zero velocity and is dropped
into the evacuated shaft, we may next write:
dv
= g dt = lr dt = - (lr/v) dr,
where dv is differential
velocity = - dr/dt. This expression yields:
v2
dv = -lr dr.
Integrating the
left-hand side between 0 and v, and the right-hand side between rE
and r (where rE is the earth's radius, and v is the velocity
at r), we find that:
v2
= l(rE2 - r2) or v = Öl
(rE2 - r2)½.
We now write further:
dr
= -v dt = - l
(rE2 - r2)½ dt
where dr is differential
distance measured radially; or:
dr/[(rE2
- r2)½] = - l
dt.
Treating this integral
as an indefinite integral, we find:
arcsin
(r/rE) = l t, or
t = [arcsin (r/rE)]/l.
The time it takes
the projectile to reach the center of the Earth is found by setting
r = rE. Now, l
= 2.15 x 10-3 (MKS units), and so:
tTo
Center = 732 seconds = 12 minutes.
A round trip through
the earth and back to the starting point would take 48 minutes = 2,880
seconds, and the oscillating frequency of the projectile would be 3.5
x 10-4 sec-1.
Q.E.D.