Geoscience Reference
In-Depth Information
Hence, we find, with the density
= constant,
U
=
G
2
π
α
=0
a
b
sdsdzdα
s
2
+(
c − z
)
2
s
=0
z
=0
(3-5)
=2
πG
a
s
=0
b
sdsdz
s
2
+(
c
z
)
2
.
−
z
=0
The integration with respect to
s
yields
a
z
)
2
z
)
2
=
s
2
+(
c
sds
s
2
+(
c
a
−
−
0
(3-6)
s
=0
=
a
2
+(
c
−
z
)
2
−
c
+
z,
so that we have
U
=2
πG
b
0
z
)
2
dz .
c
+
z
+
a
2
+(
c
−
−
(3-7)
The indefinite integral is 2
πG
times
2
a
2
ln
c
z
)
2
,
(3-8)
z
)
a
2
+(
c
z
+
a
2
+(
c
1
1
1
z
)
2
2
(
c
−
−
2
(
c
−
−
z
)
2
−
−
−
as may be verified by differentiation. Hence,
U
finally becomes
U
e
=
πG
(
c
b
)
a
2
+(
c
b
)
2
+
c
√
a
2
+
c
2
b
)
2
c
2
−
−
−
(
c
−
−
(3-9)
b
)
2
+
a
2
ln
c
+
√
a
2
+
c
2
,
b
+
a
2
+(
c
a
2
ln
c
−
−
−
where the subscript e denotes that
P
is external to the cylinder.
The vertical attraction
A
is the negative derivative of
U
with respect to
the height
c
[see Eq. (2-22)]:
∂U
∂c
A
=
−
.
(3-10)
Differentiating (3-9), we obtain
A
e
=2
πG
b
+
a
2
+(
c
a
2
+
c
2
.
−
b
)
2
−
(3-11)