Geoscience Reference
In-Depth Information
P
on cylinder
In this case we have
c
=
b
, and Eqs. (3-9) and (3-11) become
U
0
=
πG
,
b
2
+
b
a
2
+
b
2
+
a
2
ln
b
+
√
a
2
+
b
2
a
−
(3-12)
A
0
=2
πG
a
+
b
a
2
+
b
2
.
−
(3-13)
P
inside cylinder
We assume that
P
is now inside the cylinder,
c<b
. By the plane
z
=
c
we
separate the cylinder into two parts, 1 and 2 (Fig. 3.3), and compute
U
as
the sum of the contributions of these two parts:
U
i
=
U
1
+
U
2
,
(3-14)
where the subscript i denotes that
P
is now inside the cylinder. The term
U
1
is given by (3-12) with
b
replaced by
c
,and
U
2
by the same formula with
b
replaced by
b − c
.Theirsumis
U
i
=
πG
c
)
2
+
c
√
a
2
+
c
2
+(
b
c
)
a
2
+(
b
c
2
−
−
(
b
−
−
−
c
)
2
.
(3-15)
c
+
a
2
+(
b
+
a
2
ln
c
+
√
a
2
+
c
2
a
c
)
2
+
a
2
ln
b
−
−
a
It is easily seen that the attraction is the difference
A
1
−
A
2
:
A
i
=2
πG
2
c
c
)
2
;
a
2
+
c
2
+
a
2
+(
b
−
b
−
−
(3-16)
this formula may also be obtained by differentiating (3-15) according to
(3-10).
bc
-
2
P
b
1
c
a
Fig. 3.3. Potential and attraction on an internal point
