Geoscience Reference
In-Depth Information
1.2
Potential of a solid body
Let us now assume that point masses are distributed continuously over a
volume
v
(Fig. 1.2) with density
=
dm
dv
,
(1-11)
where
dv
is an element of volume and
dm
is an element of mass. Then the
sum (1-10) becomes an integral (Newton's integral),
V
=
G
v
dm
l
=
G
v
l
dv ,
(1-12)
where
l
is the distance between the mass element
dm
=
dv
and the at-
tracted point
P
. Denoting the coordinates of the attracted point
P
by
x, y, z
and of the mass element
m
by
ξ, η, ζ
,weseethat
l
is again given by (1-5),
and we can write explicitly
V
(
x, y, z
)=
G
v
(
ξ, η, ζ
)
(
x
ζ
)
2
dξ dη dζ ,
(1-13)
−
ξ
)
2
+(
y
−
η
)
2
+(
z
−
since the element of volume is expressed by
dv
=
dξ dη dζ .
(1-14)
This is the reason for the triple integrals in (1-12).
z
P xyz
(,,)
l
v
d
³
dm(,,)
»´³
d
»
d
´
y
x
Fig. 1.2. Potential of a solid body
