Environmental Engineering Reference
In-Depth Information
d
V
y
q
x
P
Figure 9.3
A spherical distribution of matter.
The use of this formula is perhaps best illustrated in an example. Let us use it
to determine the potential arising due to a uniform sphere of radius
R
and of total
mass
M
. Figure 9.3 defines the various quantities of interest. Consider breaking
the sphere into lots of tiny volume elements. In spherical polar co-ordinates, the
volume of a general element is
r
2
sin
θ
d
r
d
θ
d
φ
=
d
V
(9.20)
and the mass of this element is just equal to
M
4
πR
3
/
3
.
d
M
=
d
V
(9.21)
Thus the potential at P arising from this element is equal to
G
d
M
y
d
=−
(9.22)
and the potential is obtained after summing over all the elements which make up
the sphere, i.e.
4
πR
3
R
d
r
2
π
0
d
φ
π
0
3
M
d
θr
2
sin
θ
y
=−
G
.
(9.23)
0
The azimuthal
φ
integral is easy enough and just gives a factor of 2
π
but the other
integrals are harder since the distance
y
varies as
θ
and
r
change. We need to
express
y
in terms of
r
and
θ
, and we can do this using the cosine rule:
y
2
r
2
x
2
=
+
−
2
xr
cos
θ.
(9.24)
We have to choose whether to do the
r
or the
θ
integral first. Let us choose the
θ
integral, in which case we must evaluate
π
sin
θ
d
θ
(9.25)
(x
2
+
r
2
−
2
xr
cos
θ)
1
/
2
0