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
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