Environmental Engineering Reference
In-Depth Information
Example 3.2.1
Obtain an expression for the potential energy of a particle in the
gravitational field of a planet.
4
Solution 3.2.1
The force is given by
GMm
r
2
F
G
=−
r
,
where M is the mass of the planet, m is the mass of the particle, r is the distance and
r
is a unit vector directed from the centre of the planet to the particle. To calculate
the potential energy we will determine the work done in bringing the particle in
from a point at infinity along a path anti-parallel to the unit vector
r
. Starting the
particle out at infinity is our choice, any other point would do equally well. In this
case
d
r
=
d
r
r
and
r
B
r
B
GMm
r
2
GMm
r
2
W
∞
B
=
−
r
·
r
d
r
=−
d
r.
∞
∞
This integral is evaluated to give
GMm
r
B
=
W
∞
B
.
Thus the work done in bringing the particle from an infinite separation to a distance
r
B
from the centre of the planet is a positive quantity. This is what we expect, since
together with the Work-Energy Theorem this implies an increase in kinetic energy,
consistent with the effect of an attractive force. To obtain the potential energy we
need to choose where it should be zero. The natural choice is to pick the potential
to be zero at infinity, in which case
GMm
r
U(r)
=−
.
(3.18)
Example 3.2.2
Calculate the escape speed of a projectile launched from the surface
of the Earth.
Solution 3.2.2
If we ignore the effects of air-resistance, which is a dissipative force,
then the mechanical energy is conserved. We want to consider a projectile launched
with an initial velocity v
0
from the Earth's surface (r
=
r
E
) that is able to reach a
separation r
. Since this is, by definition, the separation at which the potential
energy is zero in Eq. (3.18) we get
→∞
1
2
mv
0
−
GMm
r
E
=
0
since the final kinetic will be identically zero if the projectile has just enough energy
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The proof that the gravitational force is conservative can be found in Section 9.1.
