Environmental Engineering Reference
In-Depth Information
the eccentricity of the orbit. For example, we can compute the total energy of an
elliptical orbit by computing instead the energy in a circular orbit whose radius
is equal to
a
.
Example 9.5.1
Prove that the velocity of a particle in an orbit described by
Eq. (9.59) is purely tangential at the pericentre and the apocentre of the orbit.
Solution 9.5.1
The pericentre of the orbit is the point of closest approach and the
apocentre is the point of farthest approach. From Figure 9.10 we can see that these
occur at θ
=
0
and θ
=
π respectively. The motion is tangential when
d
r/
d
t
=
0
.
Differentiating Eq. (9.59) gives
d
r
d
t
αε
d
θ
d
t
=
sin
θ
ε
cos
θ)
2
+
(
1
and this vanishes whenever
sin
θ
=
0
, i.e. when θ
=
0
or π .
Finally we shall derive Kepler's Third Law by computing the period of the orbit,
T
. We already know from Kepler's Second Law, Eq. (9.43), that area is swept out
at a constant rate equal to
L/(
2
µ)
. Integrating over one cycle (and remembering
that the area of an ellipse is equal to
πab
)wehave
L
2
µ
T.
πab
=
(9.67)
Re-arranging and substituting for
L/µ
and
b
gives
2
πa
2
λ
ε
2
)
1
/
2
,
T
=
(
1
−
−
1
/
2
2
πa
2
λ
2
ηλ
2
G
2
=
.
(9.68)
M
2
M/(
2
a)
to get
We now substitute for the total energy
η
=−
G
2
πa
3
/
2
G
T
=
.
(9.69)
M
a
3
/
2
was also discovered by Kepler and
it constitutes his Third Law. Kepler thought that the constant of proportionality
was the same for all planets but as we have derived it, it is not the same since
the mass
This result, namely that the period
T
∝
M
is not exactly equal to the mass of the Sun. Nevertheless,
M
M
is
a good approximation in practice, leading to deviations from a single constant of
proportionality of no more than 0.05% for all the planets.
Note that if time is measured in years and distances in astronomical units (AU)
then Eq. (9.69) becomes
3
=
T
2
a
3
.
=
(9.70)
3
Neglecting the reduced mass correction.
