Environmental Engineering Reference
In-Depth Information
The focus is a distance
from the centre, where
=
a
−
r(
0
)
α
=
−
a
1
+
ε
=
a
−
(
1
−
ε)a
=
εa.
(9.61)
Thus the energy variable
ε
also tells us how squashed the ellipse is. For example,
a
corresponds to an ellipse with
one focus at a point on the orbit and an infinite semi-major axis, i.e. it is an ellipse
which closes at infinity. This special type of ellipse is more commonly known as a
parabola. For obvious reasons, the parameter
ε
=
0 corresponds to a circular orbit whilst
=
/a
is called the 'eccentricity'
of the ellipse. From a geometrical point of view it makes more sense to describe
an elliptical orbit in terms of the eccentricity and the length of the semi-major axis
and we might like to think of these as the two independent variables which specify
the orbit (rather than the energy and angular momentum).
The semi-minor axis can also be computed upon realising that
=
r
=−
cos
θ
(9.62)
at the point on the orbit which lies a distance
b
from the centre of the ellipse. We
can substitute this value of cos
θ
into Eq. (9.59) in order to determine the value of
r
at this point and then use Pythagoras' Theorem to determine
b
,i.e.
α
r
=
ε/r
.
(9.63)
1
−
Substituting for
and
α
in terms of
ε
and
a
and re-arranging gives
r
=
a
and
Pythagoras' Theorem then gives
b
2
a
2
2
,
=
−
ε
2
)
1
/
2
.
=
−
∴
b
a(
1
(9.64)
=
ηm
. Let us express it in terms of the
parameters that define the geometry of the ellipse,
a
and
ε
. Using Eq. (9.57) we
can write
The total energy of the orbit is
E
ε
2
)
G M
2
α
η
=−
(
1
−
(9.65)
Mµ
ε
2
)
and
but
α
=
a(
1
−
=
Mm
hence
GMm
2
a
E
=−
.
(9.66)
This is a somewhat surprising result for it tells us that the total energy in an
elliptical orbit depends only upon the length of the semi-major axis and not upon
