Begin2.DVI

Note that the Newton law of gravitation implies that the derivative given by equation

(9.107) is zero. That is, if

md

(^2)
r
dt^2
=md
v
dt
=F
=−GmM
r^2
ˆer

then
r ×

d^2 
r

dt^2

=

d

dt

(


r ×

d
r

dt

)

=−

GM

r^2

r ׈er=
 0. (9 .110)

An integration of this equation produces the result

r ×d
r

dt

=
h=Constant (9 .111)

Recall that the vector H
=
r ×m
v is defined as the angular momentum. The quantity

h=m^1 H
=
r ×drdt appearing in equation (9.111) is called the angular momentum per

unit mass. Equation (9.111) tells us that the angular momentum is a constant for

the two-body system under consideration. Since
h is a constant vector, it can be

verified that

d

dt

(


v ×
h

)

=d
v

dt

×
h=−GM

r^2

ˆer×

(


r ×d
r

dt

)

=−

GM

r^2 ˆer×

[

rˆer×

(

r

dˆer

dt +

dr

dt ˆer

)]

=−GM ˆer×

(

ˆer×dˆer

dt

)

=GM

dˆer

dt.

(9 .112)

Note that the result (9.112) was obtained by making use of the equations (9.106)

and (9.109). An integration of the result (9.112) gives us the relation

v ×
h=GM ˆer+C ,
 (9 .113)

where C
is a constant vector of integration. Using the triple scalar product formula

it is readily verified that

r ·

(


v ×
h

)

=
h·

(


r ×

d
r

dt

)

=h^2 =
r ·

(


v ×
h

)

=GM
r ·ˆer+
r ·C

or

h^2 =GMr +Cr cos θ, (9 .114)

where θis the angle between the vectors C
and
r. In the equation (9.114) one can

solve for rand find

r=

p

1 + cos θ, (9 .115)