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Osculating Circles and Trajectories
Just Kidding
Osculating Circles and Trajectories
r
v
M
m
Mechanical Energy
r
GmMmvE 2
21
Circular Orbit
M
m
v
F
How do we get the mass, M, of the gravitating object?
Apply Newton’s Second Law to the mass, mM
C
r
Cmon
C maF
r
mv
r
GmM 2
2
G
rvM
2
General Orbits, Conic Sections
circle, =0
ellipse, 0<<1
parabola, =1
hyperbola, 1
r
In polar coordinates
cos1r
p
General Orbits, Conic Sections, polar unit vector notation
In polar coordinates
cos1r
p
r
r
In terms of unit vector notation we have:
rrr ˆ
Differentiating the position vector with respect to time we have the velocity:
dt
rdrr
dt
drrr
dt
dv
ˆˆˆ
v
rrrrv ˆˆ
or, rd ˆd
r
and,
ˆˆˆˆ
ˆ
dt
dr
dt
rdr
so,
ˆˆˆˆˆˆ vrvrrrrrrrv r
rvr ˆ ˆv
General Orbits, Conic Sections, polar unit vector notation
In polar coordinates
cos1r
p
r
r
Differentiating the velocity vector with respect to time we have the acceleration:
ˆˆˆˆ rrv
dt
dvrv
dt
da rr
a
ˆˆˆˆ vvrvrva rr
or,
or,
dt
dv
dt
dv
dt
rdvr
dt
dva r
r
ˆˆˆ
ˆ
or,
ˆˆˆˆ r
dt
rdrvraa rr
so,
ˆˆˆˆ rrrrrrra
General Orbits, Conic Sections, polar unit vector notation
In polar coordinates
cos1r
p
r
r
Recall that
a
rar ˆ ˆa
ˆˆˆˆ
ˆ
dt
dr
dt
rdr
rd ˆd
r
And note that
r
dt
rd
dt
dˆ
ˆˆˆˆ
ˆˆˆˆ rrrrrrra
Then we get for the acceleration,
rrrrrrra ˆˆˆˆ
Combining terms in unit vectors,
ˆ2ˆ2 rrrrra
General Orbits, Conic Sections, normal and tangential unit vector notation
In polar coordinates
In terms of unit vector notation we have:
Differentiating the velocity vector with respect to time we have the acceleration:
dt
tdvt
dt
dvtv
dt
da
ˆˆˆ
dt
tdvtaa t
ˆˆ
or,
and,
ndt
ndt
dt
tdt ˆ
ˆˆˆˆ
so,nvtatvtaa tt ˆˆˆˆ
dn
ds
Note that
dds
tdˆnd ˆ
n t
v
rn
tvtvv tˆˆ
t cos1
r
p
and
v
dt
d
dt
ds
so,
natanv
tanvtaa nttt ˆˆˆˆˆˆ2
C
General Orbits, Conic Sections
In polar coordinates
r
n t cos1
r
pnatan
vtanvtaa nttt ˆˆˆˆˆˆ
2
tana
a
We now have the acceleration in normal and tangential unit vector notation
Compare this with the acceleration in polar coordinate unit vector notation
ˆ2ˆ2 rrrrra
r
r
a
ra a
General Orbits, Conic Sections, continued
When = 0we have
1minr
p
v
ellipse, 0<<1
circle, =0
parabola, =1hyperbola, 1
rr ˆmin r
tn
General Orbits, Conic Sections, continued part 2
When = 0we have
1minr
p
v
rr ˆmin r
t
na
At the minimum distance from the massive body the acceleration is purely radial or normal depending upon your point of view.
Let’s write the acceleration at for each basis.
ˆ2ˆ2 rrrrra
becomes
rrra ˆ2min
and
natanv
tanvtaa nttt ˆˆˆˆˆˆmin
2
becomes
nanv
nvtaa nt ˆˆˆˆmin
2
General Orbits, Conic Sections, continued part 3
When = 0we have
1minr
p
v
rr ˆmin r
t
na
rrra ˆ2min
nv
a ˆmin
2
In the r,basis we have:
In the t,nbasis we have:
Because
rn ˆˆ
min
22
min v
rr
we get
The Osculating Circle at closest approach 1minr
p
v
rr ˆmin
aC
min
min
22
min v
rr
We have
and because
vr min
we get
min
2
min
2
v
r
vr
Let’s obtain r
Polar coordinates
cos1r
p
Let’s re-write this as
pr cos1
And take two successive derivatives with respect to time to obtain
0sincos1 rr
0cossinsinsincos1 2 rrrrr
Now let = 0 to get
01 2min rr
which now gives
min
22
min1r
vrr
or
1min
2
r
vr
The Osculating Circle at closest approach
1minr
p
v
rr ˆmin
aC
min
min
2
min
2
v
r
vr
We now know
1min
2
r
vr
so
min
2
min
2
min
2
1 v
r
v
r
v
and
1
11
minmin r
or
minmin 1 r
The Osculating Circle at closest approach continued
1minr
p
v
rr ˆmin
aC
min
minmin 1 r
Compare this with the polar form of a conic section
Clearly
minp
The semi-latus rectum for conic sections is nothing more than the radius of the osculating circle at the distance of closest approach.
Polar CoordinatesThe equation for a conic section could now be written as cos1min r
v
rr ˆmin
CaC
min
mM
Apply Newton’s second law to m
Cmon
C maF
min
2
min
2
2min 1 r
mvmv
r
GmM
Solving for M we get
Gvr
M
1
2min
orG
vrM
min
22min
General Orbits, Conic Sections
When = 0we have
1minr
p
v
ellipse, 0<<1
circle, =0
parabola, =1hyperbola, 1
rr ˆmin rn
1
min
min
r
cos1min r
Gvr
M
1
2min
M
Cc
1minmin r
CeCpCh
