Page 249, 6 · · 2018-03-09Page 249, 6.13 A particle moves in a spiral orbit given by r = aθ....
11
Chapter 6 – Lecture VI Problem Discussions Page 249, 6.13 A particle moves in a spiral orbit given by . θ = a r If θ increases linearly with t, is the force a central field? If not, determine how θ must vary with t for a central force. Given θ = a r and . wt = θ So we have for the angular momentum θ = & 2 mr L 2 3 2 t w ma L = Since this says ( t L then angular momentum is not constant it is not a central force. To be a central force, we must have constant 2 = = θ l r & l dt d a = θ θ 2 2 ∫ ∫ θ θ = θ θ 0 0 2 2 t dt a l d ( t a l 2 3 0 3 3 1 = θ - θ 3 0 2 3 3 θ = θ t a l
Transcript of Page 249, 6 · · 2018-03-09Page 249, 6.13 A particle moves in a spiral orbit given by r = aθ....
Chapter 6 – Lecture VI Problem Discussions
Page 249, 6.13 A particle moves in a spiral orbit given by .θ= ar If θ increases linearly with t, is the force a central field? If not, determine how θ must vary with t for a central force. Given θ= ar and .wt=θ So we have for the angular momentum
θ= &2mrL
232 twmaL = Since this says ( )tL then angular momentum is not constant ∴ it is not a central force. To be a central force, we must have
constant2 ==θ lr &
ldtd
a =θ
θ22
∫ ∫θ
θ=θθ
0 022 t
dtal
d
( ) tal2
30
33
1 =θ−θ
302
3 3 θ+=θ tal
Page 2 of 11 10/21/2002
Page 249, 6.18 A particle moves in an elliptic orbit in an inverse-square force field. Prove that the product of the minimum and maximum speeds is equal to
( )22 τπ a , where a is the semi-major axis and τ is the periodic time.
( )θε+
ε−=cos1
1 2ar ϕ= sinrvl
0atmax =θ=V
π=θ= @minV
at 0maxmin
0 0atrland90,and0 V==ϕπ=θ
and 0minmax 180@ =θ= Vrl
The area of an ellipse is
( ) 221
22 1;1Area εεπ −=−= aba
maxminminmax2 VrVrl ⋅=
also: constant22 ==θ=timedaread
rl &
Page 3 of 11 10/21/2002
( )τ
επ 21
22 12
−=
al
where τ is period Putting these two expressions for l together, we have:
( )maxminminmax
22
122 12 VVrra =
−τ
επ
( ) ( ) maxmin11 VVaa εε −+=
( ) ( ) maxmin222
22112 VVEaEa −=−
τ
π
2
minmax2
=∴
τπa
VV
Page 4 of 11 10/21/2002
Page 249, 6.19 At a certain point in its elliptical orbit about the Sun, a planet receives a small tangential impulse so that its velocity changes from vto +δv . Find the resultant small changes in a, the semi-major axis. We know the equations for the orbit
( )θ+
ε−=cos1
1 2
Ear 1
and
Kml
KE 2
2 21 ⋅+=ε 2
Where E=Kinetic + Potential = T+V and φsinrvl =
so φ222 sin2
Trm
l = .
The problem asks for ( )dvda = . In other words, for a particle described at θandr , the velocity
changes. This changes E and Lr
, but not θorr . One could take 1, keep r and θ constant and
compute εd
da, then
dvdε
to arrive at dvda
. However,
when one starts to do this it gets very messy. A better approach is to look at equation 6.10.10. It gives the total energy in terms of “a”.
Page 5 of 11 10/21/2002
aK
E2
−= 3
For our case
aK
VTE2
−=+=
but for the planets 0≈ε and the potential energy is
about aK
− , so we can say
aK
aK
TVTETotal 2−=
−+=+=
or a
KT
2= .
Thus we have
aK
mv22
1 2 = 4
or daaK
mvdv 22−= 5
Divide 5 by 4
Page 6 of 11 10/21/2002
ada
Vdv
−=2
This is the answer we seek.
vdv
ada
2−=
A small fractional change in velocity produces twice as much fractional change in a.
Page 7 of 11 10/21/2002
Page 251, 6.30 According to the special theory of relativity, a particle moving in a central field with potential energy ( )rV will describe the same orbit that a particle with a potential energy
( ) ( )[ ]2
0
2
2 cmrVE
rV−
−
would describe according to nonrelativistic mechanics. Here E is the total energy, 0m is the rest mass of the particle, and c is the speed of light. From this, find the apsidal angle for motion in an
inverse-square force field, ( ) .rkrV −=
Equation 6.14.3 gives us the equation we need to compute the apsidal angle. To use it, we need
.anddrdf
f
( )
+
−−−=−= 20
2
2 cmrk
E
rk
drd
drdV
rf
( )2
20
2 21
++−=
rk
Edrd
cmrk
rf
Page 8 of 11 10/21/2002
−
++−= 22
02 2
21
rk
rk
Ecmr
k
−
−+−= 3
2
220
21
rk
rkE
cmrk
( ) 320
2
20
21
1rcm
kcm
kErk
rf −
+−= 1
420
2
20
31
312rcm
kcm
kErk
drdf
+
+=
21
320
2
20
2
420
2
20
3
11
1312
3
−
+
+−
+
+
+=
acmk
cmE
ak
acmk
cmkE
ak
aπψ
After simplifying, this becomes:
( )2
1
20
1
++=
cmEakπψ
Page 9 of 11 10/21/2002
Page 251, 6.31 Use diagram on page 234, example 6.10.1 A comet is observed to have a speed v when it is a distance r from the Sun, and its direction of motion makes an angle φ with the radius vector from the Sun. Show that the major axis of the elliptical orbit of the comet makes an angle θ with the initial radius vector of the comet given by
φ−φ=θ − 2csc
2tancot 2
1
RV
where earRand == evvV are dimensionless ratios as defined in Example 6.10.1. Apply the result to the numerical values of Example 6.10.1. From equation 6.10.7
θε+=
cos1
2 kmlr 1
θ is measured from the line called 0r in the figure and goes clockwise around to “r”. In the picture, θ
is a Quadrant III angle, also φsinvrm
prl =
×=
rr
( )k
vmrr φ=θε+222 sincos1
Page 10 of 11 10/21/2002
ε
−
φ=θ 1
sincos
22
kmrv
2
Now GMmk = and 2eevaGM = . (This comes from
2
2
RmGM
Rmv se = .) with this 2 becomes
Rar
vva
rv
eee≡≡−
φ=θε ,vvlet;1
sincos e2
22
3 Then 3 becomes
1sincos 22 −φ=θε Rv 4
Then θ−=θ 2cos1sin 222
2 1sin1sin
ε
−φ−=θ
Rv
( )
−−=
22222 1sin
1φε
εRv 5
2ε is given on page 235, example 6.10.1
( )( )222 sin21 φε RvRv −+= . Using this in 5
Page 11 of 11 10/21/2002
( )( ) ( )
−φ−φ−+
ε=θ
222222
2 1sinsin211
sin RvRvRv
−+
−−+=
1sin2sin
sin2sin1122442
222222
2 φφ
φφ
ε RvvR
RvvRv
( )( )φφε
42422 sinsin
1−= vR
φφε
22422 cossin
1vR=
εφφ=θ cossinsin
2Rv 6
Combining 4 & 6, we have
φφ−φ
=θεθε
=θcossin
1sinsincos
cot 2
22
RvRv
or
φφ−
φφφ
=θcossin
1cossin
sincot 22
22
RvRvRv
φ
φ2sin
12tan 2Rv
−=
( )φ⋅−φ=θ∴ 2csc2
tancot 2Rv
