Parabolic Motion and Hyperbolic Motion | PHYS 608, Study notes of Physics

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Lecture 12 Outline - In and Out Burger

The Kepler Problem (Section 3.8)

Hohmann Transfer

Parabolic motion

e = 1

Hyperbolic motion

e >

Intro to Scattering (Section 3.10)

Hohmann Transfer

Requires two velocity changes for planet transfer

which is the most energy efficient way to do so!

Parabolic Motion in time (for

V

r

k/r

Parabolic Motion in time (for

F

r

k/r

t

ℓ 3

mk

2 ∫ θ θ 0

dθ

(1+

e (^) cos(

θ − θ ))′ 2

eg. For Parabola’s

e = 1)

use

1 + cos

(^) θ

= 2 cos

2 (^) (

2 θ^ )

t

3

mk

2 ∫

θ

0

sec

4 (

2 θ^ )

dθ

3

mk

2 ∫

tan

2 θ (^) )

0

x 2 ) dx

3

mk

2 ( tan

2 θ^ )

tan

3 (

2 θ^ ))

Above used

x

= tan

(^2) θ^ )

so

dx

2 1 sec

2 (^) (

2 θ^ ) dθ^

and also

sec

2 = (1 + tan

2 )

So we have

π < θ < π

where

θ = − π → t =

r

where

θ

= 0

t

= 0(

r

= apsidal)

Intergalactic Planetary

Voyager 2 path (departing 1977)

this procedure is also known as gravity assist.

Central Force Scattering

Consider effective one-body scattering

where problem has rotational symmetry about axis

Define cross section

σ (Ω)

, of units of area, for scattering

in given direction as:

σ (Ω)

d Ω =

σ (Ω)

π (^) sin Θ

d Θ =

the # of particles scattered into solid angle (

d Ω ) per

unit time DIVIDED by the incident intensity

I

which is

of particles crossing area per unit time.

Solid Angle Element

d

The # of particles scattered into

d Ω = 2

π (^) sin Θ

d Θ

(lying between

and

d Θ

is equal to the # of particles with

s

to

s (^) +

ds

ie:

πIs

ds

| = 2

πσ

I

sin Θ

d Θ

|

Given that can consider

s (Θ

, E

, differential cross section

σ (Θ) =

s

sin Θ

d ds Θ

∣ ∣ ∣ ∣

eg. for repulsive scattering

Solid Angle Expression

Want to find an expression for

s ) , take

so we go back to the equation of orbital motion

θ ( r ) =

θ 0

r

r 0

dr

r 2 √

2 ℓ mE 2

2 ℓ mV 2

r 2 1

with

r 0 =

, θ

0

π

and for

θ ( r min

θ min

π

−

π

−

θ min

∞

r min

dr

r 2 √

2 ℓ mE 2

2 ℓ mV 2

r 2 1

rewrite using

s √

2 mE

AND

π

= Ψ + Ψ + Θ

s ) =

π

−

(^2)

∫

u min

0

s du

V (^) ( u )

E

(^) s 2 u 2