Physics 321
Hour 23
CM Motion and Orbits
Bottom Line
• We use relative and cm coordinates in the
Lagrangian. The cm coordinates are ignorable.
• We find an equation for 𝑟(𝜑)
𝑐
𝑟 𝜑 =
1 + 𝜀 cos 𝜑
• 𝜀 is the eccentricity and 𝑐 is chosen to match
boundary conditions
𝜀
Conic Section
𝜀=0
Circle
𝜀<1
Ellipse
𝜀=1
Parabola
𝜀>1
Hyperbola
Two Bodies → One Body
• The Lagrangian becomes
1
1
𝛾
ℒ = 𝑚1 𝑟1 ∙ 𝑟1 + 𝑚2 𝑟2 ∙ 𝑟2 +
2
2
|𝑟2 − 𝑟1 |
• Use the transformation to simplify:
1 −𝑚2 /𝑀 𝑅
𝑟1
=
1 𝑚1 /𝑀 𝑟
𝑟2
1
𝑚2 2 1
𝑚1 2 𝛾
ℒ = 𝑚1 𝑅 −
𝑟 + 𝑚2 𝑅 +
𝑟 +
2
𝑀
2
𝑀
𝑟
1
𝑚1 𝑚2
𝛾
ℒ = 𝑀𝑅 ∙ 𝑅 +
𝑟∙𝑟+
2
2𝑀
𝑟
• Let 𝑚1 𝑚2 /𝑀 ≡ 𝜇
Two Bodies → One Body
• The Lagrangian
1
𝑚1 𝑚2
𝛾
ℒ = 𝑀𝑅 ∙ 𝑅 +
𝑟∙𝑟+
2
2𝑀
𝑟
• Letting the orbit be in the x-y plane:
1
1
𝛾
2
2
2
2
2
ℒ = 𝑀(𝑋 +𝑌 + 𝑍 ) + 𝜇(𝑥 +𝑦 ) +
2
2
𝑟
• So
𝑀𝑋 = 𝑀𝑌 = 𝑀𝑍 = 0
The cm moves at a constant velocity
Spherical Coordinates
• We know 𝑥 = 𝑟 cos 𝜑 and 𝑦 = 𝑟 sin 𝜑
𝑥 = 𝑟 cos 𝜑 + 𝑟 sin 𝜑𝜑
𝑦 = 𝑟 sin 𝜑 − 𝑟 cos 𝜑𝜑
1
1
𝛾
2
2
2
2
2 2
ℒ = 𝑀(𝑋 +𝑌 + 𝑍 ) + 𝜇(𝑟 +𝑟 𝜑 ) +
2
2
𝑟
• The other equations of motion are:
𝛾
2
𝜇𝑟 = 𝜇𝑟𝜑 − 2
𝑟
𝑑
(𝜇𝑟 2 𝜑) = 0
𝑑𝑡
Equations of the Orbit
𝛾
𝜇𝑟 = 𝜇𝑟𝜑 − 2
𝑟
2
𝑑
𝜇𝑟 2 𝜑 = 0 → ℓ ≡ 𝜇𝑟 2 𝜑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑𝑡
ℓ
→𝜑= 2
𝜇𝑟
𝛾
ℓ2
𝜇𝑟 = − 2 + 3
𝑟
𝜇𝑟
• This is like a one-dimensional problem for a body
with the reduced mass moving in a real potential
and a centrifugal potential
Equations of the Orbit
• We want 𝑟(𝜑) and then 𝑟 𝑡 , 𝜑(𝑡).
• First, we use a trick: Let u=1/r
2
ℓ𝑢
ℓ = 𝜇𝑟 2 𝜑 → 𝜑 =
𝜇
𝑑𝑟
ℓ𝑢2 𝑑 1
ℓ𝑢2 𝑢′
ℓ𝑢′
𝑟=
𝜑=
=−
=−
2
𝑑𝜑
𝜇 𝑑𝜑 𝑢
𝜇 𝑢
𝜇
𝑑𝑟
ℓ𝑢2
ℓ ′′
ℓ2 2
𝑟=
𝜑=
− 𝑢 = − 2 𝑢 𝑢′′
𝑑𝜑
𝜇
𝜇
𝜇
Equations of the Orbit
ℓ2 2 ′′
𝛾
ℓ2
𝑟 = − 2 𝑢 𝑢 𝜇𝑟 = − 2 + 3
𝜇
𝑟
𝜇𝑟
2
ℓ2 2 ′′
ℓ
− 𝑢 𝑢 = −𝛾𝑢2 + 𝑢3
𝜇
𝜇
𝜇
′′
𝑢 = −𝑢 + 𝛾 2
ℓ
The Solution
𝜇
𝑢 = −𝑢 + 𝛾 2
ℓ
𝜇
𝑢 𝜑 = 𝐴 cos 𝜑 + 𝐵 sin 𝜑 + 𝛾 2
ℓ
Let 𝑢′ 0 = 0, then
1 + 𝜀 cos 𝜑
𝑢 𝜑 =
𝑐
or
𝑐
𝑟 𝜑 =
1 + 𝜀 cos 𝜑
′′
Example
orbit_cartesian_equations.nb
orbital_parameters.nb
orbital_equation_time.nb
Kepler’s 3rd Law
The square of the period is proportional
to the cube of the semi-major axis.