Orbit Equation Derivation PDF

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This PDF was originally a PPT, so please excuse if there is any data missing. This covers the complete derivation of the Orbit Equation, one of the most important equations used in Space Systems, through the application of Newton's Laws. It takes a derivation that is incredibly complex, and hopefull...

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The Derivation of the Orbit Equation

0 The Total Orbital energy can be calculated through simple algebra E = Kinetic Energy + Potential 𝑮𝑴𝒎 𝑮𝑴𝒎 𝑮𝑴𝒎 We can also calculate the E= =𝟐𝒓 𝒓 𝟐𝒓 radial acceleration, thus, the resultant motion of the two bodies.

𝑟ሷ = 𝑟1ሷ - 𝑟2ሷ 𝐺𝑀 𝑚 𝐺𝑀 𝑚 𝑟ሷ = - 1 𝟐2 𝑟Ƹ - 1 𝟐2 𝑟Ƹ 𝑀 𝒓 𝑚 𝒓 1

𝐹1 = 𝑀1 𝑟1ሷ = -

𝐺𝑀1 𝑚2 𝒓𝟐

𝐹1 = 𝑚2 𝑟2ሷ =

𝐺𝑀1 𝑚2 𝒓𝟐

𝑟Ƹ 𝑟Ƹ

Where 𝑟Ƹ is the unit vector pointing from 𝑚2 to 𝑀1 .

𝑟ሷ = -

𝐺(𝑀1 + 𝑚2 ) 𝒓𝟐

2

𝑟Ƹ =-

μ

𝒓𝟐

𝑟Ƹ

Consider the Rate of Change of Angular Momentum 𝐿ሶ 𝑑 r x m𝑟ሶ 𝑑𝑡 Type equation here.

𝐿ሶ =

L (Angular Momentum) = r (position vector) x P (Linear Momentum) L = r x mv = r x m𝑟ሶ This is possible due to the assumption that r is large and m is small.

= (𝑟ሶ x m𝑟)ሶ + (𝑟 𝑥 𝑚𝑟)ሷ

(𝑟ሶ x m𝑟)ሶ = 0 as these vector are parallel, leaving us with 𝐿ሶ = (𝑟 𝑥 𝑚𝑟)ሷ = r x F (Newton’s second law) r and F are also parallel, therefore 𝐿ሶ = 0. This means that the angular momentum, L, must be constant and perpendicular to r and v. This, by definition, describes the motion of the body as being in one plane.

Specific Angular Momentum, h =

𝐿 m

=

r x mv m

h = r x 𝑣θ where 𝑣θ is the tangential component of velocity. 𝑑θ From circular motion 𝑣θ = rw = r( ) 𝑑𝑡 𝑑θ h = r (r ) = 𝑟 2 θሶ 𝑑𝑡

dොr

dt

෡ dθ dt

=

෡ θ d r dො dθ dt =

θ dθ෡ d෡

= dθ

dt

෠ θሶ θ

We know that r = r 𝑟Ƹ Differentiate to find the radial velocity in polar coordinates.

rሶ = = −ොr θሶ

dොr r dt

+ rෝ rሶ = θሶ θ෠ r + rො rሶ

Differentiate again to find radial acceleration. ෡ dθ ሶ θ dt

dොr ෠ ሷ ෠ ሶ r + θ θ r + θ θ r]ሶ + [ rሶ + rො r]ሷ rሷ = [ dt ෠ θሷ r + θ෠ θሶ r]ሶ + [θ෠ θሶ rሶ + rො r]ሷ rሷ = [−ොr r θሶ 2 + θ

෠ Group the quantities acting in the same direction (ොr and θ)

rሷ = rො [rሷ - r θሶ 2 ] + θ෠ [θሷ r + 2 θሶ r]ሶ

μ Resolving radially rሷ = rሷ - r θሶ 2 = - 𝒓𝟐

Substitute r = 𝑑r 𝑑𝑡

=

1

to find a closed-form solution.

𝑢 𝑑r 𝑑𝑢 1 𝑑𝑢 = - 𝑢2 𝑑𝑡 𝑑𝑢 𝑑𝑡

dr 1 du 𝑑θ = 𝑢2 dθ 𝑑𝑡 dt dr du rሶ = = - h dt dθ 2 𝑑 𝑟 d du rሷ = 2 = - h dt dθ 𝑑𝑡

However, we want r in terms of h and θ. 𝑑θ Remember that h = 𝑟 2 θ.ሶ Rearrange so that 𝑑𝑡 = h𝑢2

With some substitution we get rሷ = -

2 2 2𝑑 𝑢 𝑢 ℎ 𝑑θ2

μ

2 2 2 𝑑 𝑢, substitute rሷ = - 𝑢 ℎ 𝑑θ2 𝒓𝟐 μ 𝑑2 𝑢 we find that 𝑑θ2 + 𝑢 = ℎ2

Returning to rሷ - r θሶ 2 = -

1 θሶ = h𝑢2 and r =

Cancelling like terms Note that this form is not dissimilar from thatof simple harmonic motion where the solution is x(t) = A cos (ωt – φ) The solution to our equation takes the form Where A is a constant to be found.

μ u(θ)= ℎ2

+ A cos (θ-θ0 )

𝑢

𝑣2

μ

𝑣2

𝑣2

μ2

μ

- = 2 - μu = 2 - 2 - μA cos (θ-θ0 ) from substituting in u= ℎ2 + Acos (θ-θ0 ) 2 𝑟 ℎ 2 Next we have to find 𝑣 dr 𝑑u 𝑑θ 𝑣 2 = vθ 2 + vr 2 = ( )2 + ( )2 = ℎ2 [( )2 + 𝑢2 ] dθ dt dt 1 When substituting our values for θ,ሶ rሶ and r = Specific Energy, ε =

𝑑u dθ 𝑢2

= - A sin(θ-θ0 ) =

μ2

ℎ

4 +

2μ ℎ2

𝑑u so ( )2 dθ

=

A2

sin2 (θ-θ

𝑢

0)

and

A cos(θ-θ0 ) + A2 𝑐𝑜𝑠 2 (θ-θ0 ) 𝑑u

2 and ( Substituting 𝑢 )2 back into the equation for 𝑣 2 and cancelling terms we find that 2 dθ μ 𝑣 2 = 4 + 2μ A cos(θ-θ0 ) + A2 ℎ2 ℎ

Substituting our equation for 𝑣 2 into the specific energy equation, ε we will return 𝐴2 ℎ 2 𝜇2 ε= − 2ℎ2 . 2 h2 𝜇 𝐴 = ℎ2 1 + 2ε 𝜇2

Eccentricity can also be expressed as e = 1 + 2ε

So A = e

𝜇 ℎ2

ℎ2 𝜇2

. Substituting this into the specific energy equation returns ε = 2

ε = (𝑒 -1) 𝜇

𝜇2 2ℎ 2

𝜇

e2 𝜇 2 2ℎ 2

-

𝜇2 2ℎ 2

Substitute the constant A into u(𝜃)= ℎ2 + A cos (𝜃-𝜃0 ) gives u = ℎ2 [1 + e cos (𝜃-𝜃0 )]

ℎ2 ൗ μ

r = [1 + e cos (θ−θ )] 0 We can substitute p for

ℎ2 μ

Taking θ to be 0 and π, we know that for an ellipse 𝑝 𝑝 2𝑝 2a = 𝑟θ=0 + 𝑟θ=π = + = 1+𝑒 1−𝑒 1−e2 Thus p = a(1 − e2 ) So for an ellipse … a(𝟏−𝒆𝟐) r= [1 + e cos (𝛉−𝜽𝟎)]...