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• test  -- Posted on 4 years ago : author Jonathan
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Journal Club for March 22

Random Ruminations

So setting up the initial solution can be accomplished by the following:

First start with the assumption that the retarted time is the current time

t_r=t

DO

Calculate the displacement vector from the primary at the retarded time

\vec{d} = \vec{x}-\vec{X}_p(t_r)

Calculate the wind normal so that

(v_w \hat{n} + \vec{V}_p(t_r)) \times \vec{d} = 0

\hat{n} \times \vec{d} = \frac{-1}{v_w}\vec{V}_p(t_r) \times \vec{d}

It is possible to solve \hat{a} \times \vec{b} = \vec{c} for the unit vector \hat{a}

\hat{a}=\frac{\vec{b} \times \vec{c} + \sqrt{\vec{b} \cdot \vec{b} - \frac{(\vec{b} \times \vec{c}) \cdot (\vec{b} \times \vec{c})}{\vec{b} \cdot \vec{b}}} \vec{b}}{\vec{b} \cdot \vec{b}}

Once we have an estimate for \hat{n} we can improve the estimate by modifying the trajectory to account for the gravity from the secondary as follows:

Calculate the wind velocity from the primary

\vec{v}(t_r, x_p(t_r))=|v_w \hat{n}+\vec{V}_p(t_r)|

Now solve for the trajectory from the primary that leaves at \vec{X_p}(t_r) at velocity \vec{v}(t_r, x_p(t_r)) taking into account the force from the secondary.

Determine the path's distance of closest approach to \vec{x} and call that \vec{x}'(t')

Estimate the change in initial velocity needed \vec{v'}(t_r,x_p(t_r) = \vec{v}(t_r,x_p(t_r))+\frac{x-x'(t')}{t'-t_r}

And then solve for the unit vector that gives that direction

\hat{n} \times \vec{v}' = \frac{-1}{v_w}\vec{V}_p(t_r) \times \vec{v}'

Cycle until x' \approx x

Update the retarded time using the new distance and wind speed

t_r=t-(t'-t_r)

END DO

The only problem occurs when there are multiple solutions for the retarded time…

This will occur once we reach distances of order v_w/\Omega

If we switch to a rotating frame that rotates counter to the orbit so the angular speed is \Omega, then

\vec{v}_r=\vec{v}-\Omega \times \vec{x}

and

\alpha=\alpha_0+\Omega t

so that

\Omega t_r+\alpha = \alpha_0 - \Omega \frac{d}{v_w}