## Sunday, May 16, 2010

This is the schematic version, if you just wanted hints.  The full solution is given in Solar Sails Addendum I.

Here we present a schematic solution of the differential equation:

$\frac{dr}{dt} = \left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r} \right)\right]^{1/2}$

This is just a separable equation, so we rearrange to get an integral equation:

$\int_{r_0}^{r_f} \frac{dr}{\left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r} \right)\right]^{1/2}} = \int_{0}^{t}dt$

From here it's nice to non-dimensionalize, so our integration variable is just a number (with no units attached).  This allows us to get the integral into a form like

$k\int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} = t$

for appropriate values of k and u. Now we are ready to get started!  Typically, when I see something with a square root in the denominator that's giving me trouble, I just blindly try trig substitutions.  After an appropriate trig substitution, we get something of the form

$\int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} = \int_{x_0}^{x_f} -2\csc^3x dx$,

So now how do we solve this "easier" problem?  As a wise man once said, "When in doubt, integrate by parts."  So let's try that.

$\left[ \mbox{HINT:} -\int_{x_0}^{x_f} \csc^3x dx = \int_{x_0}^{x_f}\csc x \left(-\csc^2x \right)dx \right]$

Remembering that integration by parts goes like

$\int u dv = uv - \int v du$

we can pick appropriate values of u and v to get something nice, which eventually leads to

$-\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |_{x_0}^{x_f} - \int_{x_0}^{x_f} \csc x dx + \int_{x_0}^{x_f} \csc^3 x dx$

But this is just what we want!  Rearranging we now have that

$-2\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |_{x_0}^{x_f} - \int_{x_0}^{x_f} \csc x dx$

Evaluating our integrals, we see that

$-2\int_{x_0}^{x_f}\csc^3x dx = \sqrt{u}\sqrt{u-1} + \ln{ | \sqrt{u} + \sqrt{u-1}|}$

And this is what we have sought from the beginning.  Plugging back in to our earlier equations and rearranging gives

$t = \left(\frac{m{r_0}^3}{2\alpha} \right)^{1/2}\left[\sqrt{u(u-1)} + \ln{\left( \sqrt{u} + \sqrt{u-1}\right)} \right]$

where u = r / r_0 is our non-dimensional distance measurement.  And this is (up to some algebra) exactly what we get in the initial post.