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I am trying to calculate the terminal velocity of a solar sail starting at distance 1AU. I know that my sail will accelerate and its acceleration will be inversely proportional to the distance squared. this means that as it moves away from the sun, there are technically fewer photons hitting the surface area and therefore the acceleration of the sail will decrease.

how is it possible to calculate terminal velocity?

I initially thought of tackling the topic through calculating the change in acceleration and I was hoping that would be constant but I strangely find that the jerk (change in acceleration or the third derivative of displacement) is also changing with time. this does not allow me to derive a proper function of velocity with time, initial acceleration, jerk and ...

I have a function of the force exerted on the sail changing with distance, as well as the acceleration changing with distance but I don't know how to get to the velocity of the sail. I know that probably calculus is needed.

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This is a pretty straightforward calculation, at least up to some level of detail. Both the gravity and the light pressure on the sail drop in an inverse square relationship to distance from the sun, so the net force and the acceleration do the same. So the equation we have to solve is $$\frac{d^2r}{dt^2} = k/r^2$$ where we could take $r$ in AU, and $k$ the initial acceleration in $AU/s^2$. The boundary conditions are $r = 1$ at $t=0$ and $\frac{dr}{dt} = 0$ at $t = 0$, and, we can assume $k > 0$ or the sail isn't much use.

We can solve this pretty easily if we take $v = \frac{dr}{dt}$ then $\frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v\frac{dv}{dr}$. So our equation becomes $$v\frac{dv}{dr} = k/r^2$$ or $$v\, dv = \frac{k\, dr}{r^2}$$ Integrating $$v^2 = \frac{-2k}{r} + c$$

Now using $v = 0$ at $r= 1$ we get $c = 2k$ and $$v = \sqrt{2k\left(1-1/r\right)}$$ which provides the answer to the question -- the limiting velocity at large $r$ is $\sqrt{2k}$.

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  • $\begingroup$ +1 Does that differential equation have an analytical solution? In this answer I cheated and just pasted what seems to be one, without solving it myself analytically. If you can figure out how that's done you can add it here and again there as a new answer. It's been too many decades since I took DiffEQs. $\endgroup$ – uhoh Dec 2 '19 at 8:42

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