There are several questions and answers here about spiraling towards or away from the Sun using solar sails tilted by roughly 45 degrees to convert the sunlight's incoming radial momentum into tangential thrust.

There's talk that these spirals may be close to logarithmic spirals depending on the specific assumptions.


  1. Are solar sail spirals logarithmic in nature or at least close to it?
  2. Can this be shown analytically?
  3. Can this be shown by dimensional analysis alone?

1 Answer 1


The answer has to be 'not necessarily', because, in general, as you go along, you're free to adjust the solar sail angle and thus the trajectory. In addition, the trajectory need not lie in a single plane since the sail can produce out of plane forces.

I posted an analysis in the comments yesterday for a weak solar sail and a shallow spiral orbit and concluded that the orbit was a logarithmic spiral. But I think the analysis extends no matter how steep the spiral.

A first point to note is that both the gravitational force and the solar wind force decay as $1/r^2$ and thus keep the same ratio irrespective of radial distance. If we assume the sail is set at a fixed angle with respect to the radial direction, then not only is the ratio of the two forces constant but their respective directions are also independent of radius. This already suggests a constant angle or logarithmic spiral. A logarithmic spiral is 'self-similar' and looks the same at any scale of radius, $r$, i.e. its features scale with $r$.

The only remaining question is whether the forces (i.e. acceleration, $a$) and velocity, $v$, change together in a commensurate manner as the radius changes. The quantity $v^2/a$ has units of distance and therefore must also scale as $r$. Since $a$ scales as $1/r^2$, it follows that $v$ scales as $1/\sqrt(r)$. The radius of curvature of the curve is given by the velocity-squared divided by the perpendicular component of acceleration. It follows that the radius of curvature is proportional to r, further confirming the logarithmic spiral.

Having concluded that exponential spiral orbits exist, we should note that they are defined by a single parameter. In that sense they are similar to circular orbits. They are a special case in that presumably you must start with exactly the right velocity at the right position in order to continue on a desired spiral. Arbitary starting conditions will not generally yield a logarithmic spiral orbit any more than they might yield an exactly circular orbit.

  • $\begingroup$ Yes this is interesting! So for a given central body and a given plane there are an infinite number of circular orbit solutions distinguished by radius (or velocity or energy) and some absolute phase with respect to $t=0$. I wonder if for the same given central body and plane and for a solar sail of fixed attitude wrt the origin there are an equally infinite number of logarithmic spirals distinguished by say true anomaly at 1 AU and the same absolute phase? (in other words, where does this rabbit hole go?) $\endgroup$
    – uhoh
    Dec 14, 2020 at 8:03
  • $\begingroup$ @uhoh I'm a bit puzzled now. For a clockwise circular orbit in a given plane, it's sufficient to specify it's 2D position at a given instant. For a clockwise logarithmic spiral it seems necessary to specify both the 2D position and curvature of the spiral. Maybe the equivalence is with the spiral that has the sail adjusted to lose orbital energy most rapidly. That would remove one degree of freedom. $\endgroup$
    – Roger Wood
    Dec 15, 2020 at 3:56
  • 1
    $\begingroup$ Hmm... perhaps "for the same given central body and plane and for a solar sail of fixed attitude wrt the origin" is incomplete as it lacks "and a given area/mass ratio"; would that do it? I'm responding quickly without thinking carefully. As area/mass ratio decreases the spiral gets denser and approaches a circle so to speak. $\endgroup$
    – uhoh
    Dec 15, 2020 at 5:23
  • 2
    $\begingroup$ @uhoh Yes, I missed that. The curvature (tightness of the spiral) will depend on the area/mass as well as which way the sail is pointing. $\endgroup$
    – Roger Wood
    Dec 15, 2020 at 6:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.