Imagine that I'm designing a space probe, that will initially be placed into an Earth-like orbit around the Sun.

My goal is to have the probe fly/fall into the Sun; it's allowed to take as much time as it needs to do that, as long as it gets there eventually.

Now the hard part: Since the brute-force rocketry-approach to doing this would take a lot of fuel, I'd like my probe to be able to accomplish its goal without using any kind of rockets (except the ones required to get it to its starting position, of course).

Is there a way to do this e.g. by using a solar sail and tacking against the solar wind? If so, how long might that take? (I know the Parker solar probe is using Venus' gravity to help it along, but presumably that technique requires an initial trajectory that will bring it near Venus, which my probe won't have)


1 Answer 1


If you're already in a solar orbit, then yes. You can use a sail at an angle and send the reflections prograde. The result is to reduce your orbital energy and you spiral in.

I recall it was a standard physics problem to find the angle that maximized the energy transfer (it's not 45 degrees).

Time of journey depends on the mass of your item and the size of your sail. It also depends on the solar flux, which will increase as you approach the Sun.

The wiki page on solar sails has a section on time to reach the inner planets from Earth with some possible craft. With reasonable payloads, it's a few years to Mercury. That's more than halfway to the Sun, and the power is much higher there. So probably less than an additional 25% of time to reach the closest approach.

Optimal sail angle

This is minor addition (because it's constant at all distances), but to add in the derivation:

The power you get from the sail is the total radiation incident on the sail times the component of the momentum change in the v-bar direction.

$$ P = I \Delta p_v$$ The inbound intensity is proportional to the cosine of the sail angle, while the v-bar component of the reflected light is proportional to the sine of double the angle.

$$ P = \cos(\theta) \sin(2\theta)$$ Maximum will be found at a root of the derivative $$\frac{dP}{d\theta} = 2\cos(\theta) \cos(2\theta) - \sin(\theta) \sin(2 \theta)$$ $$ 2\cos(\theta) \cos(2\theta) = \sin(\theta) \sin(2\theta)$$ Via Wolfram $$ \theta = 2 \pi - 2 \tan^{-1}\left( \sqrt {5 - 2\sqrt{6}} \right) = 35.26^\circ$$

  • 1
    $\begingroup$ "power is much higher there" But so is the energy difference. I haven't done the math - that could be another question. $\endgroup$ Aug 25, 2018 at 2:22
  • $\begingroup$ @KeithMcClary good idea! What is the functional form for r(t) for a solar-sail deorbit into the Sun? $\endgroup$
    – uhoh
    Aug 25, 2018 at 2:50
  • 2
    $\begingroup$ As a side note- you’d potentially be able to get to the Sun faster by increasing orbital energy at first then getting a Jupiter assist to kill most of your orbital speed and maybe a final assist on the way back in to target the Sun. This was what Parker Solar Probe’s original mission plan was! $\endgroup$
    – Jack
    Aug 25, 2018 at 7:23
  • $\begingroup$ @Jack, true for a chemical burn, but probably not here. The solar power available gets pretty weak out by Jupiter. It would take a really long time to get to Jupiter by spiraling out with solar only. The Wiki page suggests Mars and Mercury are reachable in similar times. $\endgroup$
    – BowlOfRed
    Aug 25, 2018 at 7:46
  • 1
    $\begingroup$ Why do you assume it must reach Jupiter by spiraling out? You want a high speed encounter, not a rendezvous. "Burn" prograde while near the sun to raise your apoapsis, "burn" retrograde while far out to lower your periapsis. Most of your acceleration occurs much closer in than Jupiter. $\endgroup$ Aug 26, 2018 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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