In my story interstellar travel is common, but nothing's perfect, and a mining ship returning from a deep-space run (you'd be surprised what's out there) just discovered that something is very, very wrong.

\*crackle\* Mayday... Mayday... this is the Tycho Brahe... 
Primary engine offline.  Asteroid impact in the Oort cloud.
Request assistance... over.

\*squeal*\ Tycho Brahe... this is Sol reference Alpha.  What is
your vector to the initial... over.

Sol Alpha, Tycho Brahe, solar declination +23°, right ascension -87°, 
Delta-forty-alpha, Victor-five-charlie.


Sol Alpha, Tycho Brahe, are you still there?

Tycho Brahe, Sol Alpha, negative on assistance.  Repeat,
negative on assistance.  Recommend deploying solar sails
and maintain vector to the initial... over.

Question: Given 40 Km2 solar sails (and ignoring necessary support structure), is it possible for a ship with a mass of 5,000,000 Kg traveling at 0.05c at a distance of 40AU from the sun to deploy those sails and, using only the solar wind, decelerate to 250,000 Kph before crossing the "orbital sphere" of Mercury?

Solar Declination & AscensionThis doesn't actually have anything to do with the question, but just for fun and off the top of my head (please let me know if I've plagarized a published story!), I defined the reference for solar declination and solar ascension as measured from the line drawn from the center of the sun to the center of the galaxy and otherwise used in the same way Astronomers use declination and ascension. It eliminates the position of the earth from the equation (making the reference static and applicable by math to any body in the solar system ... or any solar system). Thus, "Vector to the initial" would always be your position and speed in relation to an approach toward the center of the sun.

If I haven't plagarized from somewhere... I thought of it first! 😀

Victor-five-charlieAlso for fun, an over-the-radio way of saying "my velocity (victor) is 5% or 0.05 of the speed of light (charlie)." The percentage is always assumed.

Delta-Forty-AlphaDitto, the distance from the sun along the indicated vector in "alpha" or AU.

Vector to the InitialYes, I'm not using this in the same way today's pilots do.

Orbital SphereOut of curiosity, do astronomers today have a phrase that identifies the sphere enclosing a radius from the sun equal to a planetary orbital distance?

  • 2
    $\begingroup$ This question doesn't appear to have anything to do with worldbuilding so much as physics. In other words, how does a solar sail help you construct a world or improve some property of it? Help Center $\endgroup$
    – elemtilas
    May 1, 2018 at 0:07
  • 1
    $\begingroup$ @elemtilas, that's an interesting question, and one I've been examining with my series of 0.05c questions. Referencing the "on-topic" help page... Given that a "world" can be " larger than a multiverse or smaller than a village." and my question is "How to achieve a specified effect in a defined world, including by the use of ... technology ..., while maintaining in-universe consistency", how can my question be deemed off-topic? $\endgroup$
    – JBH
    May 1, 2018 at 0:11
  • 1
    $\begingroup$ @elemtilas, I'm changing my attitude about what's off-topic at this site. My new attitude is: so long as a question does not SPECIFICALLY violate any of the bullets in the "as long as they are not about:" section, then it is on-topic. $\endgroup$
    – JBH
    May 1, 2018 at 0:15
  • 2
    $\begingroup$ @elemtilas, Funny you should mention that. I brought the question up here. I have mixed feelings, but I'm coming to the conclusion "better here than nowhere." Those other sites can get a bit... well... uppity. $\endgroup$
    – JBH
    May 1, 2018 at 0:52
  • 1
    $\begingroup$ (cont.) How I interpret that "achieve a specified effect" line is more a matter of achieving or arriving at some aspect of the world itself rather than a specific event in a history or storyline. While I have always found your questions interesting, ones like this do kind of bug me because you're not really building the world or a culture or devising a fictional science or doing any kind of fictional world construction. Frankly I'd love to see questions like this one in the writing SE or in the physics SE. Here I like to see people trying to build and flesh out worlds & settings! $\endgroup$
    – elemtilas
    May 1, 2018 at 0:53

2 Answers 2


Edit: I did make a math error, the points below still stand though.

No. The solar wind pressure simply isn't strong enough to stop you.

Making larger sails would increase the force applied to your ship, but would well surpass known material shear strengths and would destroy itself.

Assume a constant acceleration for simplicities sake and we can use:
$DeltaX = 1/2*Acc*Time^2 + Vel_{initial}*time$
$Acc = (Vel_{final} - Vel_{initial})/time$

Rearranging those gives us $Acc = (V_f-V_i)*(v_f+v_i)/(2*deltaX)$

Using $V_f = 250,000 KpH = 67444.4 m/s$
$V_i = 0.05*c = 0.05*3*10^8 =$ 45,000,000 $15,000,000 m/s$
$DeltaX = 40 AU = 40*149,597,870,700 = 5,983,914,828,000 m$

We get an acceleration of -169 m/s^2, or about 17 g's. $18.8m/s^2$ or 1.9 g's

Humans can't really survive more than a couple g's for extended durations.

This example is assuming a constant acceleration, which is even worse for our real life use case.

The solar flux is going to fall off with the inverse square law over distance, so at 40 AU it is going to be ~1600 times weaker than it is at 1 AU.

Achieving 17 g's of average acceleration over our distance will require a gentle deceleration at the start of our journey and a massive amount of force near the end as the solar wind density increases. This will absolutely shatter any plausible sail support structure and kill your crew.

0.05 c is probably too fast for any inhabited ship to decelerate from in 40 AU. Uncomfortable but survivable.

If you can take it down by a factor of 10, to 0.005c, then the required average acceleration is -0.187m/s^2, which enters into the realm of plausibility for a large enough sail.
-0.187m/s^2 is still way outside the realm of plausibility for our situation.

Addendum: So the numbers above are a bit off, and our required acceleration would be survivable, so I decided to run even further to see what the necessary pressure from our solar wind would be.

$Force = mass*Acc$
$Pressure = Force/Area = mass*Acc / Area $ $Pressure = 5,000,000kg * 18.8m/s^2 / ( 40*1000^2 m^2) = 2.345 N/m^2$

The typical pressure from solar wind at 1 AU is $6*10^{-9} N/m^2$, our required average pressure is $2.345 N/m^2$ , almost 9 full orders of magnitude higher.

Keep in mind that the average solar wind pressure will be around $3.75*10^{-12} N/m^2$ at 40AU, thanks on the inverse square law, and our situation really looks bleak.

To achieve our necessary acceleration will require a surface area of:
$Force = Pressure*Area = Mass*Acc $
$Area = Mass*Acc / Pressure = 5,000,000kg * 18.8m/s^2 / 6*10^{-9} N/m^2$
$Area = 1.56 * 10^{16} m^2 = 1.56*10^{10} km^2 $

That's a square sail with side length 125,166 km, which would require MUCH more mass than we are accounting for.

Using 0.005 c we still need a sail that is a disk with a radius of 7054km, which is a bit bigger than the Earth's radius.

And remember thats assuming we have a constant flux equal to that at 1 AU. In reality its much worse. Solar wind is simply not dense enough to slow you down appreciably, especially when you factor that the sun's gravity will be accelerating you towards it, probably with more force than our sail can generate.

To slow down your craft, you will have to eject mass. Forcefully jettisoning your sail will probably net you more deltaV than trying to use it.

If you have a magnetic acceleration cannon of any type, start scrapping unnecessary (non-vital) modules of your ship and firing them at max power straight ahead. Venting atmosphere will help too. Good luck.

  • 2
    $\begingroup$ Good to see an answer that digs into the basic mechanics to find the deceleration. You may need to check your calculations. The initial velocity of 0.05 c is 15,000 km/s not 45,000 km/s. I did a quick calculation based on your equation, which may be wrong, and came up with a deceleration of 37.6 m/s^2 or 3.84 g. The deceleration time, assuming uniform deceleration, is 9.3 days. Hard to see a human crew surviving that. Plus one for a good answer. Better after your edit. $\endgroup$
    – a4android
    May 1, 2018 at 12:50
  • $\begingroup$ What if you combine it with orbital braking around several planets? If you brake around one and come out at the right angle to hit another, you could possibly hit several of them in succession, if the orbits put them in the right place and the ship can take the strain. $\endgroup$
    – AndyD273
    May 1, 2018 at 13:11
  • $\begingroup$ @AndyD273, that's a good idea save for one thing, per my question, my engines are dead. $\endgroup$
    – JBH
    May 1, 2018 at 14:22
  • $\begingroup$ @JBH But the sail can be used to alter course by changing the angle. At least to some extent. If you knew about the problem far enough out then you could start altering course far enough out to make a difference. It would take a lot of computer power to calculate the course, a lot of skill to fly it, and a lot of luck to not die. $\endgroup$
    – AndyD273
    May 1, 2018 at 14:29
  • $\begingroup$ @AndyD273, point well taken, but regrettably beyond the scope of the question. I'm only focusing on the ability of solar sales to act against solar wind. $\endgroup$
    – JBH
    May 1, 2018 at 14:35

Solar sail - unlikely. Here is an example of ship's project that would be able to reach 0.00264c(20 times less than you need). To increase delta-v 20 times you would need rather big technology leap.

Magnetic sail is better because it is mostly made from magnetic field, that is pretty much weightless. There are some projects that are able to decelerate from 5% c using existing technology.

  • 1
    $\begingroup$ Certainty magnetic sails would be better. The paper cited suggests taking about 29 to 40 years to decelerate a 8.25 metric ton vessel from 5% c (depending on the sail technology). This may not be practicable for the question's parameters. The paper is good & useful. Thanks for the citation. $\endgroup$
    – a4android
    May 1, 2018 at 12:59

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