10
$\begingroup$

Is there a theoretical/experimental precedent for stellar photosphere aerocapture (if it is in fact, a thing)? Specifically, would using this method of deceleration be in some way more attractive than an aerobraking or magnetoshell aerocapture around a planet if one were traveling at relativistic speeds, say .1c or a bit less? As I understand it, wrongly or not, temperatures in the sun's outer layers are about 6000K as opposed to the upwards of 100,000K ablative temperature of a heat shield during aerobraking.

Plot synopsis: a ship traveling at sub-light speed is too low on fuel to decelerate in time to keep from blowing through and past its target solar system. Radical measures are considered.

Improve this question
$\endgroup$
2
  • $\begingroup$ there is no experimental precedent for relativistic spaceflight, so no, there won't be experimental precedent for anything for which such flight is a prerequisite... $\endgroup$
    – SF.
    Commented Nov 8, 2016 at 15:31
  • $\begingroup$ "Rapid" and "relativistic speeds" require a negation somewhere to be viable with anything like anything we can actually construct. If you want both "rapid" anything and "relativistic speeds", then you are well outside of mainstream science and engineering. Try Worldbuilding. $\endgroup$
    – user
    Commented Nov 8, 2016 at 15:51

4 Answers 4

Reset to default
6
$\begingroup$

As I understand it, wrongly or not, temperatures in the sun's outer layers are about 6000K as opposed to the upwards of 100,000K ablative temperature of a heat shield during aerobraking.

The relevant temperature is not the temperature of the star's atmosphere, it's the temperature to which everything is heated by friction.

If you compare 0.1c with the kinds of velocities we have for interplanetary probes (roughly on the order of magnitude of escape velocity from the solar system), it's about 100 times greater. When you increase the velocity by a factor of 100, you increase the kinetic energy by a factor of $10^4$.

A second huge factor working against you is that the time is very short. The distance over which your probe has to stop is at most on the order of the diameter of the star, and in reality it's going to be a lot less, since you don't have any way to curve your trajectory to match the curvature of the star. (If you had a way to curve your trajectory like that, it would require a large acceleration, which is what we're trying to accomplish in the first place.) Even taking the stopping distance optimistically to be the diameter of the sun, the time to stop comes out to about 20 seconds, which is extremely short.

So you have the combination of these two factors: a huge amount of heat being dissipated, and a short time to get rid of the heat. This is a huge amount of power, and convection simply isn't going to be a fast enough process to get rid of it. If you want to make the maneuver work, you're going to need some exotic and extremely efficient method of getting rid of the heat.

would using this method of deceleration be in some way more attractive than an aerobraking or magnetoshell aerocapture around a planet

Yes. The diameter of the sun is about 100 times greater than the diameter of the earth, or 10 times the diameter of a gas giant. That means that the time considerations get much worse if you try to brake using a planet's atmosphere rather than a star's. By the same logic, a supergiant star would probably be better than a main-sequence star like the sun.

Plot synopsis: a ship traveling at sub-light speed is too low on fuel to decelerate in time to keep from blowing through and past its target solar system. Radical measures are considered.

This makes it sound like you have in mind a crewed ship. To get from 0.1c to rest in a distance equal to the diameter of the sun, you would need a deceleration of 30,000 gees. That would kill the crew.

Improve this answer
$\endgroup$
6
  • $\begingroup$ My reading of OP's question is not that he wants to achieve 0 velocity, but merely that he wants to remain in the stellar system, i.e. achieve capture by the star. $\endgroup$
    – Jerard Puckett
    Commented Jul 22, 2016 at 14:35
  • 4
    $\begingroup$ @JerardPuckett: I don't think it matters that much. Escape velocity from the sun's surface is .002c. $\endgroup$
    – user687
    Commented Jul 22, 2016 at 15:21
  • $\begingroup$ That certainly puts things in perspective. $\endgroup$
    – Jerard Puckett
    Commented Jul 22, 2016 at 15:23
  • $\begingroup$ I don't think the density of the sun's plasma at this temperature would be enough to slow it down much at all. It wouldn't be necessary to slow to a dead stop, just to below escape velocity. Every star generates tremendous magnetic fields. Maybe the ship can slow by creating its own interacting magnetic field. $\endgroup$
    – Howard Miller
    Commented Jul 22, 2016 at 20:31
  • 1
    $\begingroup$ @Ben Crowell: So, for the moment, let's assume it's travelling much slower, say, .01c. For the sake of argument, we'll also suppose there is enough fuel from the fusion drive to make course corrections with respect to the (main sequence) star's curvature and plot a trajectory through its upper atmosphere where an ideal density of plasma exists to produce enough drag for a magnetoshell aerobraking maneuver (with a more powerful magnetic field than currently exists). Would the reduction of velocity allow this to be feasible, provided the crew were suitably protected vs. g-forces? $\endgroup$
    – Brian
    Commented Jul 23, 2016 at 16:06
3
$\begingroup$

At 0.1c any gas colliding with the craft's heat shield would probably undergo nuclear fusion... which would make heating effects far worse than the usual "friction" (compression) considerations. So I'd say any normal material would be unable to survive the process.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ At 0.1c, you're only over the coulomb barrier for fusion of hydrogen with relatively light elements (Z<~10). More importantly, the cross-section is small because a nucleus is a small target. The vast majority of hydrogen nuclei will stop in the heat shield without undergoing any nuclear interaction. $\endgroup$
    – user687
    Commented Jul 22, 2016 at 12:52
2
$\begingroup$

How?

  1. Consider the probe that Galileo dropped into Jupiter. While the objective wasn't aerocapture it serves as a useful comparison. The probe burned away 1/4 of it's mass in the process. We simply can't build a heat shield that can survive the maneuver.
  2. You're going fast enough that the electromagnetic force no longer controls the interaction of matter--at the interface between probe and star "solid" ceases to be truly solid. You're going to get a lot of particles embedding themselves in the heat shield rather than being deflected. Oops--said particles generally dump their kinetic energy as heat--inside the heat shield. Some will score direct enough hits to cause fusion reactions instead.
  3. Wikipedia says the density is about 2E-4 kg/m^3. Lets take a hypothetical probe that's a cylinder 50m long with an average density equal to water. Water has a density of about 1000kg/m^3. We have 1000kg / 2E-4 kg for a stopping distance of 2,000,000 meters or 2000km. I'm getting a quarter of a million gs if you go for a complete stop. Simply going for aerocapture will be somewhat less (I'm not sure what orbital velocity you need at that point) but not enough to matter. (Note: My omission of width is deliberate, it cancels out.)
Improve this answer
$\endgroup$
4
  • $\begingroup$ Re #2... You're going fast enough that the electromagnetic force no longer controls the interaction of matter--at the interface between probe and star "solid" ceases to be truly solid Not sure what you mean by this. You're right that any proton that strikes the shield will embed itself in the shield. (The stopping distance for protons at this energy is probably less than a millimeter.) Some will score direct enough hits to cause fusion reactions instead. A negligible fraction of them do this -- see my comment on Andy's answer. $\endgroup$
    – user687
    Commented Jul 22, 2016 at 13:00
  • 1
    $\begingroup$ The calculation in #3 doesn't look right to me. You seem to be forming some ratios in order to find the stopping distance, but that's not valid. The basic limit on the stopping distance is a scale that's simply set by the size of the star, which is orders of magnitude more than 2000 km. Also, it doesn't make sense to look up a value for the density of a stellar atmosphere on Wikipedia. The density falls off exponentially with height (just like the earth's atmosphere), so it's not just one number. $\endgroup$
    – user687
    Commented Jul 22, 2016 at 13:04
  • $\begingroup$ @BenCrowell I'm not saying all will end up in the shield, I'm saying that the electromagnetic forces that make things "solid" aren't strong enough to push all of the stellar matter out of the way. If even a very small portion got through it would be a very bad thing for the ship. $\endgroup$
    – Loren Pechtel
    Commented Jul 22, 2016 at 19:13
  • 1
    $\begingroup$ @BenCrowell As for the stopping distance--go back to Newton. A high speed penetrator will be basically brought to a stop after displacing matter equal to it's own mass. You don't get to punch through the whole star because it's too much drag, you stop long before you get through it. I do agree you could aim for the thinner part of the photosphere but it's not going to help you much that it wasn't worth trying to figure out the chord length at various density points. $\endgroup$
    – Loren Pechtel
    Commented Jul 22, 2016 at 19:18
1
$\begingroup$

As the other answers have pretty well shown that stellar aerobraking from 0.1c is not really an option, I wanted to find you some alternatives. The following formula will give you the acceleration (in Earth gravitys) given the final velocity, starting velocity and stopping distance. $$\frac{V_f^2 - V_i^2)}{\frac{(2*d)}{9.8}}=a$$ For starters I used the speed you gave and Plutos greatest distance from the Sun. $$\frac{0^2 - (299792458/10)^2)}{\frac{(2*7378000000)}{9.8}}=a$$ Gives -6215Gs, that is a very dead crew. I think this realy shows just how fast a fraction of the speed of light is. If we instead do the same maneuver from 1/1000th C, we get -0.62Gs. So, by reducing the initial speed, you may be able to find a way to save your crew from being crushed.

But this is using some undefined braking mechanism. Perhaps consider:

  • Giant solar sail used the create drag on the stellar wind
  • Multiple aerobraking in the event 2 planets or the star are in line with your path
  • Electric tether propulsion, not sure this actually works in this direction of travel

These methods are meant to spread the deceleration out so that the crew could survive the experience.

Improve this answer
$\endgroup$

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