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Rigorous Science

How can a Type II civilization influence accretion rates from a debris disk to a passing star?

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A young B-type star (with a mass of about 10 M$_{\odot}$) is surrounded by a debris disk extending from about 2 AU to 1000 AU away. The disk has a mass of about 300 Earth masses - enough to form quite a lot of planetesimals. There is also an outer cloud of icy, comet-like bodies extending from 750 AU to 5,000 AU away.

The star stands out from other Sun-like stars, though, because recently a strong stellar wind has developed, which comes with a mass loss rate of about 1.0 $\times$ 10-6 M$_{\odot}$ per year. The stellar wind may die down eventually, but for now it's having quite a strong effect on the star and its disk. It is thought that much of the inner debris disk will be blown away very quickly.

A nearby red dwarf of about 0.76 M$_{\odot}$ (not a flare star, fortunately) is passing through. At its nearest point, it comes extremely close - an astounding 1500 AU away from the B-type star!

There will absolutely be some accretion by the red dwarf. But how much? An advanced civilization (Type II on the Kardashev scale) is watching closely. They're considering the red dwarf as a sort of "rest station". If planetesimals form, they could be mined for raw materials, and any icy bodies could be an excellent source of water - which can be turned into hydrogen and helium.

How can they figure out just how much dust from the debris disk and icy bodies will be captured by the red dwarf? Can they then predict if planetesimals are likely to form (though subsequent planetary formation isn't necessary)?

That bit could be construed as pure science, which it may be. But there's another, much more important question that's absolutely related to world building: Can the Type II civilization do anything to influence accretion?

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Kepler's Laws give us the velocity of everything in the debris field and comet cloud--

At any radius, $i$, the velocity of the circular orbits (and a good rough definition of a "belt", although this is excluding highly elliptical objects) -- $ v_{b,i} = \sqrt{{{GM_{b}} \over {R_{b,i}}}} $

Where $v_{b,i}$, $M_{b}$, and $R_{b,i}$ are the circular orbital velocity, mass, and radius of/from the blue star

Also important is $T_{b,i}$, which is equal to $ { {2 \pi R_{b,i}} \over {v_{b,i}} } $, which is orbital period of this "belt". This will be helpful for determining how much of each belt passes next to the red dwarf.

The escape velocity equation around the red dwarf is $ v_{r} = \sqrt{ {2 G M_r} \over {R_r}} $ . Notice the "2" for escape velocity, compared to orbital velocity. This will include all things captured by the red dwarf, even if not in tidy circular orbits.

In the capture scenarios, the red dwarf has gotten close enough that the velocity of the object being evaluated $ v_{b,i} $ is less than the escape velocity of the red dwarf $ \sqrt{{{GM_{b}} \over {R_{b,i}}}} $ $\Big \lt $ $ \sqrt{ {2 G M_r} \over {R_r} } $

How fast is the red dwarf flyby? The minimum velocity is $ \sqrt{ {2 G M_b} \over {R_b}} $, where $ R_b $ = 1,500 AU or 224,396,806,050,000 meters. That'd be a velocity of 6.8 kilometers per second. Any slower than that, and the red giant would be captured. Typical delta-vs between stars are in the order of single-digit km/s values, so this minimum value is also ballpark average. Although it could be much higher.

The relative speed of the red dwarf gives us how much time it is spending inside the accretion shell (-5,000 AU to +5,000 AU). At a fly-by speed of 6.8 km/s, that will be $ 2.19 \times 10^{11} $ seconds

How much of the 5,000 AU accretion disk is captured? $T_{b,5000} = 3,52 \times 10^{12}$ seconds; $T_{b, 1500} = 5.78 \times 10^{11} $ seconds.

Summarizing --

  • roughly 10% of the 5,000 AU accretion disk, and
  • roughly 50% of the 1,500 AU accretion disk will be captured by the passing red dwarf.

At the closest approach, the red dwarf will capture anything further from the blue star than $ R_r $ $\Big \lt $ $ 2 R_{b,i} { { M_r } \over { M_{b} } } $ ... (puts this all into a spreadsheet to solve) ... 1,303 AU

So, because the blue star is so massive, the red dwarf mostly poaches things that are passing very closely to it.

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