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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 th...
Answer
#2: Post edited
by
James McLellan
·
2022-11-30T16:28:39Z (almost 2 years ago)
- 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} = $$ {{ R_r} } $ $\Big \lt $ $ 2 {{ M_r } \over {M_{b}}} R_{b,i} $Work in Progress
- 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.
#1: Initial revision
by
James McLellan
·
2022-11-30T15:59:34Z (almost 2 years ago)
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} = $
$ {{ R_r} } $ $\Big \lt $ $ 2 {{ M_r } \over {M_{b}}} R_{b,i} $
Work in Progress