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Posts posted by tdolowy
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If I have done my sums correctly
[latex]
m = \sqrt{\frac{3 \cdot c^6}{32 \cdot \pi \cdot G^3 \cdot \rho}}
[/latex]
Is the mass at which an spherical object becomes a blackhole for any given density rho
tdolowy - a pleasure. Keep investigating and "doing science"
this is exactly what I was looking for thank you!
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I think the OP might be confused and the key point to make here is that black holes (singularities) are incredibly dense objects occupying ridiculously small area of space. I would also like to say that it is a pleasure to read and answer your question tdolowy...you dont have to worry about us having to bare with you. I remember being 14 and I was very much into black holes then
Exactly. But the op is making the slightly less usual point that any density object can be a black hole provided it is big enough. For constant density the mass increases with the radius cubed - so bearing in mind that in the equation for schwarzchild radius the mass and the radius are directly related as things get higher radius thn eventually the mass within the radius is high enough to exceed that required for blackhole. It is a little contrived as such a huge sphere would compress to a higher density
You have slipped up - earth is more dense than glass and water (1 gram / cm^3) thus it will need a smaller radius to make a black hole with the DENSITY (not mass as you have put) of earth.
I make it 14.1LM, 22.3LM , and 9.4LM respectively for Glass, Water, Earth.
More interestingly - a black hole with the MASS of the earth would be about a 3rd of an inch across (9mm)
This is the mass of a sphere (the curly p is a rho which stands for density, and r is the radius)
[latex]Mass_{sphere} = \rho \cdot \frac{4}{3} \pi r^3[/latex]
This is the radius of a blackhole (G and c are constants and M is the mass)
[latex]r_s=\frac{2GM}{c^2}[/latex]
You can work out all the figures from just these two equations - note that they both have radius and mass in them
Thank you for all the answers, I was very curious and love all of the explanations you guys gave, really made me have to think
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The sun is 330,000 times heavier than the Earth and you would need at least 25 suns to make a blackhole; 330,000 x 25 = 8,250,000 Earths..Laying that many Earths side-by-side gives 7918 (rounded) x 8,250,000 = 6,532,350,000 miles. Note that about 21 suns worth of that mass would get blown away in the supernova and the rest would constitute the mass of the blackhole.
Thank you for the response and this way seems much much easier
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(Disclaimer I am a very curious 14 year old so bare with me)
I found an article saying glass with 2.5 g/cc would have to be 14 light minutes(Lm) in radius to create a black hole, I am now trying to find out how many Earths' that would have to be in a row to create a black hole. I just need help with the math.
Would this mean 1 g/cc would have to be 23.333 light minutes(Lm) large to create a black hole considering 14Lm + (14Lm ÷ 1.5g/cc) = 23.333Lm/g/cc (Lm being Light Minutes)?
1 g/cc density would take 23.333Lm to make a black hole ?
Considering this that would mean Earth (Which all together has a density of 5.513 g/cc) would be 5.513 g/cc
5.513 g/cc x 23.333Lm = apx 128.63666Lm (Light minutes) or (7.562067334e+14 Miles) wide to create a black hole with Earths' mass
If this is correct thus far what would be my next step considering Earth has a Diameter of 7,917.5 Miles while also taking in the whole volume (Width, Length, and Height)
Keep in mind I am 14 so if I am doing this all wrong im sorry!
Thanks and all I would like is positive reinforcement please!
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How many Earths' would it take to form a black hole from mass
in Mathematics
Posted
Thanks for all the answers