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what's the difference between star level and solar system level?

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I don't understand the difference
The sun has about the same mass as our solar system
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Its mass is almost the same, why would it take more to destroy a solar system than it would be to destroy the sun?
if a large star started blowing up would also create a supernova
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and some stars are even bigger than 10x the size of the sun. some stars have way more mass than our solar system and have the power to destroy it.
What about huge stars like the uy scuti?
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this should have WAYY more than enough power to destroy the solar system and has far more mass than the solar system
why is large star level below solar system when a large star would have the power to destroy a solar system, and even beyond that?
 
Wouldn't it like take a lot more energy to also cover the entire distance of a solar system
some stars have enough energy to cover the entire distance of the solar system if they were to blow up
 
wouldn't that make the star itself solar system level?
No, just Star Level as we treated destroying a solar system that involves the planets, a asteroid field, and so on as one huge attack.


Also we are applying irl logic to fiction so…
 
No, just Star Level as we treated destroying a solar system that involves the planets, a asteroid field, and so on as one huge attack.


Also we are applying irl logic to fiction so…
then what would be the difference between a star that has more mass than the solar system and has the power to destroy it, vs solar system level
 
No, just Star Level as we treated destroying a solar system that involves the planets, a asteroid field, and so on as one huge attack.


Also we are applying irl logic to fiction so…
also don't people who calc use real life logic and kinetic energy as well?
 
Because star level is overpowering the gravitational binding energy of the sun, but solar system level is destroying everything from the sun to neptune through the inverse square law.
 
then what would be the difference between a star that has more mass than the solar system and has the power to destroy it, vs solar system level
It is because the star is already undergoing a process of what is essentially dying and turns into a super nova as super novas ain’t a common event IIRC and because not all stars had the mass when they reach to the point they go boom.
 
It is because the star is already undergoing a process of what is essentially dying and turns into a super nova as super novas ain’t a common event IIRC and because not all stars had the mass when they reach to the point they go boom.
so what would happen if a character were to suddenly blow up a star, wouldn't it create a supernova
 
also don't people who calc use real life logic and kinetic energy as well?
They use the maths part to do so and that is fair.
so what would happen if a character were to suddenly blow up a star, wouldn't it create a supernova
Supernovas occurrence take time as in the case of Supernova is a over time process, but not all star goes super novas. Some turn into white dwarfs and some other things. https://www.esa.int/kids/en/learn/Our_Universe/Stars_and_galaxies/Star_death
 
but what would happen if a character just busted a large star
No.

Like Fluffy and DarkDragonMedeus said.

INVERSE-SQUARE LAW

Star level is merely overpowering the GBE of the sun. Solar System level is destroying every patch of land from the epicenter of the solar system to the orbit of Neptune via inverse-square law.

You can read more about our values for 4-B here
 
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No.

Like Fluffy and DarkDragonMedeus said.

INVERSE-SQUARE LAW

Star level is merely overpowering the GBE of the sun. Solar System level is destroying every patch of land from the epicenter of the solar system to the orbit of Neptune via inverse-square law.
btw the solar system is actually far beyond neptune
 
Yeah, Inverse Square Law is used when it comes to destroying a solar system and beyond that like the other says.

I think it also applies to Lower tiers like Tier 9s, 8s, 7s, and so on.

Anyway the OP has been addressed.
 
Yeah I never knew that, he's actually right according to google: Our solar system consists of our star, the Sun, and everything bound to it by gravity – the planets Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune; dwarf planets such as Pluto; dozens of moons; and millions of asteroids, comets, and meteoroids
 

"After Pluto was discovered in 1930, it was declared to be the ninth planet from the Sun. Beginning in the 1990s, its status as a planet was questioned following the discovery of several objects of similar size in the Kuiper belt and the scattered disc, including the dwarf planet Eris. This led the International Astronomical Union (IAU) in 2006 to formally define the term "planet"—excluding Pluto and reclassifying it as a dwarf planet."

But... even if we accounted for Pluto and all the other dwarf planets, the method for finding out the yield would be the same: Inverse square law
 
"After Pluto was discovered in 1930, it was declared to be the ninth planet from the Sun. Beginning in the 1990s, its status as a planet was questioned following the discovery of several objects of similar size in the Kuiper belt and the scattered disc, including the dwarf planet Eris. This led the International Astronomical Union (IAU) in 2006 to formally define the term "planet"—excluding Pluto and reclassifying it as a dwarf planet."
Yeah, we know Pluto is classified as dwarf planet. The thing, it still count as being part of the solar system, not outside of the solar system as far as I am aware.
 

"After Pluto was discovered in 1930, it was declared to be the ninth planet from the Sun. Beginning in the 1990s, its status as a planet was questioned following the discovery of several objects of similar size in the Kuiper belt and the scattered disc, including the dwarf planet Eris. This led the International Astronomical Union (IAU) in 2006 to formally define the term "planet"—excluding Pluto and reclassifying it as a dwarf planet."

But... even if we accounted for Pluto and all the other dwarf planets, the method for finding out the yield would be the same: Inverse square law
Yeah the point is that our definition for Solar system level is "The star system known as the solar system", but the justification for the energy output needed is " Neptune through Inverse Square Law at 4.498x10^12 meter radius (Distance from the Sun to Neptune)". The Sun to Neptune isn't actually the entire solar system, it's much larger than that
 
Problem is, it would take a lot less energy to destroy Pluto with an omnidirectional explosion from the Sun than the amount of energy it'd take to whack Neptune from the same epicenter.

Formula is this (I derived it from Assalt's inverse-square law here):

4 * (3 * (Gravitational Constant) * (Mass of the planet you want to destroy)^2) / (5 * Radius of the planet you want to blow up) * ((Distance between the planet and the epicenter of the explosion, usually the semi-major axis)/(Radius of the Planet))^2

Gravitational constant is always 6.67408e-11

Pluto Mass: 1.303e+22 kg

Pluto Radius: 1188300 m

Farthest distance between Pluto and Sun (Semi-Major Axis): 39.482 AU or 5.906423e+12 m

Energy needed to blow up Pluto from the Sun: 4 * (3 * (6.67408e-11) * (1.303e+22 kg)^2)/(5 * 1188300 meters) * ((5.906423e+12 meters)/(1188300 meters))^2= 5.6540843e+41 J (Low 4-C+, just a bit below baseline 4-C)

Neptune Mass: 1.02413e+26 kg

Neptune Radius: 24622000 m

Farthest distance between Neptune and Sun (Semi-Major Axis): 30.07 AU or 4.49841e+12 m

Energy needed to blow up Neptune from the Sun: 4 * (3 * (6.67408e-11) * (1.02413e+26 kg)^2)/(5 * 24622000 meters) * ((4.49841e+12 meters)/(24622000 meters))^2= 2.2775133e+45 J (Our current value for Solar System level)
 
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What would be the result if instead of using Pluto we used Sedna which is 937 AU away from the sun?
 
What would be the result if instead of using Pluto we used Sedna which is 937 AU away from the sun?
We have no mass values or radius values ATM, so it would currently be impossible to determine.

But I doubt it'd penetrate anything above the Low 4-C to 4-C range
 
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Star level is the energy needed to destroy a star, Solar System level is the energy needed to destroy not just the star but the entire solar system at once
 
What would be the result if instead of using Pluto we used Sedna which is 937 AU away from the sun?
Probably lower because Sedna requires so much less force to destroy compared to Neptune, so even though it's farther the result wouldn't change much

Sedna's mass is unknownm, but it's smaller than Charon, which have a mass of 1.586e+21 kg

ISo:

E = 4 * (3 * (6.67408e-11) * (1.586e+21 kg)^2)/(5 * 512,500 meters) * ((1.455608e+14 meters)/(512,500 meters))^2 = 6.3418545045694068144599288716066e+43 joules, or High 4-C

Yeah this is way lower than Solar System level
 
Interesting. And what would be the energy required to destroy the entire Oort cloud? If I'm not wrong, it is also part of the solar system
 
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