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What AP does it take to remove the core of a star?
- Thread starter GyroNutz
- Start date
Mephistus
Username Only- 939
- 700
I might be able to give you a hard number if you have a more specific timeframe / time range this action takes place.
https://en.wikipedia.org/wiki/Solar_core
https://en.wikipedia.org/wiki/Solar_radius
But for the method i'd imagine you could just use the mass of the core of the sun (34% of the sun's mass is 6.7626E29 kg) and just use the distance the core is from the outside of the sun (our sun's core extends from .2 to 25 of its solar radius from the center, lets go with .25 just to be safe here) which means its travelling .75 solar radius or 521,775,000 meters or if you want a yield range the other end would be 556,560,000 meters for .8 solar radius. Fully extracting and moving the entire thing out of the sun's radius + the core's own diameter would be 1.4 or 1.5 solar radius travelled; 973,980,000 meters or 1,043,550,000 meters, respectively.
Divide the distance above by the timeframe in seconds and you got the velocity.
From there you could just do a kinetic energy calc; presuming the velocity doesn't hit faster than light speeds you should be golden.
https://en.wikipedia.org/wiki/Solar_core
https://en.wikipedia.org/wiki/Solar_radius
But for the method i'd imagine you could just use the mass of the core of the sun (34% of the sun's mass is 6.7626E29 kg) and just use the distance the core is from the outside of the sun (our sun's core extends from .2 to 25 of its solar radius from the center, lets go with .25 just to be safe here) which means its travelling .75 solar radius or 521,775,000 meters or if you want a yield range the other end would be 556,560,000 meters for .8 solar radius. Fully extracting and moving the entire thing out of the sun's radius + the core's own diameter would be 1.4 or 1.5 solar radius travelled; 973,980,000 meters or 1,043,550,000 meters, respectively.
Divide the distance above by the timeframe in seconds and you got the velocity.
From there you could just do a kinetic energy calc; presuming the velocity doesn't hit faster than light speeds you should be golden.
- 12,552
- 2,508
- Thread starter
- #9
Alright. So in the context, Artemis was struggling for survival in the vacuum of space, so the feat was pretty urgent. I'll take 10 seconds as a high-end, 30 seconds as a mid-end, 60 seconds as a low-end and 300 seconds as a super low-end.
Assuming the distance is 1.45 solar radius, and using 10 seconds, (973,980,000/10)^2 * (6.7626E29)/2 = (100,876,500)^2 * 3.3813E29 = 3.44083E45 J
For 30 seconds, we get 3.82315E44 J
For 60 seconds we get 9.55787E43 J
For 300 seconds, we get 3.82315E42 J
First one is Solar System level, rest are Large Star level
Assuming the distance is 1.45 solar radius, and using 10 seconds, (973,980,000/10)^2 * (6.7626E29)/2 = (100,876,500)^2 * 3.3813E29 = 3.44083E45 J
For 30 seconds, we get 3.82315E44 J
For 60 seconds we get 9.55787E43 J
For 300 seconds, we get 3.82315E42 J
First one is Solar System level, rest are Large Star level
- 61,134
- 14,533
Star Wars actually has a feat liks this too
- 10,748
- 6,476
Solar system level for removing a part of a star. Why am I even surprised?
- 10,748
- 6,476
That's exactly what I mean when I say "Why am I even surprised"
- 1,305
- 613
Something something Large Planet level by lifting a mountain.
- 10,748
- 6,476
*Moon sized mountain. So essentially Large Planet level by lifting a moon.
- 10,748
- 6,476
Although I did find out that lifting Everest could yield Low 5-B if moved really close to the speed of light
