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How fast/powerful is this?
- Thread starter Eficiente
- Start date
- 64,039
- 16,083
1) Easily FTL
2) Moon level, its hitting the moon so hard it goes FTL
3) Hypersonic
2) Moon level, its hitting the moon so hard it goes FTL
3) Hypersonic
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By the looks of the feat, he moved both the Earth and the sun from 153 pixels apart until they are 4 pixels apart. To work out how far that is, I'll use the Earth as a scale.
Earth=132 pixels from North to South pole or 40,008 km IRL.
1 pixel measurement=303.1 kilometers. Earth and the sun are therefor 1212.4 kilometers apart at the end of movement and 46,374.3 kilometers apart at the start of the movement. The distance travelled is therefor 45,161.9 kilometers. And, because they meet in the middle, that mean that each planetary body moved 22,580.95 kilometers.
The movement takes a total of 0.6 seconds. So with the calulation speed=distance x time, that tells me that each planetary body was moved at 37,634,900 m/s or Mach 109,722.74052 or 12.5% of the speed of light. (So the speed is Relativistic, not FTL)
To calculate the force required to move the Earth and the Sun at that speed, I will calulate their force in KE seperately.
KE=0.5mv^2.
Substituting 5.972 × 10^24 kg and 37,634,900 m/s for mass and velocity squared into my calculation, I find that he expended 4.279 × 10^39 joules.
Doing the same for the sun, which weighs 1.989 × 10^30 kg gives me the ridiculous number of 1.425 × 10^45 joules.
The total force applied to move these two both is... actually no different from the force required to move the sun. Well, it is, but it makes no difference to the overall total, or tier that this feat will get placed in. So, final answer to your question?
1.425 × 10^45 joules, or Large Star level.
Earth=132 pixels from North to South pole or 40,008 km IRL.
1 pixel measurement=303.1 kilometers. Earth and the sun are therefor 1212.4 kilometers apart at the end of movement and 46,374.3 kilometers apart at the start of the movement. The distance travelled is therefor 45,161.9 kilometers. And, because they meet in the middle, that mean that each planetary body moved 22,580.95 kilometers.
The movement takes a total of 0.6 seconds. So with the calulation speed=distance x time, that tells me that each planetary body was moved at 37,634,900 m/s or Mach 109,722.74052 or 12.5% of the speed of light. (So the speed is Relativistic, not FTL)
To calculate the force required to move the Earth and the Sun at that speed, I will calulate their force in KE seperately.
KE=0.5mv^2.
Substituting 5.972 × 10^24 kg and 37,634,900 m/s for mass and velocity squared into my calculation, I find that he expended 4.279 × 10^39 joules.
Doing the same for the sun, which weighs 1.989 × 10^30 kg gives me the ridiculous number of 1.425 × 10^45 joules.
The total force applied to move these two both is... actually no different from the force required to move the sun. Well, it is, but it makes no difference to the overall total, or tier that this feat will get placed in. So, final answer to your question?
1.425 × 10^45 joules, or Large Star level.
- 3,464
- 891
And, if someone mentions that the distance between the Earth and the Sun is 149.6 million km, not 46,374.3 km, I tell them to watch the feat. It very clearly isn't that far apart in this particular universe.
- 3,464
- 891
You are welcome, good sir.
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