this post was submitted on 09 Jul 2025
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[–] x00z@lemmy.world 5 points 1 week ago (7 children)
[–] floquant@lemmy.dbzer0.com 5 points 1 week ago (4 children)

I wonder if you could tell by measuring g today

[–] elbarto777@lemmy.world 4 points 1 week ago (1 children)
[–] floquant@lemmy.dbzer0.com 2 points 1 week ago (2 children)

I mean, I know you theoretically could, but would it be a "noticeable" difference? Like to the second or third decimal?

[–] suicidaleggroll@lemmy.world 8 points 1 week ago* (last edited 1 week ago)

If my math is right...it will have an effect of approximately 0.0000000001 g

Angular acceleration is r*w^2, so for a normal day that would be 6371000*(2*pi/86400)^2/g = 0.0034345580g

On this accelerated day, it becomes 6371000*(2*pi/86399.9987)^2/g = 0.0034345581g

That's at the equator assuming a radius of 6371 km, which is a decent ballpark, the specific number won't change the result much.

[–] elbarto777@lemmy.world 1 points 1 week ago

Well, that's a totally different question.

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