▲ 159 ▼ Tomorrow The Earth Will Spin Faster Than Normal Making The Day 1.30 Milliseconds Shorter (www.iflscience.com) submitted 1 year ago by vxx@lemmy.world to c/science@lemmy.world 19 comments fedilink hide all child comments
[–] floquant@lemmy.dbzer0.com 5 points 1 year ago (4 children) I wonder if you could tell by measuring g today permalink fedilink source parent hideshow 4 child comments replies: [+] elbarto777@lemmy.world 4 points 1 year ago (3 children) [deleted] permalink fedilink source parent hideshow 3 child comments replies: [–] floquant@lemmy.dbzer0.com 2 points 1 year ago (2 children) I mean, I know you theoretically could, but would it be a "noticeable" difference? Like to the second or third decimal? permalink fedilink source parent hideshow 2 child comments replies: [–] suicidaleggroll@lemmy.world 7 points 1 year 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. permalink fedilink source parent load more comments (1 reply)
[+] elbarto777@lemmy.world 4 points 1 year ago (3 children) [deleted] permalink fedilink source parent hideshow 3 child comments replies: [–] floquant@lemmy.dbzer0.com 2 points 1 year ago (2 children) I mean, I know you theoretically could, but would it be a "noticeable" difference? Like to the second or third decimal? permalink fedilink source parent hideshow 2 child comments replies: [–] suicidaleggroll@lemmy.world 7 points 1 year 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. permalink fedilink source parent load more comments (1 reply)
[–] floquant@lemmy.dbzer0.com 2 points 1 year ago (2 children) I mean, I know you theoretically could, but would it be a "noticeable" difference? Like to the second or third decimal? permalink fedilink source parent hideshow 2 child comments replies: [–] suicidaleggroll@lemmy.world 7 points 1 year 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. permalink fedilink source parent load more comments (1 reply)
[–] suicidaleggroll@lemmy.world 7 points 1 year 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. permalink fedilink source parent