Jeopardy: Centrigugal Force Makes You Lighter?

[iamthewalrus_3]

@Malacandra

We already did. We’ve done it 0! times.

Are you sure it wasn’t .99999… times?

[Smeghead]

@iamthewalrus_3

Are you sure it wasn’t .99999… times?

That’s only if you put the 9 on a treadmill.

[GargoyleWB]

@Bootis

Would you weigh less at the peak of Chimborazo, which is 1 degree south, than at any point at 0 degrees latitude?

This is the one I’m wondering about…I’ll have to get my TI-89 and a couple glasses of wine tonight and pound out the numbers…

[Chronos]

Your TI-89 probably doesn’t have enough horsepower to answer the question. For questions like this, the spherical approximation often isn’t good enough, so you have to take into account the shape of the Earth including mountains.

[GargoyleWB]

@Chronos

Your TI-89 probably doesn’t have enough horsepower to answer the question. For questions like this, the spherical approximation often isn’t good enough, so you have to take into account the shape of the Earth including mountains.

,but if I assume a spherical earth in a vacuum…

[Bootis]

@Chronos

Your TI-89 probably doesn’t have enough horsepower to answer the question. For questions like this, the spherical approximation often isn’t good enough, so you have to take into account the shape of the Earth including mountains.

You’d have to figure out exactly how much less centrifugal force is at 1 degree south than at the equator, and also what is the difference between the height of Chimborazo and the highest point at 0 degrees,and factor those numbers into whatever equation exists that defines the relationship of gravity, altitude, and centrifugal force in respect to weight. I want to believe that you weigh less on top of Chimbarazo, just so Jeopardy can be wrong.

[ZenBeam]

…

[ZenBeam]

@Chronos

Your TI-89 probably doesn’t have enough horsepower to answer the question. For questions like this, the spherical approximation often isn’t good enough, so you have to take into account the shape of the Earth including mountains.

Don’t go discouraging him, Chronos!

GargoyleWB, if you go to the spherical harmonics section of the geoid link, they have the representation or the geoid, in all its gory detail, but if all you want is to model it as an oblate spheroid, the only term you need to include is the C[sub]20[/sub] = -0.484165E-03 term from here. All the rest are at least 2 orders of magnitude smaller.

Since you’re almost at the equator, you could just go with

V = (1+ C[sub]20[/sub] (a/r)[sup]2[/sup])*GM/r
with
GM=3986004.415E+8 m3/s2
a=6378136.3 m

Then you can just assert that the mountains don’t matter. No better way to trick someone else into figuring out if they do.