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Ok, so this is a culmination of a few questions into one final question: am I doing this right? I am trying to figure out the gravitational difference from one end of a continent to another. There is a lot of math here, sorry.

My world exists within a spinning sphere. The distance from the equator to either pole is 12,429 miles. That means each angle of degree is (roughly) 138 miles. My continent is 400 miles, starting at the equator. That means at the the most extreme angle is 2.9, I will round up to 3.

Based on the answer by @Daron, in this question, using the graph he provided below, that's a cosine of (again, roughly) 0.991.

graph

Which, I think, means at the weakest point of the continent, the gravity is 0.991g.

Did I do this right?

Thanks guys!

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  • $\begingroup$ Keep in mind that if your world has water, it will collect at the equator. $\endgroup$
    – Alexander
    Commented Apr 30, 2020 at 0:12
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    $\begingroup$ @AdrianColomitchi I take that means my calculations were off by a bit. I don't know what the cos (I assume cosine) or the pi means $\endgroup$
    – NRJohnson
    Commented Apr 30, 2020 at 14:35
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    $\begingroup$ @AdrianColomitchi - if you will lay out in an answer the reasons for your math, it will be educational, and I will give you the fattest upvote I can give. Icing on top if reasons are well larded with superfluous but witty repartee. $\endgroup$
    – Willk
    Commented Apr 30, 2020 at 16:23
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    $\begingroup$ @NRJohnson pi is π, cos is indeed cosine - standard math notation. Centrifugal force = ω²·r where ω is the angular rotation speed (in radians/second - and it's constant because your sphere is rigid) and r is the radius of rotation. The r - radius of rotation - is cos(latitude_angle_in_radians)·Rₚₗₐₙₑₜ - a latitude of 3° corresponds to a 3°/360°·2π, so the centrifugal force at 3° latitude is cos(3°/360°·2π)·centrifugal_force_at_equator. So cos(3/360*2π)=0.99862953475. $\endgroup$ Commented Apr 30, 2020 at 23:04
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    $\begingroup$ Minor comment on the title: 'Specific Gravity' is a well established term in some contexts (especially geology and the brewing industry) for relative density. You might want to update your title to avoid this so it's less ambiguous what your question is about for someone skimming through questions. $\endgroup$ Commented May 4, 2020 at 2:05

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Yes, you do it mostly right, I just would suggest a higher degree of detail.
That fact was already analysed in the comments but never got a proper answer, so I will repeat your math and the math from the comments with some more detail, summarize and comment it:

The distance from the pole to the equator is 12,429 miles. The angle for this distance is 90° so for every angle of degree we get $12,429miles / 90° = 138,1 miles/°$.
The continent is 400 miles in length (even it isn't deffinetly said in the question we can assume thats the distance north to south), so that means the angle of the continent is $400miles / (138,1miles/°) = 2,8965°$.
As @Adrian_Colomitchi stated in the comments, centrifugal force is $F = ω²·r$ where ω is the angular rotation speed (in radians/second) and r is the radius of rotation. Using the trigonometric functions we can get $r_{north}=cos(latitudeangle)·R_{equator}$ , so $F_{north} = ω²·r_{north} = ω²·cos(latitudeangle)·R_{equator} = ω²·R_{equator}·cos(latitudeangle)=F_{equator}·cos(latitudeangle)$
$cos(2.8965°)=0,99872$ and if we assume $F_{equator}=1g$ that means you will have a gravitational force of $0,99872g$ at the most northern point of your continent.

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