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My textbook derives the formula for apparent weight at the equator, which is just equating the centripetal force to the difference between the true weight and normal reaction. $$ mg' = mg + m \omega^2 R_e$$

However, it just states the formula for apparent weight at any latitude as-

$$ mg' = mg + m \omega^2 R_e \cos^2\lambda , $$ where $\lambda $ is the latitude in degrees.

I tried proving this by following the same method- equating the forces, but as far as I could see, the $ \omega $ remains the same as in the case when it was at the equator, only the $ R_e$ changes to $ R_e \cos \lambda$. So I am arriving at

$$ mg' = mg + m \omega^2 R_e \cos\lambda , $$

Could someone please point me to what I am missing here?

I am a high school student, so would appreciate simpler answers.

Shub
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Astrophile
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2 Answers2

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What you're neglecting is that the gravitational force, the normal force, and the centripetal acceleration are vectors. This means that you must have $$ m\vec{g} + \vec{F}_N = (m \omega^2 R_e \cos \lambda) \hat{e} $$ where $\hat{e}$ is a unit vector pointing towards the Earth's rotation axis. On the equator, this distinction doesn't matter, because all three vectors point along the same axis, and so you can just add or subtract their magnitudes. But anywhere else along the Earth's surface, the vectors are not colinear and so you have to take their directions into account. In particular, the angle between $\vec{g}$ and $\hat{e}$ is $\lambda$, and this ends up being the reason that there's a second factor of $\cos \lambda$ involved.

If you want to try to derive the result given in the book, you will need to start with the equation $$ \vec{F}_N = m (\omega^2 R_e \cos \lambda \hat{e} - \vec{g}). $$ Draw the vector diagram and use the law of cosines to find the magnitude of $\vec{F}_N$. You'll end up with a nasty equation involving a square root. Using the binomial approximation, you can show that this nasty expression is approximately equal to the expression from your textbook so long as $\omega R_e^2 \ll g$ (which it is.) If you succeed in this derivation, you will understand why it wasn't included in an introductory textbook. :-)

Michael Seifert
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For the apparent reduction in weight, you need to consider the acceleration vector "anti-parrallel" to $g$.

Centrifugal acceleration $a = ω^2r = ω^2R\cosλ $

Acceleration "anti-parrallel" to g $= a \cosλ = ω^2R\cos^2λ$

So apparent weight $= mg - mω^2R\cos^2λ$ (the normal reaction will be equal and opposite to this)

Note: $a$ will also have a tangential component that leads to equatorial buldge.

Shub
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