this post was submitted on 19 Nov 2024
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[–] wanderer@lemmy.world 1 points 1 day ago (1 children)

A mass weighing 100kg at the north pole would only weigh 99.5kg at the equator

That assumes a perfectly spherical earth. The earth is not perfectly spherical.

[–] gentooer@programming.dev 0 points 1 day ago (1 children)

This. Planets are in hydrostatic equilibrium, meaning that the combined acceleration by gravity and the centrifugal "force" is equal all over the world (except for local differences due to mountains and dense crust).

[–] Deme@sopuli.xyz 5 points 1 day ago* (last edited 1 day ago) (1 children)

Hydrostatic equilibrium yes, but equal? No. We agree that centrifugal force is a factor. Now ask yourself, why would gravity suddenly strengthen at the equator to get the surface acceleration to stay equal to that of the poles?

It doesn't. As a result the Earth seeks a new hydrostatic equilibrium, bulging out at the equator. This in turn strengthens the centrifugal force a bit while also slightly diminishing the force of gravity (because more of the planet's mass is farther away). So the same effect is taken even further. Local differences add a layer of noise on top of this, but the end result is that the net surface acceleration is measured to average slightly less at equatorial regions than at the poles, with for example Singapore getting 9.7639 m/s2 of downward acceleration, while Helsinki gets 9.825 m/s2.

[–] gentooer@programming.dev 4 points 1 day ago

You're right, I had it backwards. Hydrostatic equilibrium makes it that the combined force vector of gravity and the centrifugal force is perpendicular to the planet surface everywhere.