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At the equator at sea level you are about 17 miles higher up than at the poles ie 17 miles further from the center of the earth. Yet the air is not the same as it would be 17 miles up at a higher latitude. It seems to average the same as at sea level anywhere else on the earth. Is there one reason for this or is it that a number of different factors balance out to give this equivalence?

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  • $\begingroup$ Gravity at the equator is slightly less than at the poles for the same reason, but you'd think this would cause lower air pressure. Good question. $\endgroup$ Commented Apr 26, 2023 at 21:41
  • $\begingroup$ Well, you also have more mass under you, so $g$ is slightly higher at the equator. $\endgroup$
    – Jon Custer
    Commented Apr 26, 2023 at 22:04
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    $\begingroup$ The surface of the earth is roughly a gravitational equipotential, so the pressure shoujld be the same everywhere (if it were not for the heat of the sun causing the weather). $\endgroup$
    – mike stone
    Commented Apr 26, 2023 at 23:59
  • $\begingroup$ @JonCuster Note that although there is more mass "under" you at the Equator you are further away from the "centre" of the Earth. $70\%$ of the reduction in the free fall acceleration as compared with the poles is due to the rotation of the Earth and the other $30\%$ is due to the equatorial bulge. $\endgroup$
    – Farcher
    Commented Apr 27, 2023 at 7:49
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    $\begingroup$ Atmospheric pressure does not depend on how close you are to Earth center, but rather on how high you are with respect to mean sea level at that location. At the poles and equator you will be exactly at same sea level,- $h=0$ above Earth mean surface, hence same pressure $p_0$. $\endgroup$ Commented Apr 27, 2023 at 7:50

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As someone said in a comment, the surface of the earth is an equipotential. In fact, that's why it bulges - the rotation of the earth would nominally decrease the potential on a spherical body, so the earth bulges, increasing the potential at the equator until the gravitational+centrifugal potential at the equator is the same as the potential at the poles.

This is all fancy-physics-speak for "the poles cannot be downhill from the equator, or mass would move to fix that situation." For the very same reason, the pressure of the atmosphere at the poles is the same as the pressure at the equator. Otherwise air would move to fix that situation.

Seen another way, if the atmosphere was in thermal equilibrium (it isn't but hang with me), then the density would be proportional to $e^{-\Phi/k_B T}$, where $\Phi$ is the gravitational plus centrifugal potential - another reason the atmosphere isn't less dense at the equator. It's not $e^{-h/k_B T}$ (h being the height from the center of the earth). Of course, the influence of the sun fricks this all up, but suffice to say this explains why the atmosphere at the equator isn't as low pressure as the atmosphere 17km above the north pole.

--for experts-- yes I know that the atmosphere is better approximated as an adiabatic invariant rather than thermal equilibrium, but for the sake of the above argument it doesn't matter

One more interesting note here... for a very similar reason, one might ask a question like "why does the equatorial bulge exactly agree with a calculation I do assuming the earth is uniform in density, although the earth isn't uniform in density." The answer is, because each layer of the earth bulges by the same fraction, and ultimately, that calculation just calculates said fraction. The iron core bulges by a fraction $f=17\text{ km}/R_e$, the mantle bulges by that fraction, and the crust bulges by that fraction. And indeed, the atmosphere also bulges by that same fraction. Because all these things rotate at the same angular frequency.

See also, my favorite question on this topic, which clarifies a subtle point about the equatorial bulge.

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Could it be that the atmosphere is in the same shape as the Earth where it keeps the same distance from sea level to the atmosphere.

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    $\begingroup$ Is this supposed to be an answer? It seems more like a question. $\endgroup$
    – PM 2Ring
    Commented Apr 27, 2023 at 7:09

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