Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons:
- The centrifugal force cancels out the gravity minimally, more so at the equator than at the poles.
- The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field.
TL;DR version: There are three reasons. In order of magnitude,
The poles are closer to the center of the Earth due to the equatorial bulge. This strengthens gravitation at the poles and weakens it at the equator.
The equatorial bulge modifies how the Earth the gravitates. This weakens gravitation at the poles and strengthens it at the equator.
The Earth is rotating, so an Earth-bound observer sees a centrifugal force. This has no effect at the poles and weakens gravitation at the equator.
Let's see how the two explanations in the question compare to observation. The following table compares what a spherical gravity model less centrifugal acceleration predicts for gravitational acceleration at sea level at the equator ($g_{\text{eq}}$) and the north pole ($g_{\text{p}}$) versus the values computed using the well-established Somigliana gravity formula $g = g_{\text{eq}}(1+\kappa \sin^2\lambda)/\sqrt{1-e^2\sin^2 \lambda}$.
$\begin{matrix}
\text{Quantity} & GM/r^2 & r\omega^2 & \text{Total} & \text{Somigliana} & \text{Error} \\
g_\text{eq} & 9.79828 & -0.03392 & 9.76436 & 9.78033 & -0.01596 \\
g_\text{p} & 9.86431 & 0 & 9.86431 & 9.83219 & \phantom{-}0.03213 \\
g_\text{p} - g_\text{eq} & 0.06604 & \phantom{-}0.03392 & 0.09995 & 0.05186 & \phantom{-}0.04809
\end{matrix}$
This simple model works in a qualitative sense. It shows that gravitation at the north pole is higher than at the equator. Quantitatively, this simple model is not very good. It considerably overstates the difference between gravitation at the north pole versus the equator, almost by a factor of two.
The problem is that this simple model does not account for the gravitational influence of the equatorial bulge. A simple way to think of that bulge is that it adds positive mass at the equator but adds negative mass at the poles, for a zero net change in mass. The negative mass at the pole will reduce gravitation in the vicinity of the pole, while the positive mass at the equator will increase equatorial gravitation. That's exactly what the doctor ordered.
Mathematically, what that moving around of masses does is to create a quadrupole moment in the Earth's gravity field. Without going into the details of spherical harmonics, this adds a term equal to $3 J_2 \frac {GMa^2}{r^4}\left(\frac 3 2 \cos^2 \lambda - 1\right)$ to the gravitational force, where $\lambda$ is the geocentric latitude and $J_2$ is the Earth's second dynamic form. Adding this quadrupole term to the above table yields the following:
$\begin{matrix}
\text{Quantity} & GM/r^2 & r\omega^2 & J_2\,\text{term} & \text{Total} & \text{Somigliana} & \text{Error} \\
g_\text{eq} & 9.79828 & -0.03392 & \phantom{-}0.01591 & 9.78027 & 9.78033 & -0.00005 \\
g_\text{p} & 9.86431 & 0 & -0.03225 & 9.83206 & 9.83219 &
-0.00013 \\
g_\text{p} - g_\text{eq} & 0.06604 & \phantom{-}0.03392 & -0.04817 & 0.05179 & 0.05186 & -0.00007
\end{matrix}$
This simple addition of the quadrupole now makes for a very nice match.
The numbers I used in the above:
$\mu_E = 398600.0982\,\text{km}^3/\text{s}^2$, the Earth's gravitational parameter less the atmospheric contribution.
$R_\text{eq} = 6378.13672\,\text{km}$, the Earth's equatorial radius (mean tide value).
$1/f = 298.25231$, the Earth's flattening (mean tide value).
$\omega = 7.292115855 \times 10^{-5}\,\text{rad}/\text{s}$, the Earth's rotation rate.
$J_2 = 0.0010826359$, the Earth's second dynamic form factor.
$g_{\text{eq}} = 9.7803267714\,\text{m}/\text{s}^2$, gravitation at sea level at the equator.
$\kappa = 0.00193185138639$, which reflects the observed difference between gravitation at the equator versus the poles.
$e^2 = 0.00669437999013$, the square of the eccentricity of the figure of the Earth.
These values are mostly from Groten, "Fundamental parameters and current (2004) best estimates of the parameters of common relevance to astronomy, geodesy, and geodynamics." Journal of Geodesy, 77:10-11 724-797 (2004), with the standard gravitational parameter modified to exclude the mass of the atmosphere. The Earth's atmosphere has a gravitational effect on the Moon and on satellites, but not so much on people standing on the surface of the Earth.
No, the main reason the poles are colder is because the surface is angled with respect to the sun rays.
That is the same reason why in winter it is colder than in summer. You can think that in the poles the winters are harder and the summers are softer, while in the equator it is the other way around.
That is also the reason why some solar panel are motorized: to keep them perpendicular to the sun rays and get the most out of them.
About your hypothesis:
If the atmosphere were to absorb the energy, there still would be heat at the poles, maybe in the air instead of in the ground, but still there.
In polar winter there is constant darkness (24-hour night) up to 6 months in the geometrical pole. But in polar summer there is constant sun and it is still quite cold. So while the shadow actually makes a difference (polar winter is way colder than polar summer), that is not why the poles are colder than equator.
The Earth is 150000000 km from the sun, and the radious of the Earth is about 6500 km, way to small to make a difference.
Best Answer
There are a couple of ways the atmosphere can be viewed as being "thicker" at the equator than at the poles. One sense is the mass per unit area of all of the air overhead, from sea level to outer space, at the equator vs at the North Pole. Atmospheric pressure at sea level, being essentially a potential, is on average pretty much constant worldwide when averaged out over the course of years.
Since atmospheric pressure is the weight per unit area of all of the air overhead, those places where acceleration due to gravity is low must necessarily support more air in terms of mass to achieve the same pressure than areas where gravitational acceleration is high. Gravitational acceleration at sea level is a bit less at the equator than at the poles, making the atmosphere at the equator a tiny bit "thicker" than the atmosphere over the North Pole, by about half a percent.
This is a tiny effect. It is not what people mean when they say the atmosphere is thicker at the equator. The tropopause is considerably higher over equatorial regions (more specifically, over the intertropical convergence zone, or ITCZ) than it is over polar regions. The tropopause can 20 km above sea level or higher at the ITCZ and only 4 km above sea level at the poles. The boundary between the troposphere and stratosphere is essentially non-existent at the South Pole during parts of the winter.
The small decrease in gravitation (including centrifugal acceleration) from the poles to the equator accounts for only a tiny bit of that large change in troposphere height. In fact, Of the five reasons listed in the question, none is primary. The key reason for the tropopause height reaching a maximum at the ITCZ is the general circulation of the atmosphere, with the ITCZ itself playing a key role.
The ITCZ is where the moisture-laden northern and southern hemisphere trade winds meet at the surface, are driven to great heights due to convection and latent heat, and finally separate to flow northward and southward near the top of the troposphere. It is the strong convection and associated very strong thunderstorms created by this meeting of these atmospheres that pushes the tropopause to great heights at the ITCZ. Tropical cyclones and CAPE-driven thunderstorms elsewhere can also push the troposphere to great heights, but these are transient local effects compared to globe-spanning ITCZ.
The polar regions generally have very little, if any convection. Oftentimes they suffer descending rather than rising air, pulling the tropopause downward. The tropopause is also pulled downward at the jet streams, particularly the polar jet.