So I saw this article stating that gravity is stronger on the top on the mountain due to there being more mass under you however I have read some questions other people have asked and most of the responses state that the mass is concentrated at the middle of the earth meaning gravity doesn't get stronger the higher up you go. I would like to know which one of these it is as the article is a pretty reliable source. Here is the link to the article https://nasaviz.gsfc.nasa.gov/11234
Newtonian Gravity – Does Gravity Get Stronger Higher Up on a Mountain?
earthestimationgeophysicsnewtonian-gravityplanets
Related Solutions
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.
Forget about force. Force is a bit much irrelevant here. The answer to this question lies in energy, thermodynamics, pressure, temperature, chemistry, and stellar physics.
Potential energy and force go hand in hand. The gravitational force at some point inside the Earth is the rate at which gravitational potential energy changes with respect to distance. Force is the gradient of energy. Gravitational potential energy is at it's lowest at the center of the Earth.
This is where thermodynamics comes into play. The principle of minimum total potential energy is a consequence of the second law of thermodynamics. If a system is not in its minimum potential energy state and there's a pathway to that state, the system will try to follow that pathway. A planet with iron and nickel (and other dense elements) equally mixed with lighter elements is not the minimum potential energy condition. To minimize total potential energy, the iron, nickel, and other dense elements should be at the center of a planet, with lighter elements outside the core.
A pathway has to exist to that minimum potential energy state, and this is where pressure, temperature, and chemistry come into play. These are what create the conditions that allow the second law of thermodynamics to differentiate a planet. As a counterexample, uranium is rather dense, but yet uranium is depleted in the Earth's core, slightly depleted in the Earth's mantle, and strongly enhanced in the Earth's crust. Chemistry is important!
Uranium is fairly reactive chemically. It has a strong affinity to combine with other elements. Uranium is a lithophile ("rock-loving") element per the Goldschmidt classification of elements. In fact, uranium is an "incompatible element", which explains the relative abundance of uranium in the Earth's crust.
Nickel, cobalt, manganese, and molybdenum, along with the most extremely rare and precious metals such as gold, iridium, osmium, palladium, platinum, rhenium, rhodium and ruthenium, are rather inert chemically, but they do dissolve readily in molten iron. These (along with iron itself) are the siderophile (iron-loving) elements. In fact, iron is not near as siderophilic as the precious metals. It rusts (making iron is a bit lithophilic) and it readily combines with sulfur (making iron a bit chalcophilic).
This is where pressure and temperature come into play. Pressure and temperature are extremely high inside the Earth. High pressure and high temperature force iron to relinquish its bonds with other compounds. So now we have pure iron and nickel, plus trace amounts of precious metals, and thermodynamics wants very much to have those dense elements settle towards the center. The conditions are now right for that to happen, and that's exactly what happened shortly after the Earth formed.
Finally, there's stellar physics. The Earth would have a tiny little core of rare but dense elements if iron and nickel were as rare as gold and platinum. That's not the case. Iron and nickel are surprisingly abundant elements in the universe. There's a general tendency for heavier elements to be less abundant. Iron (and to a lesser extent, nickel) are two exceptions to this rule; see the graph below. Iron and nickel are where the alpha process in stellar physics stops. Everything heavier than iron requires exotic processes such as the s-process or those that occur in a supernova to create them. Moreover, supernova, particularly type Ia supernovae, are prolific producers of iron. Despite their relatively heavy masses, iron and nickel are quite abundant elements in our aging universe.
(source: virginia.edu)
Best Answer
You are getting different answers from NASA and from other sources, as they are talking about slightly different things.
NASA is talking about the acceleration of the GRACE satellite towards the earth, as it orbited over different regions. When it went over the Himalayas, for example, the acceleration (gravity) was higher than average.
Other sources are talking about the difference in acceleration due to gravity at ground level, compared to if you were to walk up the Himalayas, then the acceleration would decrease. That's because even though there would be more mass underneath, you've increased the distance from the earth.
More detail:
At the bottom of a cone shaped mountain of mass $m$, radius $r$ and height $r$, the acceleration due to gravity is $g$, due to the earth of mass $M$, radius $R$.
$$g=\frac{GM}{R^2}\tag1$$
the difference in gravity after climbing the mountain is
$$\frac{GM}{{(R+r)}^2}+\frac{Gm}{{(\frac{3}{4}r)}^2} - g\tag2$$
The 3/4 is due to the position of the COM of a cone. Using 1) it's
$$g\bigl((1+\frac{r}{R})^{-2}+\frac{16mR^2}{9Mr^2}-1\bigr)\tag3$$
From formulae for the volume of a sphere and a cone and assuming equal density
$$\frac{m}{M} = \frac{r^3}{4R^3}\tag4$$
so 3) becomes, in terms of $g$
$$y= (1+\frac{r}{R})^{-2}+\frac{4r}{9R}-1\tag5$$,
putting $x = \frac{r}{R}$
$$y= (1+x)^{-2}+\frac{4}{9}x-1\tag6$$
plotting this
shows that there is a decrease in the acceleration due to gravity for all realistic cone shaped mountains.
For Everest, if it were a cone, $x=0.0014$ and the reduction in gravity is $y=0.002g$, so the usual $9.81$ becomes about $9.79 \;\text{m}\,\text{s}^{-2}$.