Does gravity get stronger the higher up you are on a mountain?

The strength of gravity does vary from point to point on the Earth's surface, because the earth is not of even mass distribution. The images in the link you have provided are evidence of this.

The strength of the earth’s gravitational field is given by $$a=frac{GM}{r^2}$$ where $M$ is the mass of the earth, $G$ is the gravitational constant, and $r$ is the distance from the center of the earth.

Since it is inversely proportional to the distance from center squared, the further you move away from earth, the weaker this field strength is.

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.

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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

$$frac{GM}{{(R+r)}^2}+frac{Gm}{{(frac{3}{4}r)}^2} - gtag2$$

The 3/4 is due to the position of the COM of a cone. Using 1) it's

$$gbigl((1+frac{r}{R})^{-2}+frac{16mR^2}{9Mr^2}-1bigr)tag3$$

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 $g$

$$y= (1+frac{r}{R})^{-2}+frac{4r}{9R}-1tag5$$,

putting $x = frac{r}{R}$

$$y= (1+x)^{-2}+frac{4}{9}x-1tag6$$

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 $9.79 ;text{m},text{s}^{-2}$.

I looked at the link you gave, I think it may not mean to say the higher up you go on a mountain, the stronger the gravitational field. The link's meaning, I suppose (because they mentioned it is measured by satellites, I suppose it was measuring gravity fields at the altitude of the satellites, which supposedly stayed the same for the whole measurement) is something like this: they measured gravity at the same altitude around the globe, and found that at this altitude and at spots with Mountain ranges right below, such as the Himalayas, due to the high concentration of mass, the gravity field is stronger. While at the same altitude and at spots with Ocean trenches right below, such as the Mariana Trench, due to the low concentration of mass, they measure a weak gravity field, as expected.

There are two factors here working against each other. One is the distribution of mass of earth near the spot, which is higher near the mountains and tend to increase the acceleration due to gravity at that spot. And the other factor is the distance from the all the other mass from the other parts of the earth, which is also higher near the mountain and hence tends to decrease the acceleration due to gravity.

The net result will depend on the magnitude of these two counteracting effects, and will depend on the particular details of the mountain in question.

But the point raised in the other reply is correct about the particular NASA page you have linked. That is not a map of acceleration due to gravity on the surface of the earth. That is one measured by a satellite in orbit. For a satellite in orbit, obviously the second factor is negligible, thus, only the first factor dominates, and hence the mountain areas show higher acceleration due to gravity.

Probably not by itself.

The gravitational acceleration of a spherically symmetric body is given by: $$ g = frac{GM}{r^2} $$ where $r$ is the distance from the center. Assuming the uniform density, this would be: $$ g = frac{4pi}3frac{Grho R_0^3}{r^2} $$ where $R_0$ is the radius of the body. When standing on the surface ($r=R_0$), this becomes $$ g_0 = frac{4pi}3 Grho R_0 $$

Let's do some math for two extreme cases:

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1) First, let's consider a spherical "mountain" of radius $R_1$ ($R_1 ll R_0$) and the same density $rho$. The gravitational acceleration when standing on the top of that "mountain" is: $$ begin{split} g&=frac{4pi}3Grholeft(frac{R_0^3}{(R_0+2R_1)^2} + R_1right) = left(frac{R_0^2}{(R_0+2R_1)^2} + frac{R_1}{R_0}right) g_0 &equiv left(frac1{(1+2epsilon)^2} + epsilonright) g_0 =left(1 + frac{(epsilon - 1)(1+2epsilon)^2 + 1}{(1+2epsilon)^2}right)g_0 =left(1 + epsilonfrac{4epsilon^2 - 3}{(1+2epsilon)^2}right)g_0 < g_0 end{split} $$ where $epsilon = R_1/R_0 ll 1$. The inequality $g<g_0$ follows from $0<epsilon<sqrt3/2$.

Therefore, a "spiky" mountain, which could be roughly approximated by a sphere, will likely have a lower surface gravity on its top than the planetary average.

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2) Now, let's consider a high plateau (also of the same density $rho$) that extends long enough that its gravity on the middle of its surface can be approximated by the gravity of an infinite plate, but still negligible compared to the whole planet, i.e. a plateau of height $h$ and horizontal dimension $l$ such that $h ll l ll R_0$.

Gravitational acceleration of an infinite plane is $g_{text{plane}} = 2pi Gsigma$, where $sigma$ is the surface density, which for the plateau is $rho h$. The gravitational acceleration when standing on top of the plateau is then approximately: $$ begin{split} g&=frac{4pi}3 Grho frac{R_0^3}{(R_0+h)^2} + 2pi Grho h = left( frac{R_0^2}{(R_0+h)^2} + frac{3h}2 right)g_0 &equiv left(frac1{(1+2epsilon)^2} + 3epsilonright) g_0 =left(1 + frac{(3epsilon - 1)(1+2epsilon)^2 + 1}{(1+2epsilon)^2}right)g_0 &=left(1 + epsilonfrac{12epsilon^2+8epsilon-1}{(1+2epsilon)^2}right)g_0 < g_0 end{split} $$ where $epsilon = 2R_0/h ll 1$. Again, the inequality follows from $epsilon$ being positive and sufficiently small. In this case, less than $(sqrt7-2)/3$.

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In both cases it turns out that the surface gravity will be lower than the planetary average. It is reasonable to assume that any reasonable mountains will be small bumps that are, regarding their gravitational acceleration, between a ball and (infinite) plateau. Therefore, if uniform densities are assumed, they will

However, all of the above considerations were under the assumption of an uniform density. If the mountain has sufficiently higher density compared to the rest of the planet (less likely) or if non-uniform density in the planet is distributed in a convenient way (more likely), then gravity will be stronger at the top of the mountain. But note that this will be because of densities, not because of the mountain. In a typical case, the surface gravity is actually likely to be lower.

If you are a satellite 6870 km above the center of the Earth, and directly below you there is flat land, you will experience some gravity. If you move on to another point, also 6870 km above the center, but this time there is a huge mountain below you, then this time you will feel a gravity which is slightly larger.

If you are a person standing on the surface of the Earth, 6370 km above its center, you will feel some gravity. If from there, you climb a 4 km high mountain, after that you will be 6374 km above Earth's center. Because your distance from most of Earth increases, the gravity you feel is slightly smaller.

(In both examples, the latitude should be the same before and after, otherwise the oblateness of Earth's shape (and in the second example also the related centrifugal force due to Earth's rotation) will influence the result.)