<p>I gave this one some thought and ran some numbers.</p>

<p>The real answer to your question is that you would be lighter because the material that makes up continental plates is lighter than the basalt rock under it.  It might not look that way but continents, mountains and oceans are actually balanced with about equal weight under them.  Continents are above sea level because of larger amounts of crust which is lighter than basalt.  The earth, which is pretty malleable as a whole, adjusts for significant unbalanced changes in mass, it's still rebounding from the melted glaciers from the last ice age for example.</p>

<p>This happens with Mountains too, also slowly, but mountains are formed over millions of years so the adjustment has time to take place.  Mountains are formed when two continental plates push against each other, one slides under the other, so both what forms the mountain, and what goes under the mountain is lighter than basalt under the Earth's crust.   When you climb a mountain, you're not actually on top of more mass because much of the rock under you is lighter rock.  Mountains are actually puffy (I always wanted to say that).    If mountains were actually greater mass relative to the (relatively) spherical Earth, then they'd sink into the Earth, which they do to some extent, but, as I said above, Mountains are <strike>puffy</strike> buoyant compared to the Earth's crust, so the mass is pretty much the same on top of the mountain and at the foot of the mountain.  </p>

<p><a href="https://i.sstatic.net/tBBcO.png" rel="noreferrer"><img src="https://i.sstatic.net/tBBcO.png" alt="enter image description here"></a></p>

<p>That's not to say the Earth is perfectly balanced, it's definitely not, but the imbalance doesn't appear to follow mountain ranges.  Google "Earth Gravity Map" for more on that.  Article with pretty picture <a href="http://news.bbc.co.uk/2/hi/sci/tech/8767763.stm" rel="noreferrer">here</a>.</p>

<p>But I suspect, your question wasn't just real world but also a mathematical query, so, moving on.   If we create a cone shaped mountain, a large flat one, lets say it's 6.371 km tall (reason I picked that number will become clear soon), and lets give it 4 times that as a radius or 25.484 km radius to it's base and a specific density of 2.7, the mass of our mountain, using 1/3 base times height is:  3.1416 * .333333 * 2.7 * 25,484^2 * 6,371 * 1000  (the 1,000 is for density of 1,000 KG per cubic meter at specific density of 1) = about 1.17x10^16 KG.</p>

<p>So, on top of this artificial mountain that, if it was actually added to the earth, it would push the earth down and the earth would adjust it's mass around it over time, but ignoring that, the mountain is 1/1000th the radius of the earth so the distance from the Earth's center is increased by 0.1%</p>

<p>The mass of the mountain, 1.17E16kg is about 1/510,000,000 the mass of the Earth, 5.97E24kg, So we have a mass increase of about 0.0000002% and a corresponding reduction in gravitation by increased distance to the center of the earth (the square of 0.1%), or 0.0001.   or, 510 times as much is lost by the higher altitude than is gained by the increase in mass.  Now if you make the mountain flat enough, you'd probably be able to increase gravity the tiniest bit, but past a certain point the foot of the mountain is so spread out that the calculation grows inaccurate.  The short answer is simply, no, as the other two posters have said.  The mass of any mountain is simply insufficient to increase the force of gravity more than it loses by what's lost in the higher altitude.</p>

<p>A 3rd real factor, completely unrelated to your question is air density.  Heavier air does create some buoyancy which reduces our weight and the gravity we feel, this effect varies with high and low pressure and temperature, but on an average day at sea level air buoyancy reduces our mass by about .12%, so if you weighed yourself in a vacuum chamber vs your regular bathroom routine, you'd find you weigh about 2 or 3 ounces more for an average person.   Climb a mountain, the thinner air, you might gain an ounce due to loss of air buoyancy and that more than makes up for any small changes in gravity, so if you measure gravity by a scale and a weight, things weigh more on top of a mountain, because of thinner air, even though the G force is slightly lower.</p>

<p>The rotation of the earth and the centrifugal force also decreases mass, depending on how close the mountain is to the equator, (see picture).  as does the bulge of the Earth around it's equator, but all these changes are tiny.  </p>

<p><a href="https://i.sstatic.net/t30rF.gif" rel="noreferrer"><img src="https://i.sstatic.net/t30rF.gif" alt="https://qph.ec.quoracdn.net/main-qimg-58036f61af8f1d01722b7149dda1bc5a"></a></p>

CC BY-SA 4.0
2021-01-14T00:40:56.970Z
Great answer, you built a hypothetical mountain! And atmospheric buoyancy very interesting too. One correction: earth gravity map images do seem to show that the positive gravity anomalies line up with mountain ranges
Amphibio
CC BY-SA 3.0
2016-06-21T09:03:39.420Z
2016-06-21T22:36:30.063Z
2016-06-21T22:36:30.063Z
userLTK
<p>While it is true that, at least over time, Earth will equilibrate so the mountain sinks down, displacing the denser material below, this would also mean that the material just beside the mountain is less dense, and thus there is a similar reduction in g due to this when standing beside it.</p>
<p>And while the increased mass is quite small relative to Earth, you are also a lot closer to it. So you need to consider both factors.</p>
<p>If we do a first order approximation for the change in gravity due to the height of the mountain, we get;
<span class="math-container">$$
dg=\frac{2 G M h}{R^3}
$$</span>
where <span class="math-container">$R$</span> is the radius of Earth, <span class="math-container">$M$</span> is the mass of Earth and <span class="math-container">$h$</span> is the height of the mountain.</p>
<p>If we treat the mountain as a sphere with a diameter of <span class="math-container">$h$</span> and a density of <span class="math-container">$\rho$</span>, then the acceleration due to gravity when standing on top of it is:
<span class="math-container">$$
a=\frac{G \rho (4/3) \pi (h/2)^3}{(h/2)^2}
=\frac{2 G \pi \rho h}{3}
$$</span></p>
<p>And with that we end up with a ratio of the 2 forces dependent upon the density of the mountain (as both are linearly dependent upon the height)
<span class="math-container">$$
\frac{a}{dg}=\frac{\pi \rho R^3}{3M}
$$</span>
So for a spherical mountain you need a density of roughly 22000 kg/m^3, e.g. a mountain of osmium, to have the gravity due to the mountain increase your weight more than the extra height decreases it.</p>
<p>Having mass in addition to that sphere (without taking away from that sphere) would reduce that density requirement.
But even if you modelled it as a cylinder where h=d to calculate the mass, and overestimated the force due to gravity by treating it as a sphere/point mass, you end up needing a density of roughly 15000 kg/m^3.</p>
<p>However the effect is much greater than the 0.2% suggested by userLTK. With a specific gravity of 2.7, the effect of the mass of the spherical mountain is 12% that of the effect of the increased distance from Earth.</p>

CC BY-SA 4.0
2021-01-14T00:46:03.937Z
Mount Osmium it is then
Amphibio
CC BY-SA 4.0
2020-11-17T21:11:42.367Z
2020-11-17T21:11:42.367Z
Jack Black
<p>A map of measurements can be found <a href="https://en.wikipedia.org/wiki/Gravity_of_Earth" rel="nofollow noreferrer">here</a>.</p>
<p>This appears to show that some of the highest values are found in the Andes and in parts of the Hymalayas.</p>
<p>I find this more than unexpected, so I suspect that this is the gravity referenced to some fixed height above sea-level rather than the surface gravity.</p>
<p>Jack Black's answer using a small sphere outside the surface seemed to me too extreme to establish the maximal case for real hills. It seems to me that an extreme model that looked at the maximum effect of a non-equilibrium covering be more definitive. So I have looked at the effect of a &quot;mountain&quot; that was so wide that it could be treated as a covering of uniform thickness across the entire globe.</p>
<p>Taking the field of the Earth at height h above the surface, and without covering we have</p>
<p><span class="math-container">$$g_0 = 5510 \cdot G \cdot \frac{4 \pi}{3} \frac{R^3}{(R+h)^2} \ \ \ \ \ \ \ \ \ \ \ \frac{m}{s^2}$$</span></p>
<p>(5510 <span class="math-container">$\frac{kg}{m^3}$</span> being the mean density of the Earth, other symbols taking their usual meanings). The rate of change just above the surface is <span class="math-container">$$-2 \cdot 5510 \cdot G \cdot \frac{4 \pi}{3} \ \ \ \ \ \ \ \ \ \ \frac{m}{s^2m}$$</span> The rate of change of g at the surface of sphere with covering density <span class="math-container">$ \rho$</span> must then be <span class="math-container">$$3 \cdot \rho \cdot G \cdot \frac{4 \pi}{3} \ \ \ \ \ \ \ \ \ \frac{m}{s^2m}$$</span>
It seems that the uniform surface covering must have a density more than 2/3 of the mean density of the Earth beneath it if gravity is to increase. This would be a rock with specific gravity of more than 3.67 if there is to be any chance that the gravity at the top of the &quot;mountain&quot; exceeds that at the bottom.  The S.G. of the densest basalt is less than 3.1, and granite is even lighter (below 2.8).</p>
<p>This means that there is no realistic chance that a mountainous coating could result in an increase in gravity.</p>
<p>So the situation where hills/mountains might provide gravitational increases would be where we have steep inclines rising out of deep sea, as these provide an increase in density at levels below the sea-level surface. They would still not provide increases above standard gravity unless the total crust thickness in the region was less than average.</p>

CC BY-SA 4.0
2021-01-13T21:41:20.053Z
Welcome on the ES SE! We have quite good latex support here, type in `$\frac{4\pi}{3}$` and you will get $\frac{4\pi}{3}$. I could not understand some of your formulas at the middle of your answer, I only partially latexified them. Please fix them, they absolutely do not look well, not in dimensionality and not numerically.
peterh
CC BY-SA 4.0
2021-01-14T08:32:16.400Z
Perhaps $\frac{3 \rho G}{\frac{4}{3}\pi}$? is the unresolved formula, not sure.
JeopardyTempest
CC BY-SA 4.0
2021-01-14T23:08:10.473Z
Thank you both for your assistance - I'm new to Latex.  I've left the units for rate-of-change of gravity as acceleration-change per unit height rather than cancelling - I hope this makes things clearer.   Please let me know if anything more needs clarification.  
Thanks again
george storm
CC BY-SA 4.0
2021-01-14T23:25:28.600Z
PS  You probably deduced that I am accustomed to stating units only where they are not already implied.  I now appreciate that this is not appropriate here,
george storm
CC BY-SA 4.0
2021-01-13T18:04:24.477Z
2021-01-14T23:25:09.443Z
2021-01-14T23:25:09.443Z
george storm
<p>According to the gravity map below from NASA's GRACE, the mountains you'd want to try for this experiment would be the Andes - they have the highest positive gravity anomaly with sea level quite nearby.</p>
<p>However, lets do a bit of calculation based on this data: The map shows that the coast near the Andes is approximately -20 miligal (a unit of gravitational acceleration) and the centre of the Andes mountain range is around +70 miligal - an overall differential of +90 miligal at the middle of the mountains.</p>
<p>The high point on the Andes is about 7000 metres from sea level. The standard value used in geodesy (from a physics stack exchange answer, link below) is a reduction of 0.3086 miligals per meter of increased altitude. At 7000 metres, that gives us a value of -2160 miligal, or -2.16 gal, which is equivalent to a change of -0.0216m/s/s - significantly outstripping the gravity anomaly on the map (which is modelled on a theorised homogenous spherical earth)</p>
<p>So the answer to your question is no- the mountain tops in areas with highest gravity anomalies will still have (very slightly!) lower gravity than nearby sea level areas</p>
<p><img src="https://i.sstatic.net/8Uqgg.jpg" alt="GRACE gravity anomaly map" /></p>
<p>image source: <a href="https://www.jpl.nasa.gov/images/new-views-of-earths-gravity-field-from-grace/" rel="nofollow noreferrer">https://www.jpl.nasa.gov/images/new-views-of-earths-gravity-field-from-grace/</a></p>
<p>gravity calculation: <a href="https://physics.stackexchange.com/questions/232878/do-we-feel-0-4-less-body-weight-on-mount-everest#:%7E:text=That%27s%20a%20reduction%20of%200.28,Everest">https://physics.stackexchange.com/questions/232878/do-we-feel-0-4-less-body-weight-on-mount-everest#:~:text=That's%20a%20reduction%20of%200.28,Everest</a>.</p>

CC BY-SA 4.0
2021-01-13T22:08:13.750Z
2021-01-14T08:22:44.187Z
2021-01-14T08:22:44.187Z
Amphibio
<p>Gravity is strictly a function of mass, distance, and gravitational constants.  The higher the mountain, the less effect gravity will have.  The amount is relatively trivial, but can be computed and measured as a distance from the center of gravity.  </p>

<p>As compared to sea level, yes, their are mountain tops with gravity effects trivially higher, mountains with peaks below sea level.  There are entire mountain ranges under the oceans.</p>

CC BY-SA 3.0
2016-06-20T21:21:29.857Z
... but what about the increased mass of the mountain underneath?
Daniel Griscom
CC BY-SA 3.0
2016-06-21T01:16:39.403Z
@DanielGriscom  Running the math gets impossibly complicated, but while there's more mass under you on top of the mountain as opposed to on the side of the mountain, you're also further from the Earth's center of mass.   Taken in combination, gravity really does go down as you climb up, but the change is very small.
userLTK
CC BY-SA 3.0
2016-06-21T01:37:17.607Z
@userLTK It must depend on just how much mountain is under you. You can imagine a thin tower of a mountain where it would be just like hanging in air, and clearly the gravity will be reduced. But, what about being on a broad shield volcano?
Daniel Griscom
CC BY-SA 3.0
2016-06-21T03:07:48.467Z
@DanielGriscom  I'll try to run the math, in a very simplified way.
userLTK
CC BY-SA 3.0
2016-06-22T05:43:47.313Z
I think what 'Marijn' was getting at is not just the mountain underneath, but the gravitational affect of the mountain to the side, as in the case of a Hawaiian type shield volcano. Maybe 'userLTK' can crunch the numbers for that scenario?
Gordon Stanger
CC BY-SA 3.0
2016-06-27T22:23:01.190Z
Effects of being beside, in around mountains or other masses are real, but trivial and can be ignored in scale.  The mass of Mount Everest, though it is large is our mind, is negligible when you are standing next to it as compared to the mass of the Earth which is under you.  The geometry and associated mathematics of treating such scenarios is complex, and becomes a futile exercise which in the end yields an almost identical answer to doing the calculation as a point mass problem with you, the center of the Earth and the distance between.
dlb
CC BY-SA 3.0
2016-06-20T20:37:53.027Z
2016-06-20T20:37:53.027Z
dlb
<p>The gravitational acceleration decreases with altitude. A person for example; flying at an altitudde of 30,000 feet; above sea level over say............ mountains will feel more gravity than someone at the same elevation but over the sea. On the other hand, a person standing on the Earth's surface feels less gravity when the elevation is higher than say Gravitational acceleration at height above sea level. Though minute, the further from the Earth's center of mass.</p>

CC BY-SA 4.0
2021-01-17T01:11:00.720Z
2021-01-17T01:11:00.720Z
LazyReader
<p>The most effective geometry for gravity is a sphere. Mountain top is farther from center of earth and the density of basalt or granite is less than the mantle or the earth core. Imagine an oblong earth ultimately similar to a rectangular shape but same mass.<br>
If you would measure the gravity on the tip of this rectangle it would be lees even though it is on top of a huge mass.<br>
 The fact that planets are round indicates all the heavy primitive crest unevenness has crashed down to surface but kept afloat by the lateral pressure if adjacent material compressed by higher gravity than what was available at the altitude.</p>

CC BY-SA 3.0
2016-06-21T06:56:25.837Z
2016-06-21T06:56:25.837Z
kamran
<p>Regardless the latitude and centrifugial 'force' is the gravity on mountains higher or lower than at sealevel? </p>

<p>Gravity is depending on the radius to the centre and the mass. Now if you are on a mountain you are further from centre but it looks like you have more mass under you. Probably the differences would be tiny but is it likely that on mountains the gravity is higher or lower than at see level? As far as I could see till now where there are mountains the gravity is lower, but are there mountains where the gravity is higher?</p>

<geophysics>
Does a mountain top have higher gravity than a nearby sea-level surface?
Marijn 
# Does a mountain top have higher gravity than a nearby sea-level surface?

(Asked by: username_0 on 2016-06-20)

Regardless the latitude and centrifugial 'force' is the gravity on mountains higher or lower than at sealevel?

Gravity is depending on the radius to the centre and the mass. Now if you are on a mountain you are further from centre but it looks like you have more mass under you. Probably the differences would be tiny but is it likely that on mountains the gravity is higher or lower than at see level? As far as I could see till now where there are mountains the gravity is lower, but are there mountains where the gravity is higher?

**Tags:** geophysics


---

## 7 answers

### Answer by username_1 on 2016-06-21. Score: 9

I gave this one some thought and ran some numbers.

The real answer to your question is that you would be lighter because the material that makes up continental plates is lighter than the basalt rock under it. It might not look that way but continents, mountains and oceans are actually balanced with about equal weight under them. Continents are above sea level because of larger amounts of crust which is lighter than basalt. The earth, which is pretty malleable as a whole, adjusts for significant unbalanced changes in mass, it's still rebounding from the melted glaciers from the last ice age for example.

This happens with Mountains too, also slowly, but mountains are formed over millions of years so the adjustment has time to take place. Mountains are formed when two continental plates push against each other, one slides under the other, so both what forms the mountain, and what goes under the mountain is lighter than basalt under the Earth's crust. When you climb a mountain, you're not actually on top of more mass because much of the rock under you is lighter rock. Mountains are actually puffy (I always wanted to say that). If mountains were actually greater mass relative to the (relatively) spherical Earth, then they'd sink into the Earth, which they do to some extent, but, as I said above, Mountains are puffy buoyant compared to the Earth's crust, so the mass is pretty much the same on top of the mountain and at the foot of the mountain.

[![enter image description here](https://i.sstatic.net/tBBcO.png)](https://i.sstatic.net/tBBcO.png)

That's not to say the Earth is perfectly balanced, it's definitely not, but the imbalance doesn't appear to follow mountain ranges. Google "Earth Gravity Map" for more on that. Article with pretty picture [here](http://news.bbc.co.uk/2/hi/sci/tech/8767763.stm).

But I suspect, your question wasn't just real world but also a mathematical query, so, moving on. If we create a cone shaped mountain, a large flat one, lets say it's 6.371 km tall (reason I picked that number will become clear soon), and lets give it 4 times that as a radius or 25.484 km radius to it's base and a specific density of 2.7, the mass of our mountain, using 1/3 base times height is: 3.1416 \* .333333 \* 2.7 \* 25,484^2 \* 6,371 \* 1000 (the 1,000 is for density of 1,000 KG per cubic meter at specific density of 1) = about 1.17x10^16 KG.

So, on top of this artificial mountain that, if it was actually added to the earth, it would push the earth down and the earth would adjust it's mass around it over time, but ignoring that, the mountain is 1/1000th the radius of the earth so the distance from the Earth's center is increased by 0.1%

The mass of the mountain, 1.17E16kg is about 1/510,000,000 the mass of the Earth, 5.97E24kg, So we have a mass increase of about 0.0000002% and a corresponding reduction in gravitation by increased distance to the center of the earth (the square of 0.1%), or 0.0001. or, 510 times as much is lost by the higher altitude than is gained by the increase in mass. Now if you make the mountain flat enough, you'd probably be able to increase gravity the tiniest bit, but past a certain point the foot of the mountain is so spread out that the calculation grows inaccurate. The short answer is simply, no, as the other two posters have said. The mass of any mountain is simply insufficient to increase the force of gravity more than it loses by what's lost in the higher altitude.

A 3rd real factor, completely unrelated to your question is air density. Heavier air does create some buoyancy which reduces our weight and the gravity we feel, this effect varies with high and low pressure and temperature, but on an average day at sea level air buoyancy reduces our mass by about .12%, so if you weighed yourself in a vacuum chamber vs your regular bathroom routine, you'd find you weigh about 2 or 3 ounces more for an average person. Climb a mountain, the thinner air, you might gain an ounce due to loss of air buoyancy and that more than makes up for any small changes in gravity, so if you measure gravity by a scale and a weight, things weigh more on top of a mountain, because of thinner air, even though the G force is slightly lower.

The rotation of the earth and the centrifugal force also decreases mass, depending on how close the mountain is to the equator, (see picture). as does the bulge of the Earth around it's equator, but all these changes are tiny.

[![https://qph.ec.quoracdn.net/main-qimg-58036f61af8f1d01722b7149dda1bc5a](https://i.sstatic.net/t30rF.gif)](https://i.sstatic.net/t30rF.gif)

> **Comments:**
> **username_2**: Great answer, you built a hypothetical mountain! And atmospheric buoyancy very interesting too. One correction: earth gravity map images do seem to show that the positive gravity anomalies line up with mountain ranges  

---

### Answer by username_2 on 2021-01-13. Score: 5

According to the gravity map below from NASA's GRACE, the mountains you'd want to try for this experiment would be the Andes - they have the highest positive gravity anomaly with sea level quite nearby.

However, lets do a bit of calculation based on this data: The map shows that the coast near the Andes is approximately -20 miligal (a unit of gravitational acceleration) and the centre of the Andes mountain range is around +70 miligal - an overall differential of +90 miligal at the middle of the mountains.

The high point on the Andes is about 7000 metres from sea level. The standard value used in geodesy (from a physics stack exchange answer, link below) is a reduction of 0.3086 miligals per meter of increased altitude. At 7000 metres, that gives us a value of -2160 miligal, or -2.16 gal, which is equivalent to a change of -0.0216m/s/s - significantly outstripping the gravity anomaly on the map (which is modelled on a theorised homogenous spherical earth)

So the answer to your question is no- the mountain tops in areas with highest gravity anomalies will still have (very slightly!) lower gravity than nearby sea level areas

![GRACE gravity anomaly map](https://i.sstatic.net/8Uqgg.jpg)

image source: <https://www.jpl.nasa.gov/images/new-views-of-earths-gravity-field-from-grace/>

gravity calculation: [https://physics.stackexchange.com/questions/232878/do-we-feel-0-4-less-body-weight-on-mount-everest#:~:text=That's%20a%20reduction%20of%200.28,Everest](https://physics.stackexchange.com/questions/232878/do-we-feel-0-4-less-body-weight-on-mount-everest#:%7E:text=That%27s%20a%20reduction%20of%200.28,Everest).


---

### Answer by username_3 on 2020-11-17. Score: 4

While it is true that, at least over time, Earth will equilibrate so the mountain sinks down, displacing the denser material below, this would also mean that the material just beside the mountain is less dense, and thus there is a similar reduction in g due to this when standing beside it.

And while the increased mass is quite small relative to Earth, you are also a lot closer to it. So you need to consider both factors.

If we do a first order approximation for the change in gravity due to the height of the mountain, we get;
$$
dg=\frac{2 G M h}{R^3}
$$
where $R$ is the radius of Earth, $M$ is the mass of Earth and $h$ is the height of the mountain.

If we treat the mountain as a sphere with a diameter of $h$ and a density of $\rho$, then the acceleration due to gravity when standing on top of it is:
$$
a=\frac{G \rho (4/3) \pi (h/2)^3}{(h/2)^2}
=\frac{2 G \pi \rho h}{3}
$$

And with that we end up with a ratio of the 2 forces dependent upon the density of the mountain (as both are linearly dependent upon the height)
$$
\frac{a}{dg}=\frac{\pi \rho R^3}{3M}
$$
So for a spherical mountain you need a density of roughly 22000 kg/m^3, e.g. a mountain of osmium, to have the gravity due to the mountain increase your weight more than the extra height decreases it.

Having mass in addition to that sphere (without taking away from that sphere) would reduce that density requirement.
But even if you modelled it as a cylinder where h=d to calculate the mass, and overestimated the force due to gravity by treating it as a sphere/point mass, you end up needing a density of roughly 15000 kg/m^3.

However the effect is much greater than the 0.2% suggested by username_1. With a specific gravity of 2.7, the effect of the mass of the spherical mountain is 12% that of the effect of the increased distance from Earth.

> **Comments:**
> **username_2**: Mount Osmium it is then  

---

### Answer by username_4 on 2021-01-13. Score: 3

A map of measurements can be found [here](https://en.wikipedia.org/wiki/Gravity_of_Earth).

This appears to show that some of the highest values are found in the Andes and in parts of the Hymalayas.

I find this more than unexpected, so I suspect that this is the gravity referenced to some fixed height above sea-level rather than the surface gravity.

username_3 username_3's answer using a small sphere outside the surface seemed to me too extreme to establish the maximal case for real hills. It seems to me that an extreme model that looked at the maximum effect of a non-equilibrium covering be more definitive. So I have looked at the effect of a "mountain" that was so wide that it could be treated as a covering of uniform thickness across the entire globe.

Taking the field of the Earth at height h above the surface, and without covering we have

$$g\_0 = 5510 \cdot G \cdot \frac{4 \pi}{3} \frac{R^3}{(R+h)^2} \ \ \ \ \ \ \ \ \ \ \ \frac{m}{s^2}$$

(5510 $\frac{kg}{m^3}$ being the mean density of the Earth, other symbols taking their usual meanings). The rate of change just above the surface is $$-2 \cdot 5510 \cdot G \cdot \frac{4 \pi}{3} \ \ \ \ \ \ \ \ \ \ \frac{m}{s^2m}$$ The rate of change of g at the surface of sphere with covering density $ \rho$ must then be $$3 \cdot \rho \cdot G \cdot \frac{4 \pi}{3} \ \ \ \ \ \ \ \ \ \frac{m}{s^2m}$$
It seems that the uniform surface covering must have a density more than 2/3 of the mean density of the Earth beneath it if gravity is to increase. This would be a rock with specific gravity of more than 3.67 if there is to be any chance that the gravity at the top of the "mountain" exceeds that at the bottom. The S.G. of the densest basalt is less than 3.1, and granite is even lighter (below 2.8).

This means that there is no realistic chance that a mountainous coating could result in an increase in gravity.

So the situation where hills/mountains might provide gravitational increases would be where we have steep inclines rising out of deep sea, as these provide an increase in density at levels below the sea-level surface. They would still not provide increases above standard gravity unless the total crust thickness in the region was less than average.

> **Comments:**
> **username_5**: Welcome on the ES SE! We have quite good latex support here, type in `$\frac{4\pi}{3}$` and you will get $\frac{4\pi}{3}$. I could not understand some of your formulas at the middle of your answer, I only partially latexified them. Please fix them, they absolutely do not look well, not in dimensionality and not numerically.  
> **username_6**: Perhaps $\frac{3 \rho G}{\frac{4}{3}\pi}$? is the unresolved formula, not sure.  
> **username_4**: Thank you both for your assistance - I'm new to Latex.  I've left the units for rate-of-change of gravity as acceleration-change per unit height rather than cancelling - I hope this makes things clearer.   Please let me know if anything more needs clarification.   Thanks again  
> **username_4**: PS  You probably deduced that I am accustomed to stating units only where they are not already implied.  I now appreciate that this is not appropriate here,  

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### Answer by username_7 on 2016-06-20. Score: 0

Gravity is strictly a function of mass, distance, and gravitational constants. The higher the mountain, the less effect gravity will have. The amount is relatively trivial, but can be computed and measured as a distance from the center of gravity.

As compared to sea level, yes, their are mountain tops with gravity effects trivially higher, mountains with peaks below sea level. There are entire mountain ranges under the oceans.

> **Comments:**
> **username_8**: ... but what about the increased mass of the mountain underneath?  
> **username_1**: @username_8  Running the math gets impossibly complicated, but while there's more mass under you on top of the mountain as opposed to on the side of the mountain, you're also further from the Earth's center of mass.   Taken in combination, gravity really does go down as you climb up, but the change is very small.  
> **username_8**: @username_1 It must depend on just how much mountain is under you. You can imagine a thin tower of a mountain where it would be just like hanging in air, and clearly the gravity will be reduced. But, what about being on a broad shield volcano?  
> **username_1**: @username_8  I'll try to run the math, in a very simplified way.  
> **username_9**: I think what 'username_0' was getting at is not just the mountain underneath, but the gravitational affect of the mountain to the side, as in the case of a Hawaiian type shield volcano. Maybe 'username_1' can crunch the numbers for that scenario?  
> **username_7**: Effects of being beside, in around mountains or other masses are real, but trivial and can be ignored in scale.  The mass of Mount Everest, though it is large is our mind, is negligible when you are standing next to it as compared to the mass of the Earth which is under you.  The geometry and associated mathematics of treating such scenarios is complex, and becomes a futile exercise which in the end yields an almost identical answer to doing the calculation as a point mass problem with you, the center of the Earth and the distance between.  

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### Answer by username_10 on 2021-01-17. Score: 0

The gravitational acceleration decreases with altitude. A person for example; flying at an altitudde of 30,000 feet; above sea level over say............ mountains will feel more gravity than someone at the same elevation but over the sea. On the other hand, a person standing on the Earth's surface feels less gravity when the elevation is higher than say Gravitational acceleration at height above sea level. Though minute, the further from the Earth's center of mass.


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### Answer by username_11 on 2016-06-21. Score: 0

The most effective geometry for gravity is a sphere. Mountain top is farther from center of earth and the density of basalt or granite is less than the mantle or the earth core. Imagine an oblong earth ultimately similar to a rectangular shape but same mass.  
If you would measure the gravity on the tip of this rectangle it would be lees even though it is on top of a huge mass.  
The fact that planets are round indicates all the heavy primitive crest unevenness has crashed down to surface but kept afloat by the lateral pressure if adjacent material compressed by higher gravity than what was available at the altitude.


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