Does gravity get stronger when you climb a mountain?

Question

As stated in the question title, what happens to the strength of the gravitational field (or equivalently, your weight) as you climb a hill or mountain? Would a weighing scale show that you were lighter or heavier at the summit than at sea level?

I've been wrangling with this question ever since a walking trip (up and down hills unsurprisingly). Looking at the equation for gravitational field strength: $$ g = \frac{GM}{r^2} $$

Intuitively, I think of it in 2 way - there's more stuff (mass) beneath you when you're at the top of a mountain, pulling you down with its gravity - so a larger $M$ means larger $g$ and you would appear heavier. Conversely, moving up a hill means moving away from the centre of mass of the Earth, so $r$ increases and $g$ drops. Surely therefore there's some combination of the two factors that would leave your weight unchanged?

Thinking about edge cases;

• On the one hand you could imagine standing on a really tall, really narrow column. Clearly gravity is weaker at the top, because the column has negligible mass but you've moved significantly further away from the centre of the Earth
• On the other hand, imagine adding another layer of crust to the Earth. That's a bit like a really wide mountain. We know super-Earths - planets with similar composition to, but larger radius than Earth- have stronger surface gravity. So in this way your weight would be seen to increase.

Some combination of those two edge cases, a wide flat plateau perhaps, or a shallow gradient mountain, feels like it should preserve your weight once you get to the top.

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In my head I've come up with 2 simplifying assumptions:

1. The Earth is a sphere of uniform density
2. Any mountain added is made of the same stuff, and is the same density too

And for the sake of this I'm defining a mountain as anything that breaks the spherical symmetry, but you can think about it however you want (and ignore those assumptions)

So 2 scenarios I'm wondering about:

1. If an object is on the surface of an Earth with no mountains, and then a mountain magically appears and the object is placed on top of it (the very top), for what shape and height of mountain does the object's weight not change?
2. If the mountain is already there, does the answer change (much)

Answer 1

Does gravity get stronger when you climb a mountain?

Gravity gets weaker when you climb a mountain

When you look at force of gravity due to the mass $M$ of the earth, all of the mass $M$ can be considered to be concentrated at the center of the earth. I would think that irregularities at the earth’s surface in the form of individual mountains, and even mountain ranges, would contribute a very small percentage to the total mass $M$ concentrated at the center.

On the other hand since the gravitational force varies inversely with the square of the distance from the center of the earth, the force of gravity will be less on top of mountains because of that. But even then, the differences are not that greate.

According to one source, the acceleration due to gravity is greatest at $9.8337\frac{m}{s^2}$ at the surface of the Arctic Ocean and lowest at $9.7639 \frac{m}{s^2}$ on Mount Nevado Huascaran in Peru. That's a 0.7% difference.

But the source also makes an interesting comment that tends to support your thoughts on local anomalies. It states:

“Nevado was a bit surprising because it is about 1000 kilometres south of the equator,” says Hirt. “The increase in gravity away from the equator is more than compensated by the effect of the mountain’s height and local anomalies.”

Hope this helps.

Answer 2

Let us for simplicity consider a spherical mountain (!) of radius $r$ on top of Earth's surface, having radius $R$. The mass of the mountain will be $m=(4\pi/3)\rho r^3$. The gravity on top of it is $$g=\frac{GM}{(R+2r)^2}+\frac{Gm}{r^2}=G\left (\frac{M}{(R+2r)^2}+(4\pi/3)\rho r\right ).$$ If we take the derivative by $r$ we get $$\frac{dg}{dr}=G\left ( -\frac{4M}{(R+2r)^3} + (4\pi/3)\rho \right).$$ The first term is -92,069 for $r=0$, while the second is 11,519 if we use granite density. So clearly for mountains small compared to the radius of the Earth gravity decreases as they get larger since $dg/dr<0$. (To get an increase you need a spherical mountain of slightly over half earth radius.)

Now, real mountains are not spherical as far as I know, so this result needs to be corrected for extra mass further down. But that will not change the qualitative result since it reduces the gravitational influence on the top of the mountain. Tall mountains tend to have lower gravity.

Answer 3

I guess your alluding to the nature of change in the density of a path through the earth's core and an effect on the nature of gravity that would not be accounted for in the simplification of the equation and your right.