I Where does gravity switch from stronger to weaker, if it does?

[freeelectron]

TL;DR Summary

Force of gravity based on distance from celestial body.

I read everywhere that gravity is slightly stronger at the bottom of a mountain than at the top because the place is closer to the celestial body that causes the "attraction".
What I don't understand is that, as you get closer to the center of the planet, there's less matter below you and, therefore, gravity should decrease, because less matter attracts you from below.
I also guess that, at the center of the planet, you shouldn't feel any gravity in any direction.
Now, imagine you are in an elevator that starts from a high altitude (see attached image) and goes down through the Earth's crust down to the center of the planet, is there an height where gravity switches from getting stronger and stronger to getting weaker and weaker? If yes, can we calculate how high?

Thanks

 

[gleem]

Summary: Force of gravity based on distance from celestial body.

What I don't understand is that, as you get closer to the center of the planet, there's less matter below you and, therefore, gravity should decrease, because less matter attracts you from below.

That is correct

Summary: Force of gravity based on distance from celestial body.

I also guess that, at the center of the planet, you shouldn't feel any gravity in any direction.

Correct

Summary: Force of gravity based on distance from celestial body.

Now, imagine you are in an elevator that starts from a high altitude (see attached image) and goes down through the Earth's crust down to the center of the planet, is there an height where gravity switches from getting stronger and stronger to getting weaker and weaker? If yes, can we calculate how high?

Yes. It becomes weaker as soon as you get below the surface. This is exactly true for a perfect sphere As for the Earth which is an oblate spheroid (bulging at the equator) and has and irregular surface and non uniform composition and density the transition distance to below the surface is not be well defined so as we approach the surface gravity may fluctuate a little until we are well under the surface before it decrease significantly.

There is about a 1% difference in gravity between the equator and the poles being weaker at the equator. For a perfect sphere gravity should decrease in a linear fashion from the surface to the center.

 

[phinds]

I read everywhere that gravity is slightly stronger at the bottom of a mountain than at the top because the place is closer to the celestial body that causes the "attraction".

Correct

What I don't understand is that, as you get closer to the center of the planet, there's less matter below you and, therefore, gravity should decrease, because less matter attracts you from below.

correct. It does.

I also guess that, at the center of the planet, you shouldn't feel any gravity in any direction.

correct

Now, imagine you are in an elevator that starts from a high altitude (see attached image) and goes down through the Earth's crust down to the center of the planet, is there an height where gravity switches from getting stronger and stronger to getting weaker and weaker?

no, that would contradict what you have already pointed out as correct. It just gets weaker and weaker as you go down. Why would you expect otherwise?

EDIT: I see gleem beat me to it.

 

[FactChecker]

Suppose you are approaching a perfect sphere of constant density. The gravity would increase till you get to the surface and it would decrease once you were beneath the surface.

NOTE. The term "gravity" often includes the centrifugal force on an object which is stationary on the rotating Earth. To answer your question simply, the centrifugal force is being ignored.

 

[Orodruin]

Summary: Force of gravity based on distance from celestial body.

Now, imagine you are in an elevator that starts from a high altitude (see attached image) and goes down through the Earth's crust down to the center of the planet, is there an height where gravity switches from getting stronger and stronger to getting weaker and weaker? If yes, can we calculate how high?

Yes and yes. You in order to do the computation you need to know the density profile. In the approximation that the Earth is a homogenous sphere (which it really isn’t), the point of strongest gravity would be at the surface.

 

[Ibix]

I believe that with a realistic density profile, gravitational acceleration increases until you reach the core, then drops off rapidly to zero at the centre. This is because the core is denser than the rest of the Earth.

Wikipedia has charts of both the density profile and resulting acceleration profile - look in the Variation in Magnitude section. Edit: or look in Orodruin's next post...

 

[Orodruin]

Yes. It becomes weaker as soon as you get below the surface.

While this is true for a homogenous sphere, it is not necessarily true for spherical symmetry if the density changes with the radius. In particular, the gravitational field at radius $r$ changes as
$$
g(r) = \frac{GM(r)}{r^2},
$$
where $M(r)$ is the mass contained within radius $r$. Taking the derivative of this, we find
$$
g'(r) = \frac{GM'(r)}{r^2} - \frac{2GM(r)}{r^3}
$$
and thus we have a condition for the maximum
$$
g'(r) = 0 \quad \Longrightarrow \quad M'(r) = \frac{2M(r)}{r}.
$$
Now, assuming a density function $\rho(r)$, we would have
$$
M(r) = 4\pi \int_0^r \rho(x) x^2 dx \quad \Longrightarrow \quad M'(r) = 4\pi \rho(r) r^2
$$
and the condition for the maximum is given by
$$
\rho(r) = \frac{M(r)}{2\pi r^3}.
$$

Put in a different way, the gravitational field will decrease with $r$ if $2 \pi \rho(r) r^3 < M(r)$ and increase with $r$ if $2\pi \rho(r) r^3 > M(r)$. Inside a homogeneous sphere, $\rho = 3M(r)/4\pi r^3$ and so
$$
2\pi \rho(r) r^3 = \frac{3M(r)}{2} > M(r)
$$
and so the gravitational field increases until the surface. However, the Earth is not a homogeneous sphere. A better description of the variation of the gravitational field of the Earth with radius can be found on Wikipedia:

As can be seen from this image, the Earth's gravitational field reaches its maximum where the outer core ends, around 3500 km from the center.

Edit: Note that PREM stands for "Preliminary Reference Earth Model", which is a rather realistic spherically symmetric approximation for how density varies with radius. An interesting factoid is that after the slight drop in the gravitational strength after the core-mantle boundary, the gravitational field remains fairly constant around 10 m/s^2 throughout the mantle.

 

[sophiecentaur]

@Orodruin has supplied an interesting graph which has a further interesting bit of information in it. The straight line (dark green) shows that the attractive force is proportional to the distance from the centre. This implies that (assuming a perfectly spherical and non rotating planet) an object would fall through a hole through the centre and follow Simple Harmonic Motion (the force variation is the same as for a spring) traveling from one side to the other, endlessly. The period of the oscillation would be exactly the same as for a satellite in low orbit.
There are number of practicalities that make this idea nonsense. The temperature and pressure near the centre plus the existence of an atmosphere would prevent it happening for a start. But you could imagine a small scale model with an inert spherical asteroid of the same average density as the Earth. The 'pendulum' experiment could work and so would the orbiting (pebble?) asteroid with a period the same as for a low Earth orbit (around 90 minutes). One day in the future, a version of Elon Musk could finance such an experiment, perhaps?