Does the inverse square law equalize gravity at different spots?

[syfry]

TL;DR

For example does the inverse square law reduce the gravity from Earth's center enough that the ground directly beneath our feet now would exert a greater amount of gravity on each of us?

Inverse square law would reduce the gravity from the parts of Earth that are farthest from our feet.

It'll also reduce the gravity from Earth's center by a lesser amount, but would that be lesser enough so the gravity 20 kilometers under our feet is stronger than the core's gravity or even the rest of Earth's gravity?

I know their gravity strengths would add together, but I'm interested in each area's contribution.

So does the ground directly beneath each of our feet exert the greatest amount of gravity on each of us? Or is the core's gravity (or rest of Earth) still stronger on their own even at its distance away from us?

 

[Drakkith]

Here a look at the acceleration of an object in freefall under gravity at different radial distances from the center of the Earth for various models. The blue line (PREM) is what you'll want to look at. You'll notice that there's a small increase in acceleration (corresponding to a greater net force) as we go deeper until we reach the outer core. This is mostly because the core is somewhat denser than the rest of the Earth, so as we approach it the strength of gravity increases even though we have a bit less 'stuff' beneath us. Note that the acceleration goes to zero at the core because gravity is pulling equally in all directions, resulting in no net force, and thus no acceleration.

So does the ground directly beneath each of our feet exert the greatest amount of gravity on each of us? Or is the core's gravity (or rest of Earth) still stronger on their own even at its distance away from us?

It's hard to put into words. The ground directly beneath you exerts a greater force in terms of force per mass of stuff, but only because it is really, really close. A kg of concrete under my house exerts a lot more force on me than a kg of material in the core of the Earth. Does that answer your question?

 

[Nugatory]

So does the ground directly beneath each of our feet exert the greatest amount of gravity on each of us? Or is the core's gravity (or rest of Earth) still stronger on their own even at its distance away from us?

A ton of rock right under your feet will exert a stronger gravitational force on you than a ton of rock ten meters away…. But there’s a lot more rock ten meters away than just under your feet.

If we assume a spherically symmetric mass distribution (that is, the density is a function of just the distance from the core) and add up all the contributions from all the different parts of the earth at different distances from us, we will find that total is the same as if all the mass were concentrated in a single point at the center - proving this is a good exercise.

 

[Dullard]

Additionally:

The mass distribution of the Earth is not 'perfectly' represented by any geometric model - inconsistencies in density and shape make that impossible. Simple models are 'good enough' for most purposes. Several spacecraft have been orbited to measure the 'actual' mass distribution.

Ocean Tides are an excellent example of how gravity from distributed mass actually works.

 

[jbriggs444]

If I understand what you are trying to get at, you are trying to imagine the gravitational effect of a hemisphere of earth and stone within, let's say, 10 meters of your feet and compare that with the gravitational effect of the rest of the earth.

Perhaps we could try to put a number on how large you would have to make the hemisphere beneath your feet so that half of the gravity on you would come from that hemisphere and half from the rest of the Earth.

That is a rather nasty calculation, especially when one factors in the density variations within the Earth. However, we can get a feel for how the inverse square law works with a problem that is very much more tractable.

Alternate problem:

Let us pretend that we are dealing with a planet that has constant density and the same surface gravity as the Earth, $1g$. How much smaller would the planet have to be so that its surface gravity would be $\frac{1}{2}g$?

The volume of a smaller planet goes as the cube of the radius. So if we reduce the radius of the planet by a factor of two, the volume (and mass) of the smaller planet will be down by a factor of eight.

But it will be half as far to the center. So by the inverse square law, the gravitational effect of the mass will be up by a factor of four.

Which gives us the answer we are after. A planet half the size will have half the surface gravity [assuming density is unchanged].

This is also shown by the dark green line in the graphic referenced by @Drakkith in #2 above.

As a crude approximation, we might say that half of the gravity you feel is from the $\frac{1}{8}$ of the earth's volume nearest to you and the other half is from the remaining $\frac{7}{8}$.

 

[Baluncore]

Inverse square law would reduce the gravity from the parts of Earth that are farthest from our feet.

Consider a line from your feet, through the centre of the Earth, to your antipode. Now imagine a cone, axial with that line, with the apex at your feet. As you consider the mass along the axis of the cone, the area increases with the square of distance, while the gravitational attraction is reduced by the square of the distance. They cancel neatly, so the total mass in that cone can be modelled as being at the centre of the Earth.

 

[Steve4Physics]

Hello @syfry. I want a go too.

We can get a general insight by considering a simplified situation.

You live on a planet called Qube which happens (for reasons I am not allowed to divulge) to be cubic in shape.

In the following explanation, for clarity I’m omitting units.

Qube measures 3x3x3, so Qube’s volume is 27.

Qube is perfectly uniform and has density = 1, so its mass is 27.

Since we are only interested in comparisons, we’ll assume that the universal gravitational constant in our unit-system is 1.

We can treat Qube as consisting of 27 individual 1x1x1 cubes.

You are standing at the centre of one of Qube’s faces. Your mass is 1 and you are of negligible height. (This gives you a very large BMI but it's not personal.)

The distance from Qube’s centre to you is 1.5. We don’t have spherical symmetry but roughly your weight (due to the whole planet) will be $\frac {GmM}{d^2} = \frac {1·1·27}{1.5^2} = 12$.

Similarly, the contribution to your weight from the single 1x1x1 cube on which you are standing will be $\frac {1·1·1}{0.5^2} = 4$.

So about $\frac 4{12}= \frac 13$ of your total weight is produced by the nearest $\frac 1{27}$ of the total mass.

[Minor edit.]

 

[syfry]

Hello @syfry. I want a go too.

We can get a general insight by considering a simplified situation.

You live on a planet called Qube which happens (for reasons I am not allowed to divulge) to be cubic in shape.

In the following explanation, for clarity I’m omitting units....

Love it!

Thanks everyone, great replies!

This post got me thinking about a few interesting things about gravity. For example:

Does one half of Earth cause tidal forces to any portions of rock in Earth's opposite side?

Somewhat like the moon's gravity to Earth.

Since both sides of Earth are in motion while exerting gravity on each other. And, since the opposite side is moving in a reverse direction to this side. (they're spinning the same direction, but, when we draw a straight line from one side to another, they're headed 180° away from each other's direction of motion)

 

[Baluncore]

Does one half of Earth cause tidal forces to any portions of rock in Earth's opposite side?

No.
The Earth is also distorted by the lunar and solar tidal components. The diameter of the Earth varies tidally by ≈ ±300 mm. There is no time delay to that distortion, as there is with the ocean tides, due to flows in shallow water, near land masses.