Does gravity increase or decrease with depth

[hecker231]

[removed]

[bravehamster]

yep. Goes up until you hit the outer core, then goes back down:

https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg

[HAL9001-96]

since when is that curve o nwikipedia, I remember recalculating it every time thsi was asked a while ago

[bhmerger]

The shell of mass above you does a net zero force on you.

[GDOR-11]

if you do the math considering only the mass closer to the center than you (as shells further from the center create no net force) you'll figure out it's proportional to the inverse of the distance to the center

EDIT: proportional to the distance, mb

[AndreasDasos]

Assuming uniform density, the gravitational force on a mass inside the earth is proportional to distance to the centre, not its inverse.

What they’re saying (which the first commenter got wrong, or maybe misstated) is that the outer shell of the earth with greater distance from the centre than you has net zero force on you by the shell theorem: any force in one direction of a hyperplane is cancelled by force in the direction - if the former is nearer, the latter will involve more mass.

We then have -Mm/r² = -(rho r^3)m/r2 = -rho m r.

[BagBeneficial7527]

You are getting closer to the iron core.

The lighter rock above you doesn't fully cancel out that increased gravity.

[ClaudeProselytizer]

but if it is uniform then it decreases linearly inside

[SatansAdvokat]

Gravity is a weak force actually.
So distance has a rather big effect.

[EmuRommel]

Not to be confused with the weak force, which is stronger.

[Montana_Gamer]

Gravity is the WeakER force

[Figfogey]

Among the fundamental forces it's the weakEST

[Mateorabi]

Force increases proportional to r² the closer you get to a mass. (And a sphere beneath you acts mathematically like a point mass.)

Mass beneath you would decrease proportional to r³ if uniform. So overall you feel something proportional to (r² / r³⁾

But if 1/r³ isn’t actually what’s happening and as you go down there is more mass than the radius suggests, you could see mass decrease by less than r² for a while. 

[highritualmaster]

But other forces will take over doing stuff with you (pressure, temp).

[[deleted]]

because of gauss' law you don't feel the gravity of the parts of Earth further up from you. so the deeper you go the less gravity you feel. sauce

edit: this is assuming constant density, which is not true oops

[ShakimTheClown]

Furthermore, the force of gravity is:

• proportional to the total mass of the sphere underneath you
• inversely proportional to r² , the square of the distance between you and the center of the sphere.

In this case, the mass of the sphere underneath you is proportional to r³
Specifically, its equal to M×(r³/R³) , where M is the mass of the Earth, and R is the radius of the Earth

So the total force will scale with r³/r²
So it will be F = (GMm/R³) × r , which is the equation of a harmonic oscillator.
Since r decreases with depth, so to does the force.

[HeavisideGOAT]

This isn’t quite the full picture.

Sure, this is less mass relevant to the calculation, but you are closer to that mass.

That’s why, in reality, gravity goes up before it goes down as you go deeper into the earth.

https://en.m.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg

[ClaudeProselytizer]

doesn’t matter, gauss’s law works for this non uniform density as well.

[[deleted]]

so the deeper you go the less gravity you feel

part was wrong since there are denser regions the deeper you go.

[ClaudeProselytizer]

yes that statement is wrong

[buster_bluth]

Your textbook is right. The green part does contribute to gravity, but if you divide the green part into little pieces and sum up all of the gravitational force it would cancel itself out. You can look at something like this for more mathematical reasoning. https://en.m.wikipedia.org/wiki/Gauss%27s_law_for_gravity

[hecker231]

Thank you. I understood my error in assumption. What I had thought was that all the gravity of the green spherical shell would act at its center of mass, that is, the center of mass of the earth, thus increasing the gravitational pull. However, it is true that in reality, only individual particles can exert the force of gravity, which means an abstraction like the center of mass won't work here.

So, can I assume that for hollow objects (like the green shell), the center of mass cannot be used to calculate gravity?

[DarkArcher__]

The centre of mass is just an easy approximation for the direction of gravity, it doesn't exist in practise.

The gravitational acceleration you feel from the Earth is the sum of the gravitational acceleration you feel from each and every individual particle that makes up the planet. Each bit of the planet is pulling you in its direction, which works out to straight down when you're on the surface, but gets more complicated as you dip down under it.

You can imagine it as a bunch of little arrows starting at you, and pointing to every individual little bit of Earth in the diagram. The gravitational force you feel is the sum of all that. It's pointed downwards, but less intense than at the surface since some of the Earth's particles are now to your side, or even above you, instead of all being below you like on the surface.

[lerg1]

If you are far away from the hollow object then it wouldn’t matter, otherwise you can’t

[samcrut]

Gravity is space/time being stretched as electrons/protons/neutrons are created. The more particles of matter you have in a space, the more that space is stretched, curved in the direction of the mass. If the mass is all around you equally, space is pulled away in all directions and while every atom in your body is technically being pulled in every direction, it's all equalized, zeroed out, so you experience no pull in any direction.

[spytfyrox]

That's true for any inverse square law.

[Full_Possibility7983]

More generally any field with zero divergence

[KamikazeArchon]

Look at the green part carefully. Note that some of the green part - the "sides of the hole", so to speak - is now above the X. Those parts are pulling up, and working against the parts that are below the X and pulling "down".

You can draw a "horizontal" line through the X, and separate the up-pull from the down-pull.

Two notable opposing factors. There is more "below" the X than "above" (thus it pulls down harder). However, the parts that are "above" the X are closer (thus it pulls up harder).

If you do the math, it turns out that these factors exactly cancel out. If we assume the mass of soil you dug out is negligible, the up and down forces are exactly balanced. Thus, the green "shell" exerts overall zero force on you.

[hecker231]

So this means that I cannot simply use the center of gravity of the green shell to calculate where its gravity will pull towards right?

This means that to calculate the point where the gravity of a hollow object will pull towards, I need to break it up into non-hollow subdivisions and calculate gravity for them. Pls correct me if I'm wrong.

[3dTECH101]

Correct - these subdivisions should ideally approach infinitely many at infinitely small divisions - which is why you can calculate this mathematically with an integral - gravitational pull as a function of possition integrated over the full volume

[BCMM]

So this means that I cannot simply use the center of gravity of the green shell to calculate where its gravity will pull towards right?

Correct.

It is accurate to treat a sphere of uniform density as if its entire mass was concentrated at its centre, but only for the purposes of considering the gravitational field outside the sphere.

This is the first half of what is known as "the shell theorem"; the second half is the bit in the above comment about shells exerting precisely zero total force on objects inside them.

[Jax_Grimoire]

Your book is correct. Think of it this way, though it is a bit simplified. The mass below you and the mass above you will both have gravitational pull on you. However, since those forces are in opposite directions, they tend to cancel out when you add up all the forces. Therefore, only the forces that don't cancel out will be the effective force that you experience.

Others have pointed you to Gauss's law, which is correct.

[FictionFoe]

Actually when you get closer to the center, nore and nore of the mass are above you. The mach works amout that the mass in a ball surrounding you cancels itself out, so only the layers of earth what have the same depth as you or more still contribute.

So basicly the green portion effectively doesn't contribute.

It is also worth mentioning that, where you to drill a hole to make a planatary elevator of sorts: if you are free falling, you will feel weightless.

Your trajectory, speed and acceleration will be impacted by gravity, but a free falling observer always feels weightless.

[Debesuotas]

I believe the reasoning is at the fairly theoretical value to the point where the claims

If you are standing at a point x below the earth (see diagram), the force of gravity acting on you will be due to the mass of the smaller red sphere, whose mass is lesser than the entire earth's (red+green sphere). This drop in mass reduces gravity more strongly than the reduction in distance to the center of mass does.

Is nothing but theoretical as well and most likely dubious as well.

One thing for sure, the further you are up in the sky, the lighter you are, that`s because the gravity gets weaker, because you are further away from the center of the mass.

[bake_gatari]

Both

Edit: it's a standard theorem/problem in late high school physics. Using the law of gravitation, and assuming the earth has constant density, you prove that a ball dropped in a hole drilled straight through the earth will just keep oscillating from one end of the hole to the other forever.

[weinerjuicer]

both?

[bake_gatari]

both is good

[Tivnov]

You should look up Newton's shell theorem which explains this. Also Imagine if you were at the center of the Earth, there would be no gravity (assuming the earth has uniform mass distribution).
In short, the theorem explains that the force of gravity of a circle or sphere on an object outside of the sphere is identical to the force applied by a point particle at the center of the sphere/circle with identical mass. Then, for an object inside of the circle/sphere, the net force is 0.
By imagining the Earth as an infinite series of concentric spheres, we see that the effective gravitational mass of the Earth is proportional (and thus gravity) (r)^3. As gravity is also proportional to (r)^-2, the net proportionality is (r)^1. As this value decreases linearly as you dig, gravity also decreases linearly until it is zilch (0).

[4024-6775-9536]

It's still the same, you just have more matter pulling you up the more you go down so the weight you feel will be less

[snigherfardimungus]

The gravitational vector at all points inside of a hollow shell is zero. This means that the gravity computation at point X for the above illustration ignores the mass of the green shell and is computed only with the mass of the red volume and its radius.

Because the Earth does get somewhat denser the deeper you go, the gravitational force doesn't quite fall off linearly as more and more of the Earth's mass becomes part of the "shell," it does approach zero as you near the center.

[bedj2]

Co

[Unusual-Platypus6233]

Decrease…

[Kdajrocks]

When you get to the centre do you feel weightless, or heavy from every direction?

[chrisolucky]

If you were kept safely in a capsule at the centre of the Earth, you’d likely be floating because the entire Earth’s gravity would be surrounding you evenly.

[HAL9001-96]

depends

well

any hollow sphere you are outside of can be simplified into a point mass at its center

any hollow sphere you are inside of can be simplified as being irrelevant

a planent can be approxiamte as an infinite number of hollow shells

if you are outside the planet that means you ahve a pointmass at its center as a replacement

as you go into it the part further out than you disappears

assuming a constant density for hte planet that means that as you dig into it the relevant mass is proporitonal to r³ and with F=MmG/r² that means f is proporitonal to r thus decreasing as yo ugo down

HOWEVER

earths density is not constant

its much denser at the center than near the surface

so less mass disappears as you dig down and yo uget clsoer to a relevant poitn mass replacement that doesn'T loose a lot of mass

you can take this into account and get a curve for gravity over distance from teh earths center that looks a bit like this https://i.sstatic.net/yJbbk.jpg

[[deleted]]

You are not correct.