Suppose one were to go really deep underground until he or she is only a few kilometres from the centre of the Earth (provided that he or she wears super antiheat suits). As the point of attraction of gravity is at Earth's center of gravity, you will be attracted by gravity towards the center of the Earth. But should we take into consideration of the huge bulk of mass above the explorer (which is the layers of earth and magma he or she had climbed through), would the matter have gavity pull on him or her?

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Integral
Staff Emeritus
Gold Member
The way it works out you are essentially "weigh less" at the center of the earth. The only mass which contributes to the earths gravitational pull is the mass between you and the center of the earth. Anything further from the center then you does not count.

drag
Like Integral said, only what's close to the center
matters. If you approximate the density of the Earth
to be the same, the mass is a function of density
multiplied by volume which means that it's a function
of radius^3 (third power). The force of gravity depends
on mass devided by distance - radius in this case, square
(second power). So you when you devide mass by radius^2
you still see that the force is proportionate to the radius
(again, at constant density) - so the higher inside a
planet you are - the greater the gravity.

Of course, when you're closer to the center you gain greater
potential energy.

Live long and prosper.

Gokul43201
Staff Emeritus
Gold Member
For any point inside the earth you can draw a spherical shell centered on the earth's center, and passing through that point. This shell divides the earth into 2 parts, an inner and an outer (relative to you, who is standing at this point). It so happens, that the outer part has no net gravitational pull on you. So you can imagine that part as not there at all. Only the inner part contributes to the gravitational force. So, its just as if you scaled the earth down, but are still standing on its surface.

And as explained above F(at some r) = Kr, where r is the radial distance to the center. So you become weightless when you are exactly at the center...but this is obvious from the symmetry of the situation.

LURCH
Gokul43201 said:
For any point inside the earth you can draw a spherical shell centered on the earth's center, and passing through that point. This shell divides the earth into 2 parts, an inner and an outer (relative to you, who is standing at this point). It so happens, that the outer part has no net gravitational pull on you. So you can imagine that part as not there at all. Only the inner part contributes to the gravitational force. So, its just as if you scaled the earth down, but are still standing on its surface.
Is that true, or only a simplification? I would have thought that the mass "above" you (further from the center) would have gravitational pull away from the center, correct?

jcsd
Gold Member
LURCH said:
Is that true, or only a simplification? I would have thought that the mass "above" you (further from the center) would have gravitational pull away from the center, correct?
It's a simplification in that it assumes perfect spherical symmetry and uniform density, but it's still not a bad approximation, as the Earth is pretty close to spherically symmetric (and gravity being so weak doesn't mind slight peturbations) and it's 'shells' are also pretty clos to uniform density

Gokul43201
Staff Emeritus
Gold Member
But for the case of a true sphere, where the density varies, at most radially, it will be perfectly true that the outer shell exerts no net force.

You notice that we are talking about spherical shells here. The outer sperical shell is not just above you, but also way, way below on the other side. If you add up the forces from the different parts of this shell, they end up canceling out to give zero force.

Yes, the mass directly above you does pull you away. But the problem with looking at this is that it doesn't help (in a mathematical way) to divide the earth into (i) the portion just above you and (ii) the rest of the earth. You will find it hard to calculate the total force on you if you divided the earth up this way. On the other hand, breaking it up into the inner sphere and the outer shell makes the calculation mathematically "doable", as explained before.

I disagree with Drag's view on saying that the force of gravity actually increase when you get closer to the center of the Earth! Drag, you said that "the force of gravity depends on mass devided by distance - radius in this case, square (second power)", and you also wrote this:

"So you when you devide mass by radius^2
you still see that the force is proportionate to the radius
(again, at constant density) - so the higher inside a
planet you are - the greater the gravity.

"Of course, when you're closer to the center you gain greater
potential energy."

Yeah, you are correct in terms of maths, but you must really remember that the poor explorer is only several kilometres from the center of the Earth. So the mass should not be the mass of the entire Earth but instead, is the "sphere of matter that is below the explorer (a.k.a: Earth's core)"! Hooray! I'm a genius!

(this is one of the good thing about studying physics. The 'aha!' sensation you get when you understand or find out something is really cool.)

The answer to how much is the gravitational force on the explorer remains unknown until some guy finds out what is the mass of Earth's core.

But now, the main question is: does the layer of earth and magma which the explorer had climbed through exert a gravitational pull on him? I think I have the answer.

Gokul43201 wrote: "If you add up the forces from the different parts of this shell, they end up canceling out to give zero force. " Well, what I had found out is that the resulting force isn't zero.

I had worked out to find the single resultant grav. pull from the matter that surrounds the explorer, using lines that are relative to the thickness of the Earth, in various directions from the explorer. It turned out that all the forces cancel out each other except a pair of opposite 'pulls'. The pair is along a straight line that crosses both the explorer and the center of the Earth. Deriving from this pair of opposite pulls, the resultant force is a pull towards the center of the Earth!

Regardless of whether you get me or not (I find it better if I explain with a diagram), you just keep in mind that the resultant pull from the outside layer of Earth is actually pulling towards the center of the Earth!!

Doc Al
Mentor
physicskid said:
I disagree with Drag's view on saying that the force of gravity actually increase when you get closer to the center of the Earth!
I think you misread what drag meant when he said: "so the higher inside a planet you are - the greater the gravity".

He meant (and said) that within the Earth the gravitational field is proportional to the distance from the center (assuming a uniform Earth model). So the further you are from the center (the higher you are), the greater the gravitional field.

You are in violent agreement.

Doc Al
Mentor
physicskid said:
Gokul43201 wrote: "If you add up the forces from the different parts of this shell, they end up canceling out to give zero force. " Well, what I had found out is that the resulting force isn't zero.
Sorry, but Gokul43201 is correct. The gravitational field is zero everywhere inside a uniform shell of mass.
I had worked out to find the single resultant grav. pull from the matter that surrounds the explorer, using lines that are relative to the thickness of the Earth, in various directions from the explorer. It turned out that all the forces cancel out each other except a pair of opposite 'pulls'. The pair is along a straight line that crosses both the explorer and the center of the Earth. Deriving from this pair of opposite pulls, the resultant force is a pull towards the center of the Earth!
Try again. One way to see this (without doing any math) is to divide up the shell using opposite cones of equal spherical angle. You will see that the mass subtended by the cone is proportional to distance squared, but the force is proportional to inverse distance squared. So all opposite "cones" of force cancel out, no matter where you are.

Thanks, Doc Al! But anyway, what do you mean by "violent agreement"? Never heard of this phrase before!

Doc Al
Mentor
"in violent agreement" :-)

physicskid said:
... what do you mean by "violent agreement"? Never heard of this phrase before!
By saying that you were "in violent agreement" I just meant that while you thought you were saying something different (than drag, in this case), in fact the two of you were saying the same thing.

(I see this syndrome many times over in meetings where I work. I'll often see two people (sometimes one is me!) getting really emotional and screaming at each other. And then someone points out that they were both arguing for the same position, but they didn't realize it since they each expressed it slightly differently. D'oh!)

(And then a long silence of embarrassment follows. Happens all too often in political arguments as well.)

An easy explaination is:

Above the earth surface gravity decreases by 1/R^2 (compared to surface gravity):
Below the surface (assuming uniform density), gravity decreases by 1/R.

So, at 2 radii gravity is 1/4th that of the surface;
But at 1/2 radii (halfway to the center), gravity decreases by 1/2.

Creator

drag
Greetings !

You may wan'na try the definitions of "gravity"
and "Gauss's Law" on hyperphysics to understand
this even further (Gauss's Law is for electrical charges,
but as an approximation it fits gravity as well, where the
"charge" is mass).
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Violent agreement...

Live long and prosper.