# Gravity inside the earth

atom888
If one pass the Earth's surface and move inward to the center. What would happens?

1/ would he get stretch
2/ would he get compress
3/ or would he feel nothing at the center?

Gold Member
There's quite an extensive discussion of this around here somewhere, if you want to do a forums search.
Basically, the gravity pulling from all sides at the centre would render one effectively weightless.

Feldoh
If one could fall from the Earth's surface to the center, you would accelerate downward at a decreasing rate, until you reached the center at which point the would be no force on you.

atom888
sorry to bring back the old tocpic. I couldn't find the search topic fuction. Anyway, my wonder for for this question is if this is our assumption. how can we explain collapse of stars under its own gravity? Not to mention hard metals like rocks and diamond, which form under extreme temperature and pressure. However I think I see it. Though gravity is zero inside, a pressure with value of integral from 0-r of density x gravity dr where gravity is a function of radius .btw, how do you put in math sympbol and expression. Thx

Homework Helper
As you know, matter tends to clump together due to gravitation. But other forces resist this clumping once the matter particles come close enough to each other. This gives rise to pressure at every self-gravitating body like Earth or a star, though the forces resisting gravity are different in the Earth and in a star. The pressure is highest at the centre because the weight of all the outer layers are acting on it, and decreases as you go toward the surface, as you have rightly pointed out.

In a star, this force against gravity is provided by hot gases which are kept hot due to nuclear reactions. When all the nuclear reactions stop because the fuel has run out, the star may tend to shrink due to self gravity, until it can shrink no more due to some other force. For example, in a white dwarf, it is the electron degeneracy pressure which resists gravity. If the mass is too much, then no force may be able to stop it from shrinking, and a black hole may form.

Inside the earth, the pressure is very high compared to our standards, and is enough to form rocks or diamonds or fossil fuels. If you were put inside, you would get really compressed and burnt to a crisp.

If a small tunnel was drilled through a solid sphere passing through the centre, and you fell into it, then the motion would be an SHM, but I don’t think that was your question.

For math symbols, you can use latex. Click on the ∑ sign above.

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Homework Helper
If one could fall from the Earth's surface to the center, you would accelerate downward at a decreasing rate, until you reached the center at which point the would be no force on you.

Just to clarify- you would not stop at the center of the earth: once you had reached the center of the earth, you would have built up a pretty good speed and would continue through, having just enough energy to (neglecting friction) reach the other side of the earth. As Feldoh says, it is the acceleration that would be decreasing, not the speed.

Feldoh
Just to clarify- you would not stop at the center of the earth: once you had reached the center of the earth, you would have built up a pretty good speed and would continue through, having just enough energy to (neglecting friction) reach the other side of the earth. As Feldoh says, it is the acceleration that would be decreasing, not the speed.

Yeah, to add a bit more, if we pretend there is a tunnel straight through the center of the Earth and you jump into it, you would actually oscillate back-and-forth through the earth. (assuming no air resistance )

But that's not really what the OP is asking

atom888
Yeah, to add a bit more, if we pretend there is a tunnel straight through the center of the Earth and you jump into it, you would actually oscillate back-and-forth through the earth. (assuming no air resistance )

This... is definitely a good project the world should invest in! \$ 200 USD for each oscillation!

Staff Emeritus
The pressure at the center of the Earth is estimated to be about 350-360 GPa (or about 3.6 million atm) from the gravity pulling on the matter of the Earth.

Gold Member
Hmmm... you'd have to pop your ears a couple of times on the way down.

atom888
The pressure at the center of the Earth is estimated to be about 350-360 GPa (or about 3.6 million atm) from the gravity pulling on the matter of the Earth.

Dang, some good pressure. I'm trying to make some synthetic diamond so I can get marry with less cost. I'm encountering problem with heat source. Where can I get 4000K ? no metal would hold that temp even ceramic. I'm thinking concentrate sunlight intersecting beams +conventional fire. Crap, it seems easier buy that stone.

Gold Member
Keep in mind, Atom, that the concept of engagement rings was instigated by the diamond industry in the Great Depression. They are not something that has ever been either socially or religiously sanctioned. Not diamond-wise, that is. My mother's wedding band was a pasitc coil formerly used as a leg-band on a chicken. A couple of years later, she put on her late mother's gold band, which she is still wearing at the age of 95.

atom888
Keep in mind, Atom, that the concept of engagement rings was instigated by the diamond industry in the Great Depression. They are not something that has ever been either socially or religiously sanctioned. Not diamond-wise, that is. My mother's wedding band was a pasitc coil formerly used as a leg-band on a chicken. A couple of years later, she put on her late mother's gold band, which she is still wearing at the age of 95.

Haha it's useless if we're the only one get this message. Oh well, I guess it's a challenge to see if we can concentrate energy. I think that's science never ending break through point; concentrate the forces of nature into a fine point and control it.

kasse
If all the Earth's matter was concentrated on it's surface, and distributed evenly, we wouldn't have a gravitational field inside the earth.

pixchips
Yeah, to add a bit more, if we pretend there is a tunnel straight through the center of the Earth and you jump into it, you would actually oscillate back-and-forth through the earth. (assuming no air resistance )

But that's not really what the OP is asking

Okay, but it's a fun question. If you drill your tunnel through the poles I think you might just oscillate, but what if you do it at the equator? I think on the way down you'd get whomped to one side, and then on the other side on the way out. So you'd need a lot of lubricant. Oh, and is your tunnel a sealed vacuum? If not, what does the pressure profile look like? Just in case we ever build this thing ;).

atom888
Okay, but it's a fun question. If you drill your tunnel through the poles I think you might just oscillate, but what if you do it at the equator? I think on the way down you'd get whomped to one side, and then on the other side on the way out. So you'd need a lot of lubricant. Oh, and is your tunnel a sealed vacuum? If not, what does the pressure profile look like? Just in case we ever build this thing ;).

You're worry too much. The worst case is we forget about dampen and leave somone hanging at the center. Just get a big rope and pull him up. lol

pixchips
Okay, gotcha ... but I'm still worried that he might be a bloody pulp being squished by a gazillion atmospheres. But you're right, we should just give it a try ... ummm ... you first! ;).

Domenicaccio
The pressure at the center of the Earth is estimated to be about 350-360 GPa (or about 3.6 million atm) from the gravity pulling on the matter of the Earth.

Interesting...

So at the centre of the earth, despite zero gravity, there is still a huge pressure caused by the fact that everything around the centre indeed is accelerated towards it?

Can we say that in practice what really crushes you is not so much what is immediately around you, but rather what is far away from the centre? That the soil (or whatever) a few meters around you would not really want to hurt you, but is pushed down by what is above which pushes a little more due to higher G, and is in turned pushed even more by the next layer above etcetera?

dmt740
If you could build a tube-like structure that would negate the temperature and pressure at the core, and jumped in, would you then oscillate back and forth continuously, or would you eventually come to rest at the center?

swapnilster
In ideal conditions, you would continue to oscillate forever.That is because force of gravity increases as you move away from the center and so the force pulling you towards the center would increase with distance from the center.Hence you would get a condition like F=-kx(k is a constant and x is displacement) similar to eqn for simple harmonic motion.
g'=g(1-d/r)
g'=apparent force of gravity
d=depth

pzlded
There is a difference between zero gravity in space and zero gravity within the center of a spherical mass. For a sphere within a mass, the surface of the sphere has net gravitational attraction for nearby mass. A point at the center of the sphere has not net gravitational attraction.

dmt740
Ok, using my tube from above, if you scaled down it (as opposed to jumping in) what happens when you get to the center?

DavidWhitbeck
If one pass the Earth's surface and move inward to the center. What would happens?

1/ would he get stretch
2/ would he get compress
3/ or would he feel nothing at the center?

The center-of-mass undergoes simple harmonic motion as discussed by previous posters. An extended body should have a non-zero stress since the force acting on the body is non-uniform. I would think that the object should be compressed at the end away from the center (closer to the surface) because the end accelerates faster than the middle since the force is greater the further away from the center. the object should also be compressed at the other end because the end being closer to the center doesn't accelerate as fast as the middle.

So I think that an extended body should be compressed for the entire motion. Is that what you are asking? This is of course assuming that the body is compressible.

Domenicaccio
If you could build a tube-like structure that would negate the temperature and pressure at the core, and jumped in, would you then oscillate back and forth continuously, or would you eventually come to rest at the center?

Assuming no friction with air, you'd be looping between your starting point on the surface and an opposite point at the other side of earth.

I might be wrong, but I think it would not be a harmonic cycle. Acceleration (gravity) is probably linear with depth inside the Earth (max on the surface, the linearly going down to zero at the centre, then up again with opposite direction), hence speed would be going up "parabolically" (I guess it would look like an inverted parabole) until a maximum in the centre then decrease to zero again.

DavidWhitbeck
I might be wrong, but I think it would not be a harmonic cycle. Acceleration (gravity) is probably linear with depth inside the Earth (max on the surface, the linearly going down to zero at the centre, then up again with opposite direction), hence speed would be going up "parabolically" (I guess it would look like an inverted parabole) until a maximum in the centre then decrease to zero again.

How is that not harmonic?? What you described with the acceleration is precisely what simple harmonic motion does. A block on a spring would also have acceleration proportional to length (from Hooke's Law). I don't understand the parabolic bit. The speed oscillates in time (by sin or cos) with the maximum at the center of the Earth.

Domenicaccio
Where's the edit button?

Gold Member
Interesting...

So at the centre of the earth, despite zero gravity, there is still a huge pressure caused by the fact that everything around the centre indeed is accelerated towards it?

Can we say that in practice what really crushes you is not so much what is immediately around you, but rather what is far away from the centre? That the soil (or whatever) a few meters around you would not really want to hurt you, but is pushed down by what is above which pushes a little more due to higher G, and is in turned pushed even more by the next layer above etcetera?

That's kind of like saying "it's not the tires of the 18-wheeler that crush you, it's what's on top of the tires..."

jvicens
Okay, but it's a fun question. If you drill your tunnel through the poles I think you might just oscillate, but what if you do it at the equator? I think on the way down you'd get whomped to one side, and then on the other side on the way out. So you'd need a lot of lubricant. Oh, and is your tunnel a sealed vacuum? If not, what does the pressure profile look like? Just in case we ever build this thing ;).

Leaving aside all the technical reasons why such enterprise is not possible and using Newton's Law of Gravity shows that something very interesting would happen if you in fact manage to dig such a hole.

If such a hole exists and you jump into it, you will experience accelerated motion towards the center of the Earth. As you get closer to the center of the Earth, the force of gravity pulling on you will become weaker and weaker (because there is less and less mass under your feet) and it actually becomes zero in the center. At this point the acceleration is also zero but you have maximum speed. You will not stop here but you will continue (due to inertia) moving towards the exit hole on the other end of the hole. As you leave the center of the Earth behind you, the force of gravity (pulling action) will start increasing again (because there is more and more mass under your feet) and this force will slow you down. Due to the symmetry of the problem, your speed will be zero whenever you reach the exit hole and your acceleration will be maximum but always pointing towards the center of the Earth. At this moment, you will fall back inside the hole moving towards the center of the Earth, which you will pass again at maximum speed, reaching the other end of the hole. The same process will take place over and over and will not stop if we assume that there is no friction between you and the wall of the hole or between you and the air filling the hole. In conclusion you will move from one end to another end with S.H.M. (Simple Harmonic Motion), which period will be around 90 minutes.

Something curious here: ~ 90 minutes is also the period of any satellite (for example the International Space Station, ISS) as it orbits Earth. This is not a coincidence. The reason for this fact is that a particle's position undergoing S.H.M. (in a circular orbit) can actually be mathematically expressed as the combination of two movements perpendicular to each other (plotted on horizontal and vertical axis) and the particle's position versus time plotted on each axis responds to S.H.M as well.

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"Never underestimate the joy people derive from hearing something they already know."
Attributed to Enrico Fermi - Nobel Prize in Physics in 1938

pixchips
Very interesting. So it's like this: if you had two holes through the center at 90 degrees to each other, and started one mass oscillating 15 minutes after the other, the center of mass of the two masses would describe a circular path around the circuference with a period of 90 minutes. I have to say, it's extremely obvious now that you point it out. Thanks jvicens ... an interesting point.

Gold Member
Something curious here: ~ 90 minutes is also the period of any satellite (for example the International Space Station, ISS) as it orbits Earth.
This is somewhat misleading. Satellites have all sorts of periods, all the way up to 28 days! It is dependent on their altitude.

However, what is true is that the harmonic motion you speak of is the same as the orbital period of a hypothetical satellite orbiting at zero altitude.

pixchips
Not so misleading if you drop a mass from the surface into a hole through the planet. It will oscillate under ideal conditions between the opposit surfaces, which would correspond to a zero height orbit. Real low Earth orbits are 60 to 90 miles, which is a small fraction of a percent of the radius. So it really does model a low Earth orbit in some sense.

Now here's an interesting realization (to me anyway): according to Newtown, the gravitational force felt beneath the surface of a homogenous planet is due to the mass contained in a spherical radius between you and the center. That makes the force proportional to x^3, and inversely proportional to x^2 ... therefore, so ma=-kx, and we actually do get harmonic motion.

So how does it change if the amplitude exceeds the radius? Once the mass leaves the surface the rule reverts to inverse square. The nice harmonic oscillation breaks, and the nifty model of a circular motion projected onto the center line fails too. The projection of the orbit will always be sinusoidal, but the oscillating mass won't remain sinusoidal once the amplitude exceeds the radius. Or did I miss something?

blackwing1
Dang, some good pressure. I'm trying to make some synthetic diamond so I can get marry with less cost. I'm encountering problem with heat source. Where can I get 4000K ? no metal would hold that temp even ceramic. I'm thinking concentrate sunlight intersecting beams +conventional fire. Crap, it seems easier buy that stone.
doesn't a diamond evaporate somewhere 2000- 3000 degrees fahrenheit? or was it somewhere in the 4000s, i forgot...

Feldoh
Not so misleading if you drop a mass from the surface into a hole through the planet. It will oscillate under ideal conditions between the opposit surfaces, which would correspond to a zero height orbit. Real low Earth orbits are 60 to 90 miles, which is a small fraction of a percent of the radius. So it really does model a low Earth orbit in some sense.

Now here's an interesting realization (to me anyway): according to Newtown, the gravitational force felt beneath the surface of a homogenous planet is due to the mass contained in a spherical radius between you and the center. That makes the force proportional to x^3, and inversely proportional to x^2 ... therefore, so ma=-kx, and we actually do get harmonic motion.

So how does it change if the amplitude exceeds the radius? Once the mass leaves the surface the rule reverts to inverse square. The nice harmonic oscillation breaks, and the nifty model of a circular motion projected onto the center line fails too. The projection of the orbit will always be sinusoidal, but the oscillating mass won't remain sinusoidal once the amplitude exceeds the radius. Or did I miss something?

There's a conservation of energy, so unless you started with an velocity >= escape velocity you would continue to oscillate, I believe.