How much force a person must exert to escape the center of the earth?

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Discussion Overview

The discussion revolves around the question of how much force a person would need to exert to escape from the center of the Earth, particularly focusing on the weight of the rock above them and the energy required to reach the surface. The scope includes theoretical considerations of gravitational forces, pressure, and energy calculations related to this scenario.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants question the clarity of the original question, asking whether it pertains to the weight of the rock column above a point at the center or the work needed to climb to the surface.
  • One participant explains that at the exact center of the Earth, gravitational forces cancel out, leading to a state of weightlessness, but this changes as one moves away from the center.
  • Another participant discusses the energy required to raise a mass from the center to the surface, suggesting that the energy needed can be calculated based on varying gravitational forces with depth.
  • Some participants express frustration with the question, labeling it as "dumb," while others defend its relevance by comparing it to other physics questions about motion through the Earth.
  • There are discussions about the calculations needed to determine the pressure and weight of the rock column, with references to external data sources for more accurate information.
  • One participant humorously mentions their lack of physics knowledge, referencing a comic book scenario, while seeking to understand the forces involved in escaping the Earth's core.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the original question, with some viewing it as legitimate and others dismissing it. There are multiple competing views on how to approach the calculations and the underlying physics concepts, indicating that the discussion remains unresolved.

Contextual Notes

There are limitations in the assumptions made regarding the uniformity of Earth's density and the simplifications involved in calculating gravitational forces and pressures at various depths. The discussion also highlights the complexity of the physics involved, which may not be fully addressed in the responses.

kbclassof09
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Hi I'm new to this forum. I've always wanted to know how much force it would take a person to escape the center of the earth. Let's say someone is buried deep into the Earth's mantle nearing the outer core. How much weight of rock (be specific if you can) would be upon that person's body and how much force must he exert to escape back to the Earth's surface. Yeah I know it's a weird question, but I would really appreciate it if someone answered this for me...
 
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kbclassof09 said:
Hi I'm new to this forum. I've always wanted to know how much force it would take a person to escape the center of the earth. Let's say someone is buried deep into the Earth's mantle nearing the outer core. How much weight of rock (be specific if you can) would be upon that person's body and how much force must he exert to escape back to the Earth's surface. Yeah I know it's a weird question, but I would really appreciate it if someone answered this for me...

Welcome to the PF.

Your question is a bit confusing (at least to me). Are you asking what the weight of the column of rock is above a point at the center of the Earth?

Or are you asking how much work a person would have to do to climb a ladder up a hole from the center of the Earth to the surface?
 
These dumb questions need to be deleted.
 
Dr_Morbius said:
These dumb questions need to be deleted.

That's a fair viewpoint, and sometimes we do delete questions like this. But I responded to see if the OP could be guided to some more reasonable version of their question. We'll see how the response goes. It may be that we can teach them some basic physics, which could be a good thing. :smile:
 
Dr_Morbius said:
These dumb questions need to be deleted.

Why don't you tell him how that question is dumb?
It is fairly relevant if you consider the amount of force required by the person to climb a hole which leads to upper crust.
 
Actually the answer is far from obvious. At the exact center of the earth, the gravitational attraction in any direction is canceled out (assuming that the Earth was purely spherical, rather than the oblate spheroid that it is) by the attraction from an equal attraction in the other direction. If the Earth was hollow save for a thin shell, this would be true at any point within the earth, as it can be shown via calculus (I'll leave that exercise to you, but you can find it in any elementary physics book), that so long as you are within the sphere, there will be no force acting on you. However, anything outside of the shell see gravity as a point source at the center of that shell.

However if the Earth is filled (as it is), the moment that you move away from that center point, things get interesting. Consider the sphere r meters in radius centered on the Earth's center of gravity. This sphere now has a mass of m_{s} = (4/3)πr^{3}ρ, where ρ is the density of the sphere, which means the gravitational attractions becomes f = Gm_{s}m_{p}/r^{2} = (4/3)Gπr^{3}m_{p}ρ/r^{2} = 4Gπrm_{p}ρ/3.

Now the mass in question (m_{p}) is not the mass of the person, but the mass of the column of magma and rock. To a first approximation, for an area A, the mass will be m_{p} = A(r_{g} - r)ρ, where r_{g} is the radius of the earth. Substitution, you get, f = 4GπrA(r_{g} - r)ρ^{2}/3, or the pressure P = f/A = 4Gπr(r_{g} - r)ρ^{2}/3.

Now there are admittedly a number of simplifications here. ρ for instance, is technically a function ρ(r) which varies depending upon the properties of the material and the pressure acting on that property, but that's probably getting more complex than its worth. What it does imply is that the greatest pressure is not, in fact, at the center, but at the midpoint between the surface and the center of the earth.
 
berkeman said:
Welcome to the PF.

Your question is a bit confusing (at least to me). Are you asking what the weight of the column of rock is above a point at the center of the Earth?

Or are you asking how much work a person would have to do to climb a ladder up a hole from the center of the Earth to the surface?

I'm asking about the weight of the column of rock. How much weight would be upon your body if you were buried in the core of the earth?
 
Who decides what a dumb question is?
Recently I have seen questions similar to this asking to find the motion of a particle dropped down an imaginary tunnel through the centre of the Earth.
That was a perfectly legitimate question
 
kbclassof09 said:
I'm asking about the weight of the column of rock. How much weight would be upon your body if you were buried in the core of the earth?

Here are the data you need to answer this question:

http://geophysics.ou.edu/solid_earth/prem.html

Just multiply the pressure P with the base area of the column.
 
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  • #10
"How much force" is a difficult question to pose. If you wanted to know some detail about coming from the centre of the Earth to the surface then the quantity you need to consider is "Energy"
If you assume that the density of the Earth is uniform then g will vary uniformely with depth from 9.81 (I will take it to be 10N/kg in what follows) to 0 at the centre of the Earth.
This means that the ENERGY required to raise 1kg from the centre of the Earth to the surface will be 0.5gR where R = radius of Earth = 6.4x10^6m
this comes out at 32x10^6 Joules to raise 1kg.
If you need to take the variation in density with depth into consideration then the data recommended by DrStupid will help.
You will then need to plot a graph of g against depth and find the area under the graph to calculate the energy to raise 1kg. I would 'count squares' but I am sure someone will tell you how to integrate the data.
Hope this encourages you to remain interested in physics.
I cannot add anything to you statement about 'pressure'...sorry
 
  • #11
DrStupid said:
Here are the data you need to answer this question:

http://geophysics.ou.edu/solid_earth/prem.html

Just multiply the pressure P with the base area of the column.

So do I multiply 363.850 GPA by 3900 miles by 3 feet of rock to get the total weight?
 
Last edited by a moderator:
  • #12
I really suck at Physics you guys. I read Superman comic books, and in one issue he was trapped in the Earth's core. I just want to know how much weight was on his body and how much energy did he have to output to escape back to the Earth's surface.
 
  • #13
:smile:
kbclassof09 said:
I really suck at Physics you guys. I read Superman comic books, and in one issue he was trapped in the Earth's core. I just want to know how much weight was on his body and how much energy did he have to output to escape back to the Earth's surface.

:smile: Certainly, you have interest for it!
 
  • #14
kbclassof09 said:
So do I multiply 363.850 GPA by 3900 miles by 3 feet of rock to get the total weight?

No, you multiply 636.85 GPa by height of Superman by width of Superman.

kbclassof09 said:
how much energy did he have to output to escape back to the Earth's surface

That does not depend on the weight of the rock above him.
 
  • #15
DrStupid said:
No, you multiply 636.85 GPa by height of Superman by width of Superman.
Actually Superman is weightless at the center of the Earth, even though the pressure is so high.
 
  • #16
TheLil'Turkey said:
Actually Superman is weightless at the center of the Earth

The column of rock upon his body is not and that's what kbclassof09 asked for.
 

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