What Happens to Gravitational Force at the Center of the Earth?

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Discussion Overview

The discussion revolves around the gravitational force experienced at the center of the Earth, particularly focusing on the implications of the gravitational formula Fg = G*m1*m2/d^2 as distance approaches zero. Participants explore theoretical aspects of gravitational attraction, the behavior of forces within a mass, and the limitations of applying certain equations in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that at the center of the Earth, gravitational forces cancel out due to symmetry, resulting in a net gravitational force of zero.
  • Others argue that the formula Fg = G*m1*m2/d^2 is not applicable when d approaches zero, as it leads to an undefined result, and that gravitational attraction must be considered in the context of mass distribution.
  • One participant notes that only the mass interior to a given radius affects gravitational force, and at the center, there is no mass below, leading to no gravitational force.
  • Another perspective suggests that as objects get very close, other forces become significant, complicating the gravitational interactions beyond the simple model.
  • Some participants challenge the use of the gravitational formula, indicating that it should be applied differently within a uniform mass density, suggesting a linear relationship with distance from the center.
  • There are claims that Gauss's law can be applied to derive gravitational force within a spherical mass, leading to a different expression for gravitational acceleration as one approaches the center.
  • One participant proposes a modification to the gravitational constant G, suggesting it should be multiplied by 4π in certain contexts.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of gravitational formulas at the center of the Earth, with no consensus reached on the correct interpretation or application of these concepts. Multiple competing models and explanations are presented, reflecting ongoing debate.

Contextual Notes

Limitations include the dependence on assumptions about mass distribution, the applicability of certain gravitational equations in different contexts, and the unresolved nature of forces at very small distances.

uranium_235
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I am sure most here are familiar with it
Fg = G*m1*m2/d^2
With this formula, if one were to venture to the centre of the earth, the distance between he and the centre of the Earth == 0.
Therefore we get the bottom equal to 0^2 or 0. If we then continue we get an undefined answer. Certainly the gravitational force between the Earth and a body and the centre of the Earth is not infinite or else everything would collapse inwards towards the centre. So does this mean the gravitational force between the Earth and a body at its centre is Null and void, or in other words equal to 0?

Please excuse my ignorance; I am very new to this.
 
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When you're at the center of the earth, gravity is balanced in every direction, so there is zero field there. In other words, the same amount of matter is above your head as below your feet, so their attractions cancel.

On your way down to the center, you should note that only the mass interior to you matters. In other words, if you're 1 km from the center, only the mass within a 1 km radius of the center has any effect on you. The exterior mass all cancels out. When you get to the center, there is no longer any mass interior to you, and thus no force.

- Warren
 
My question is not so much concerning the centre of the earth, but rather when d=0.
 
The distance (d) will never be 0, because the matter occupies some volume (it is not at one single position). In case of the earth, if you move to its center some mass will be in front of you and, if you are under ground level, some behind you. All the mass of the Earth will pull on you but not all of it will pull in the same direction, some will pull you forwards (the mass that is in front of you) and some will pull you backwards (the mass that is behind you). When you are in the center of the mass(es) you will be pulled with equal force in all directions.

If two pieces of matter get very close together other kinds of forces (other than gravity) will come into play that are much larger than the force of gravity. You can mathematically make d go to zero so that Fg goes to infinity but physically this is not possible, at very small d you will get into molecules, into atoms etc. then things get really complicated physically and Fg will be of no importance compared to the other forces.

If you think about masses getting very close together, you will need to imagine very small objects and you will get into the part of physics that has not so much to do with gravity but with atomical and subatomical forces, these will cancel out if you compare your body (an enormous complex of atoms) with the world (an even larger complex of atoms).
 
uranium_235 said:
I am sure most here are familiar with it
Fg = G*m1*m2/d^2
With this formula, if one were to venture to the centre of the earth, the distance between he and the centre of the Earth == 0.
Therefore we get the bottom equal to 0^2 or 0. If we then continue we get an undefined answer. Certainly the gravitational force between the Earth and a body and the centre of the Earth is not infinite or else everything would collapse inwards towards the centre. So does this mean the gravitational force between the Earth and a body at its centre is Null and void, or in other words equal to 0?

Please excuse my ignorance; I am very new to this.

The expression you gave is not the correct one. That holds only for a point body or outside a body with a spherically symmetric mass density. Inside a body with a uniform mass density the force is linearly proportional to the mass density and to the distance to the center of the body. Think of it like this - Inside a spherically symetrical shell of mass the gravitational field is zero. Now consider the field inside the body. You can think of this as being inside a sphericall shell and outside a body of radius r with a uniform mass density. The closer you get to the center the less mass there is inside this imaginary shell which you're standing on.

You can use Gauss's law to calculate this. Gauss's law is the surface integral of the gravitational field vector g over the sphere containing a quantity of mass M. The surface of this sphere is called a "Gaussian surface". The value of that integral is -4*pi*GM where M is the mass inside the Gaussian surface. Solving for g gives the magnitude of

g = (4/3)*pi*G*rho*r

where rho is the mass density of the body. The direction is toward the center of the sphere. As expected g -> 0 as r - > 0. To get the force on the body multiply g by the mass of the body. You can express the above relation in terms of the total mass of the body by replacing rho with rho = M/V where M is the total mass and V is the volume of the entire body

Pete
 
Last edited:
uranium_235 said:
I am sure most here are familiar with it
Fg = G*m1*m2/d^2
With this formula, if one were to venture to the centre of the earth, the distance between he and the centre of the Earth == 0.
Therefore we get the bottom equal to 0^2 or 0. If we then continue we get an undefined answer. Certainly the gravitational force between the Earth and a body and the centre of the Earth is not infinite or else everything would collapse inwards towards the centre. So does this mean the gravitational force between the Earth and a body at its centre is Null and void, or in other words equal to 0?

Please excuse my ignorance; I am very new to this.
This is wrong, because you are using the approximation of the Earth to a sizeless particle, in a situation where it cannot be correct. Instead, you must mathematically break up your massive object into a large number of small components, work out the gravitational attraction from each component, and then sum them.
 
Thank you for all your responses. I have much to learn.
 
pmb_phy said:
The expression you gave is not the correct one. That holds only for a point body or outside a body with a spherically symmetric mass density. Inside a body with a uniform mass density the force is linearly proportional to the mass density and to the distance to the center of the body. Think of it like this - Inside a spherically symetrical shell of mass the gravitational field is zero. Now consider the field inside the body. You can think of this as being inside a sphericall shell and outside a body of radius r with a uniform mass density. The closer you get to the center the less mass there is inside this imaginary shell which you're standing on.

You can use Gauss's law to calculate this. Gauss's law is the surface integral of the gravitational field vector g over the sphere containing a quantity of mass M. The surface of this sphere is called a "Gaussian surface". The value of that integral is -4*pi*GM where M is the mass inside the Gaussian surface. Solving for g gives the magnitude of

g = (4/3)*pi*G*rho*r

where rho is the mass density of the body. The direction is toward the center of the sphere. As expected g -> 0 as r - > 0. To get the force on the body multiply g by the mass of the body. You can express the above relation in terms of the total mass of the body by replacing rho with rho = M/V where M is the total mass and V is the volume of the entire body

Pete
Gravity force must be equal to
G (4pi) M1*M2/(4pi)r^2
I.e. a gravity force must be inversely to area of sphere with radius r. Then G is not correct and should be multiplied on 4pi.
On my glance it is consequence of 4D action.

Michael
 
Michael F. Dmitriyev said:
Gravity force must be equal to
G (4pi) M1*M2/(4pi)r^2
I.e. a gravity force must be inversely to area of sphere with radius r. Then G is not correct and should be multiplied on 4pi.
On my glance it is consequence of 4D action.

Michael

The gravitational force must equal GM1*M2/(4pi)r^2 only for the force of one particle (or outside a spherically symmetric mass distribution) on another. It does not have that value in all cases and especially not in the present case.

Pmb
 
  • #10
pmb_phy said:
The gravitational force must equal GM1*M2/(4pi)r^2 only for the force of one particle (or outside a spherically symmetric mass distribution) on another. It does not have that value in all cases and especially not in the present case.

Pmb
Agree, for conformity to experimental data G should be multiplied on 4pi.
There should be also a logic explanation "visualizing" this formula. Why a sphere for mass and a circle for charge or magnetic force?

Michael
 

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